Polynomial Functions — Free Algebra 2 Review Games.

The highest exponent is 4, so the degree is 4.

The leading coefficient is the coefficient of the highest-degree term: -5.

A degree-n polynomial has at most n zeros.

Odd degree with positive leading coefficient: rises to the right.

Factor: \((x+1)(x+2)/(x+1) = x+2\).

By the Factor Theorem, if f(c)=0 then (x-c) is a factor.

\(f(1) = 1 - 2 + 1 = 0\). The remainder when dividing by \((x-1)\) is \(0\).

Even degree, negative leading coefficient: both ends go to negative infinity.

Factor: \(x(x^2-1) = x(x+1)(x-1)\). Zeros: \(0, 1, -1\).

Using synthetic division with -2: the remainder is 0, so (x+2) is a factor.

Zeros \(1, 2, -3\) give factors \((x-1)(x-2)(x+3)\).

A degree-n polynomial has exactly n zeros counting multiplicity and complex zeros.

Let \(u=x^2\): \(u^2-5u+4=(u-1)(u-4)=0\), \(u=1\) or \(4\). \(x=\pm 1, \pm 2\).

Even multiplicity means the graph touches but does not cross the x-axis.

Sign changes: \(+,-,+,-\) gives 3 sign changes. Positive zeros: \(3\) or \(1\) (subtract by 2s).

The degree is 4 (even) and the leading coefficient is 2 (positive). For even-degree polynomials with a positive leading coefficient, both ends of the graph rise to +∞. As x → -∞, f(x) → +∞. Choice B would be correct only if the leading coefficient were negative.

By the Fundamental Theorem of Algebra, every polynomial of degree n has exactly n zeros counted with multiplicity in the complex number system. A degree-4 polynomial always has exactly 4 complex zeros. Choice D is wrong because the theorem guarantees exactly n zeros without exception.

The Factor Theorem states that (x - a) is a factor of f(x) if and only if f(a) = 0. Since (x - 5) is a factor, substituting x = 5 makes that factor zero, so f(5) = 0. Choice A confuses the input value with the output value — a common mix-up.

The Remainder Theorem states that dividing \(f(x)\) by \((x - a)\) gives a remainder of \(f(a)\). Here \(a = 2\), so \(f(2) = (2)^2 + 3(2) - 5 = 4 + 6 - 5 = 5\). Choice A (remainder \(0\)) would mean \((x - 2)\) is a factor, but since \(f(2) = 5 \neq 0\), that is not the case.

As \(x \to -\infty\), \(x^5 \to -\infty\) (odd powers preserve sign). Then \(-x^5 \to -(-\infty) = +\infty\). Choice B describes what happens as \(x \to +\infty\) for this function, which is a common end-behavior mix-up for odd-degree polynomials with negative leading coefficients.

The degree of a product of polynomials equals the sum of their individual degrees. The first factor has degree \(2\) and the second has degree \(3\), so the product has degree \(2 + 3 = 5\). Choice D (\(6\)) is a common error of multiplying degrees rather than adding them.

By the Fundamental Theorem of Algebra, a degree-4 polynomial has exactly 4 zeros counting multiplicity. With 2 real zeros accounted for, 4 - 2 = 2 zeros must be non-real. For polynomials with real coefficients, non-real zeros always occur in conjugate pairs, so 2 non-real zeros is the only consistent possibility. Choice B (1) is impossible because non-real zeros cannot appear alone.

The Remainder Theorem says the remainder equals \(f(1)\). Evaluating: \(f(1) = (1)^3 + 2(1)^2 - (1) + 4 = 1 + 2 - 1 + 4 = 6\). Choice A (\(4\)) is just the constant term — a common error of only reading off the last coefficient rather than fully evaluating the polynomial.

Factor as a difference of squares: \(x^4 - 16 = (x^2 - 4)(x^2 + 4) = (x - 2)(x + 2)(x^2 + 4)\). The factor \((x^2 + 4)\) yields no real zeros since \(x^2 = -4\) has no real solution. Only \(x = 2\) and \(x = -2\) are real zeros. Choice D (\(4\)) incorrectly assumes all four zeros from the Fundamental Theorem must be real.

Odd multiplicity means the graph crosses the x-axis. With multiplicity 3 (odd and greater than 1), the graph flattens out and forms an inflection-point shape as it crosses — less steep than a simple zero. Choice A describes even multiplicity (touching and bouncing). Choice C describes multiplicity 1 behavior, where the crossing is sharp.

The Complex Conjugate Zeros Theorem states that if a polynomial has real coefficients and a + bi is a zero, then its conjugate a - bi must also be a zero. The conjugate of 3 + 2i is 3 - 2i. Choice B (-3 + 2i) is the additive inverse of the real part only, not the conjugate — a frequent sign error.

Possible rational zeros have the form p/q, where p divides the constant term (6) and q divides the leading coefficient (2). For 3/2: p = 3 divides 6 and q = 2 divides 2, so 3/2 is valid. For 2/3: q = 3 does not divide 2, so it fails. For 4: 4 does not divide 6, so it is also invalid.

Evaluate \(f(-2)\): \(2(-2)^3 - (-2)^2 + 3(-2) - 5 = 2(-8) - 4 + (-6) - 5 = -16 - 4 - 6 - 5 = -31\). Choice A (\(-27\)) likely results from a sign error when squaring \(-2\) in the second term. Choice D (\(31\)) is the positive version, arising from ignoring all the negative signs.

End behavior is determined by the leading term, -3x^5. Odd degree with a negative leading coefficient means the right end falls: as x → +∞, f(x) → -∞. Choice A would apply if the leading coefficient were positive. Choice C is incorrect because polynomial functions grow without bound rather than leveling off.

Using the difference of cubes formula \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\) with \(a = x\) and \(b = 2\): \(x^3 - 8 = (x - 2)(x^2 + 2x + 4)\). So \((x - 2)\) is a factor. Choice A \((x + 2)\) is wrong — \(f(2) = 0\), not \(f(-2)\). Choice B has the wrong sign in the middle term of the quadratic factor.

Set up synthetic division with root \(2\) and coefficients \(1, -3, 0, 4\) (the missing \(x\)-term has coefficient \(0\)). Bring down \(1\). Then: \(2(1) = 2\), \(-3 + 2 = -1\); \(2(-1) = -2\), \(0 + (-2) = -2\); \(2(-2) = -4\), \(4 + (-4) = 0\). The remainder is \(0\) and the quotient is \(x^2 - x - 2\). Choice D uses \(+2\) as the constant, which would produce a nonzero remainder.

Setting each factor to zero: (x - 1)^2 = 0 gives x = 1, and the exponent 2 on the factor indicates multiplicity 2; (x + 3) = 0 gives x = -3 with multiplicity 1. Choice C has reversed signs — solving x - 1 = 0 gives x = 1, not x = -1, and solving x + 3 = 0 gives x = -3, not x = 3.

The nonzero coefficients in order are \(+1, -2, +1, -1\). Sign changes: \(+\) to \(-\) (one), \(-\) to \(+\) (two), \(+\) to \(-\) (three), giving \(3\) sign changes. By Descartes' Rule, the number of positive real zeros equals the number of sign changes or is less by an even number: \(3\) or \(1\), meaning at most \(3\). Choice D is wrong because Descartes' Rule gives a maximum, not an exact count.

Factor by grouping: \(x^3 - x^2 + x - 1 = x^2(x - 1) + 1(x - 1) = (x^2 + 1)(x - 1)\). Setting each factor to zero: \(x - 1 = 0\) gives \(x = 1\); \(x^2 + 1 = 0\) gives \(x^2 = -1\), so \(x = i\) and \(x = -i\). Choice C has the wrong real zero — the grouping yields the factor \((x - 1)\), not \((x + 1)\).

Since the polynomial has real coefficients and \(1 + i\) is a zero, its conjugate \(1 - i\) must also be a zero. The quadratic factor from this conjugate pair is \((x - (1+i))(x - (1-i)) = (x-1)^2 + 1 = x^2 - 2x + 2\). Combined with the zero at \(x = 2\) (mult 2): \(f(x) = (x - 2)^2(x^2 - 2x + 2)\). Choice B uses \(x^2 + 2x + 2\), which factors into zeros at \(-1 + i\) and \(-1 - i\), not the required pair.

For \(f(x)\): signs are \(+, -, +, +, -, -\). Changes occur at positions 1 to 2, 2 to 3, and 4 to 5: \(3\) sign changes, so at most \(3\) positive real zeros. For \(f(-x) = -x^5 - 2x^4 - x^3 + x^2 + x - 1\): signs are \(-, -, -, +, +, -\). Changes occur at positions 3 to 4 and 5 to 6: \(2\) sign changes, so at most \(2\) negative real zeros. Choice B overcounts \(f(-x)\) by missing that the first three terms all share the same sign.

Let \(u = x^2\) to get the quadratic \(u^2 + u - 12 = 0\), which factors as \((u + 4)(u - 3) = 0\). From \(u = 3\): \(x^2 = 3\), so \(x = \pm\sqrt{3}\) (real zeros). From \(u = -4\): \(x^2 = -4\), so \(x = \pm 2i\) (non-real complex zeros). Choice B is wrong — testing \(x = 2\) gives \((2)^4 + (2)^2 - 12 = 16 + 4 - 12 = 8 \neq 0\), so \(\pm 2\) are not zeros.

Crossing requires odd multiplicity; touching and turning back requires even multiplicity. Zero at x = -1 (crosses): factor (x + 1) with odd multiplicity. Zero at x = 2 (touches): factor (x - 2)^2 with even multiplicity. Zero at x = 4 (crosses): factor (x - 4) with odd multiplicity. Total degree: 1 + 2 + 1 = 4. Choice B has zero at x = 1, not x = -1. Choice C places the even-multiplicity zero at x = 4 instead of x = 2.

Tally known zeros with multiplicity: x = 0 (mult 1), x = 1 + 3i (mult 1), and since real coefficients force the conjugate, x = 1 - 3i (mult 1), plus x = -2 (mult 2). Total: 1 + 1 + 1 + 2 = 5. A degree-6 polynomial needs exactly 6 zeros, so 1 more is needed. That final zero cannot be non-real — if it were, its conjugate would also be required, giving 7 zeros total and exceeding degree 6. So it must be real. Choice D incorrectly counts only 4 zeros by omitting the required conjugate 1 - 3i.

Factor by grouping: \(x^3 + 3x^2 - 4x - 12 = x^2(x + 3) - 4(x + 3) = (x^2 - 4)(x + 3) = (x - 2)(x + 2)(x + 3)\). Setting each factor to zero: \(x - 2 = 0\) gives \(x = 2\); \(x + 2 = 0\) gives \(x = -2\); \(x + 3 = 0\) gives \(x = -3\). Choice B has \(x = 3\) instead of \(x = -3\), a sign error from incorrectly solving \(x + 3 = 0\).

The degree of a polynomial is the highest exponent of the variable. Even though 4x^3 appears first, the term -2x^5 has the highest exponent of 5, so the degree is 5. The order terms are written does not determine degree — always look for the greatest exponent. Choice A (3) is wrong because it names the exponent of the first written term, not the highest.

End behavior is determined by the leading term, which is 2x^4. The degree is even (4) and the leading coefficient is positive (2), so both ends of the graph rise toward positive infinity. Choice A describes an odd-degree polynomial with positive leading coefficient. Choice B describes an even-degree polynomial with negative leading coefficient.

The Fundamental Theorem of Algebra guarantees that every polynomial of degree n has exactly n zeros in the complex number system, counting multiplicity. For a degree-5 polynomial, this means exactly 5 zeros. Choice A is too weak — the theorem guarantees exactly 5, not just at most 5. Choice D is wrong because the theorem counts all complex zeros, not only real ones.

By the Factor Theorem, (x - c) is a factor of f(x) if and only if f(c) = 0. Here c = 4, so f(4) = 0. Choice A is wrong because (x - 4) means c = 4, not c = -4; that would correspond to the factor (x + 4). Choice C confuses the factor with its reciprocal.

The Remainder Theorem states that when f(x) is divided by the linear binomial (x - c), the remainder is equal to f(c). This means you can find the remainder by simply evaluating the polynomial at x = c, without performing the full division. Choice B is a common sign error — f(-c) would correspond to dividing by (x + c), not (x - c).

Synthetic division requires a coefficient for every degree from highest to lowest, including zero coefficients for missing terms. Since \(f(x) = x^3 + 0x^2 + 5x - 7\), the coefficient row is \(1, 0, 5, -7\). The divisor value used is \(c = 2\) (from \(x - 2\), not \(-2\)). Choices A and D omit the zero placeholder for \(x^2\), which would produce an incorrect result.

A polynomial of degree n can have at most n - 1 turning points (local maxima and minima). For example, a degree-4 polynomial can have at most 3 turning points. Choice A (n) overstates the maximum. This rule is useful for checking the reasonableness of a sketch — if a graph shows more than n - 1 turns, it cannot be a degree-n polynomial.

At a zero of odd multiplicity (1, 3, 5, ...), the graph crosses through the x-axis. At a zero of even multiplicity (2, 4, 6, ...), the graph touches the x-axis and bounces back. Vertical asymptotes occur in rational functions where the denominator is zero, not in polynomials. Recognizing crossing vs. bouncing behavior is key to sketching polynomial graphs.

Dividing by (x + 3) means using c = -3 in synthetic division. Setting up with coefficients 2, 3, -11, -6 and divisor -3: bring down 2; then 2(-3)=-6, so 3+(-6)=-3; then -3(-3)=9, so -11+9=-2; then -2(-3)=6, so -6+6=0. The remainder is 0 and the quotient is 2x^2 - 3x - 2. Choice A results from incorrectly using +3 as the divisor.

Factor as a difference of squares: \(x^4 - 16 = (x^2 - 4)(x^2 + 4) = (x - 2)(x + 2)(x^2 + 4)\). The factors \((x - 2)\) and \((x + 2)\) give real zeros \(x = 2\) and \(x = -2\). The factor \((x^2 + 4) = 0\) gives \(x^2 = -4\), which has no real solutions (only the complex zeros \(x = \pm 2i\)). So there are exactly \(2\) real zeros. Choice C (\(4\)) incorrectly assumes all four zeros are real.

The described end behavior (right side down, left side up) requires an odd-degree polynomial with a negative leading coefficient. Choice C, \(f(x) = -x^3 + x + 2\), has odd degree (\(3\)) and negative leading coefficient (\(-1\)), matching exactly. Choice A has odd degree but positive leading coefficient, giving the opposite behavior. Choice B has even degree, meaning both ends go the same direction.

Since \(x = 1\) is a zero, \((x - 1)\) is a factor. Dividing \(f(x)\) by \((x - 1)\) using synthetic division with coefficients \(1, -6, 11, -6\) and \(c = 1\) gives quotient \(x^2 - 5x + 6\). Factoring: \(x^2 - 5x + 6 = (x - 2)(x - 3)\). Therefore \(f(x) = (x - 1)(x - 2)(x - 3)\). Choice A incorrectly uses \((x + 1)\), which would correspond to a zero at \(x = -1\), not \(x = 1\).

The multiplicity of a zero is the exponent on the corresponding factor. Since the factor (x - 2) is raised to the third power, the zero x = 2 has multiplicity 3. A multiplicity of 3 is odd, which means the graph of f crosses through the x-axis at x = 2, with an S-shaped flattening. The zero x = -1 comes from (x + 1)^1 and has multiplicity 1.

At a zero of even multiplicity (2, 4, 6, ...), the factor appears an even number of times, so the function value does not change sign across that zero. This causes the graph to touch the x-axis and bounce back. At a zero of odd multiplicity, the graph crosses through. Cusps and holes are features of other types of functions, not polynomial zeros.

By the Remainder Theorem, the remainder when \(f(x)\) is divided by \((x - 1)\) equals \(f(1)\). Evaluating: \(f(1) = (1)^3 + 2(1)^2 - 5(1) + 2 = 1 + 2 - 5 + 2 = 0\). The remainder is \(0\), which also means \((x - 1)\) is a factor of \(f(x)\). Choice A (\(2\)) results from evaluating \(f(0)\) instead of \(f(1)\), and Choice D (\(1\)) may come from a calculation error.

Count sign changes in \(f(x) = x^4 - 3x^2 + 2\) by looking at consecutive nonzero coefficients: \((+1)\) to \((-3)\) is one change, then \((-3)\) to \((+2)\) is a second change. There are 2 sign changes, so by Descartes' Rule there are either 2 or 0 positive real zeros (the count decreases by 2 each step). In fact the zeros are \(x = \pm 1\) and \(x = \pm \sqrt{2}\), giving 2 positive real zeros. Choice B (3 or 1) would require an odd number of sign changes.

The Rational Zero Theorem states that any rational zero has the form p/q, where p divides the constant term (6) and q divides the leading coefficient (3). Factors of 6: 1, 2, 3, 6. Factors of 3: 1, 3. Possible rational zeros include plus or minus 1, 2, 3, 6, 1/3, 2/3. Only 2/3 appears among the choices. Choice A (4) is not a factor of 6. Choice B (5/3) uses 5, which does not divide 6. Choice C (4/3) uses 4, which does not divide 6.

Using synthetic division with \(c = \frac{1}{2}\) on coefficients \(2, -5, 4, -1\): bring down 2; then \(2(\frac{1}{2})=1\), so \(-5+1=-4\); then \(-4(\frac{1}{2})=-2\), so \(4+(-2)=2\); then \(2(\frac{1}{2})=1\), so \(-1+1=0\). The quotient is \(2x^2 - 4x + 2 = 2(x^2 - 2x + 1) = 2(x - 1)^2\). So \(x = 1\) is a zero of multiplicity 2. All zeros: \(x = \frac{1}{2}\) and \(x = 1\) (multiplicity 2). Choice C is wrong because it introduces \(x = -1\) without justification — verifying \(f(-1) = -2 - 5 - 4 - 1 = -12\), not 0.

Because the polynomial has real coefficients, complex zeros come in conjugate pairs. So \(x = 1 - i\) is also a zero. The three factors are \((x + 2)\), \((x - (1+i))\), and \((x - (1-i))\). Multiplying the complex pair: \((x-(1+i))(x-(1-i)) = x^2 - 2x + (1+1) = x^2 - 2x + 2\). Then \(f(x) = (x+2)(x^2-2x+2) = x^3 - 2x^2 + 2x + 2x^2 - 4x + 4 = x^3 - 2x + 4\). Choice A omits the conjugate pairing step and uses incorrect expansion.

Testing \(x = 1\): \(1 - 3 + 6 - 12 + 8 = 0\), so \((x - 1)\) is a factor. Synthetic division gives \(x^3 - 2x^2 + 4x - 8\). Testing \(x = 2\): \(8 - 8 + 8 - 8 = 0\), so \((x - 2)\) is a factor. Synthetic division gives \(x^2 + 4\). Setting \(x^2 + 4 = 0\) gives \(x^2 = -4\), so \(x = \pm 2i\). All four zeros are \(x = 1\), \(x = 2\), \(x = 2i\), and \(x = -2i\). Choice D incorrectly finds \(x = \pm 4i\) by solving \(x^2 = -16\) instead of \(x^2 = -4\).

The factors are \((x-2)^2\), \((x+1)\), and \((x^2+9)\) (since \((x-3i)(x+3i) = x^2+9\)). First, \((x-2)^2(x+1) = (x^2-4x+4)(x+1) = x^3 - 3x^2 + 4\). Then multiply by \((x^2+9)\): \((x^3-3x^2+4)(x^2+9) = x^5 + 9x^3 - 3x^4 - 27x^2 + 4x^2 + 36 = x^5 - 3x^4 + 9x^3 - 23x^2 + 36\). Choice B has a sign error on the \(x^2\) coefficient, resulting from forgetting to combine \(-27x^2\) and \(+4x^2\) correctly.

Possible rational zeros include plus or minus 1, 2, 1/2, 1/3, 2/3, 1/6. Testing x = 2: 6(8) - 11(4) - 6 + 2 = 48 - 44 - 6 + 2 = 0. Synthetic division with c = 2 on 6, -11, -3, 2 yields quotient 6x^2 + x - 1. Factoring: 6x^2 + x - 1 = (2x + 1)(3x - 1), giving x = -1/2 and x = 1/3. All zeros: x = 2, x = -1/2, x = 1/3. Choice A has reversed signs on the two fractional zeros, a common error when factoring the quadratic quotient.

Because the coefficients are real, \(x = -i\) is also a zero. The four factors (with leading coefficient 2) give: \(f(x) = 2(x-3)^2(x-i)(x+i) = 2(x^2-6x+9)(x^2+1)\). Expanding \((x^2-6x+9)(x^2+1) = x^4 + x^2 - 6x^3 - 6x + 9x^2 + 9 = x^4 - 6x^3 + 10x^2 - 6x + 9\). Multiplying by 2: \(2x^4 - 12x^3 + 20x^2 - 12x + 18\). Choice B has coefficient 10 on \(x^2\) before multiplying by 2 — it forgets to double the result.

Dividing \(2x^4 - 3x^3 + x^2 + 0x - 5\) by \(x^2 - 2x + 1\): Step 1 — \(2x^4 / x^2 = 2x^2\); subtract \(2x^2(x^2-2x+1) = 2x^4-4x^3+2x^2\); remaining: \(x^3 - x^2\). Step 2 — \(x^3 / x^2 = x\); subtract \(x(x^2-2x+1) = x^3-2x^2+x\); remaining: \(x^2 - x\). Step 3 — \(x^2 / x^2 = 1\); subtract \(1(x^2-2x+1) = x^2-2x+1\); remaining: \(x - 6\). Quotient is \(2x^2 + x + 1\) with remainder \(x - 6\). Choice B has sign errors in the quotient resulting from arithmetic mistakes in the subtraction steps.

Substitute \(u = x^2\) to get \(u^2 + 4u - 5 = (u + 5)(u - 1) = 0\), so \(u = -5\) or \(u = 1\). For \(u = 1\): \(x^2 = 1\) gives \(x = \pm 1\) (two real zeros). For \(u = -5\): \(x^2 = -5\) gives \(x = \pm i\sqrt{5}\) (two complex non-real zeros). Total: 2 real and 2 complex non-real zeros, consistent with the degree-4 guarantee from the Fundamental Theorem of Algebra. Choice A is wrong because \(x^2 = -5\) has no real solutions.

The Fundamental Theorem of Algebra states that a degree-\(n\) polynomial has exactly \(n\) zeros in the complex number system, counting multiplicity. Choice B ('at most \(n\)') is a common misconception — the theorem guarantees exactly \(n\), not 'at most' \(n\). For example, \(x^2 - 2x + 1 = (x-1)^2\) has two zeros counting multiplicity (both equal to 1), not just one.

The degree of the quotient equals the degree of the dividend minus the degree of the divisor: 5 - 1 = 4. Choice A (degree 3) would result from dividing by a degree-2 divisor. Choice C is incorrect because division always reduces the degree — the quotient cannot have the same degree as the dividend.

By the Factor Theorem, since \(f(3) = 0\), the corresponding factor is \((x - 3)\). Choice A, \((x + 3)\), would correspond to \(f(-3) = 0\), not \(f(3) = 0\). Choice D, \((x^2 - 9) = (x-3)(x+3)\), would only be a factor if both \(x = 3\) and \(x = -3\) are zeros — knowing only \(f(3) = 0\) is not sufficient to conclude \(x = -3\) is also a zero.

For an even-degree polynomial with a negative leading coefficient, both ends of the graph fall toward -inf (both ends go down). The leading term dominates for large |x|, and a negative even-degree term is negative for all large |x|. Choice A describes even degree with a positive leading coefficient. Choices C and D describe odd-degree behavior, where the two ends point in opposite directions.

A degree-n polynomial can have at most n - 1 turning points. For n = 4, the maximum is 4 - 1 = 3. This follows because turning points occur where f'(x) = 0, and f'(x) has degree n - 1, giving at most n - 1 real roots. Choice A (2) is too few — a degree-4 polynomial can reach 3 turning points. Choice C (4) would require degree 5.

Synthetic division only applies when the divisor is a monic linear polynomial of the form \((x - a)\). It cannot be used with quadratic or higher-degree divisors — those require polynomial long division. Choice A is incorrect because a quadratic divisor, for example, is not compatible with the synthetic division algorithm. Non-monic linear factors like \((2x - 1)\) also require an adjustment before synthetic division can be used.

For a polynomial with real coefficients, non-real complex zeros always appear in conjugate pairs, consuming 2 zeros at a time. Since a polynomial of odd degree has an odd total number of zeros (counting multiplicity), it is impossible for all of them to be paired up as complex conjugates — at least one must be real. Choice A (0) is impossible for odd degree with real coefficients. Choice C (2) is not guaranteed; a degree-3 polynomial can have exactly 1 real zero and one conjugate pair of complex zeros.

Set up synthetic division with \(c = 2\) and coefficients \(1, 0, -7, 6\) (the 0 is the placeholder for the missing \(x^2\) term). Bring down 1; \(1 \times 2 = 2\), add to 0 to get 2; \(2 \times 2 = 4\), add to \(-7\) to get \(-3\); \(-3 \times 2 = -6\), add to 6 to get 0. The quotient is \(x^2 + 2x - 3\). Choice B is wrong because the middle coefficient comes out to \(+2\), not \(-2\). Forgetting the 0 placeholder is a common error that scrambles all subsequent coefficients.

The Remainder Theorem states the remainder when dividing \(f(x)\) by \((x - c)\) equals \(f(c)\). The divisor is \((x + 1) = (x - (-1))\), so \(c = -1\). Evaluate: \(f(-1) = (-1)^4 - 3(-1)^2 + 2(-1) - 5 = 1 - 3 - 2 - 5 = -9\). Choice B (\(-7\)) results from a sign error on the \(2x\) term. Choice D (\(-5\)) is the constant term of \(f(x)\), which equals the remainder only when \(x\) is divided out — not when dividing by \((x + 1)\).

The Complex Conjugate Zeros Theorem states that if a polynomial has real coefficients and a + bi is a zero, then its complex conjugate a - bi is also a zero. The conjugate of 1 + 2i is 1 - 2i (negate only the imaginary part). Choice B (-1 + 2i) negates the real part, which is not the conjugate. Choice C (-1 - 2i) is the negation of the entire number, not the conjugate.

The leading term is determined by the highest-power parts: \(-(x^3)(x) = -x^4\). This is a degree-4 polynomial with a negative leading coefficient. Even degree with a negative leading coefficient means both ends fall toward \(-\infty\) (both ends go down). Choice A describes even degree with a positive leading coefficient. Choices C and D describe odd-degree behavior where the two ends go in opposite directions.

Choice A has zeros at \(x = 1\) (multiplicity 2), \(x = -2\), and \(x = 3\) — three distinct real values. Choice B has only one distinct real zero (\(x = 1\), multiplicity 4). Choice C has only \(x = 2\) as a real zero; \(x^2 + 1 = 0\) produces only non-real complex solutions. Choice D has exactly two distinct real zeros: \(x = 0\) and \(x = -1\).

Using synthetic division with c = 2 and coefficients 2, 3, -11, -6: bring down 2; 2 x 2 = 4, add to 3 to get 7; 7 x 2 = 14, add to -11 to get 3; 3 x 2 = 6, add to -6 to get 0 (zero remainder). Quotient is 2x^2 + 7x + 3. Choice B is wrong because the x-coefficient accumulates to +7, not -7. The remainder of 0 confirms that x = 2 is a zero of the original polynomial.

By Vieta's formulas for a monic cubic with zeros \(r_1, r_2, r_3\), the coefficient of \(x^2\) is \(a = -(r_1 + r_2 + r_3)\). Here \(a = -((-1) + 2 + 3) = -(4) = -4\). This can be confirmed by expanding \((x + 1)(x - 2)(x - 3) = x^3 - 4x^2 + x + 6\). Choice A (4) has the wrong sign. Choice C (\(-6\)) is the constant term \(c\), not \(a\).

At any zero of odd multiplicity, the graph crosses the x-axis (sign of \(f(x)\) changes). However, when the odd multiplicity exceeds 1 (e.g., 3 or 5), the graph visibly flattens near the zero, appearing to 'hesitate' before crossing — much like the curve \(y = x^3\) at the origin. Choice A describes a zero of even multiplicity, where the graph touches but does not cross. Choice D describes a simple zero (multiplicity 1), which crosses without the flattening effect.

The polynomial is f(x) = 2(x + 1)(x - 1)(x - 2)(x - 3). The constant term equals f(0) = 2(0+1)(0-1)(0-2)(0-3) = 2(1)(-1)(-2)(-3) = 2(-6) = -12. The product (-1)(-2)(-3) = -6, negative because there are three negative factors. Choice A (12) results from a sign error. Choice C (-6) omits the leading coefficient factor of 2.

Count sign changes in the sequence of coefficients: \(+1, -2, +1, +1, -3, +1\). Changes occur at positions: $(+1$ to $-2)$, $(-2$ to $+1)$, $(+1$ to \(+1\) — no change$)$, $(+1$ to $-3)$, $(-3$ to $+1)$. That is 4 sign changes. By Descartes' Rule, the number of positive real zeros equals 4, or decreases by 2 repeatedly: so 4, 2, or 0. Choice B omits 0 as a valid count. Choice D (3 or 1) corresponds to \(f(-x)\), which counts negative real zeros.

The Factor Theorem states \((x - a)\) is a factor if and only if \(f(a) = 0\). Here \(a = 2\), so compute \(f(2) = 16 - 16 - 6 + 6 = 0\). Since \(f(2) = 0\), \((x - 2)\) is indeed a factor. Choice B checks \(f(-2)\), which tests whether \((x + 2)\) is a factor, not \((x - 2)\). Choice C is logically backwards — a nonzero value of \(f(2)\) would confirm \((x - 2)\) is NOT a factor.

Divide by \((x - 2)\) synthetically to get \(x^3 - 3x - 2\). Test \(x = -1\): \((-1)^3 - 3(-1) - 2 = -1 + 3 - 2 = 0\), confirming \(x = -1\) is a zero. Divide \(x^3 - 3x - 2\) by \((x + 1)\) to obtain \(x^2 - x - 2 = (x - 2)(x + 1)\). Therefore \(f(x) = (x - 2)^2(x + 1)^2\), giving zeros \(x = 2\) and \(x = -1\), each with multiplicity 2. Choice B incorrectly lists four distinct zeros; \(x = 2\) and \(x = -1\) each appear twice. Choice C wrongly assigns multiplicity 3 to \(x = 2\).

Since the polynomial has real coefficients and \(1 + i\) is a zero, its conjugate \(1 - i\) is also a zero. The quadratic factor from these two complex zeros is \((x - (1+i))(x - (1-i)) = (x-1)^2 - (i)^2 = (x-1)^2 + 1 = x^2 - 2x + 2\). Combined with the repeated real zero at \(x = 2\): \(f(x) = (x - 2)^2(x^2 - 2x + 2)\). Choice B gives \(x^2 + 2x + 2\), which corresponds to zeros \(-1 \pm i\), not \(1 \pm i\).

Since \(x = 1\) and \(x = -3\) are zeros, \((x - 1)(x + 3) = x^2 + 2x - 3\) is a factor. Dividing \(f(x)\) by \(x^2 + 2x - 3\) (matching coefficients) gives quotient \(x^2 - x - 2 = (x - 2)(x + 1)\). The remaining zeros are therefore \(x = 2\) and \(x = -1\). All four zeros are \(x = 1\), \(x = -3\), \(x = 2\), and \(x = -1\). Choice D reverses both signs, giving \(x = -1\) and \(x = -2\), which do not satisfy the original equation.

First step: \(3x^3\) divided by \(x^2\) gives \(3x\). Multiply \(3x(x^2 - x + 1) = 3x^3 - 3x^2 + 3x\); subtract from dividend to get \(x^2 + x - 1\). Second step: \(x^2\) divided by \(x^2\) gives \(1\). Multiply \(1(x^2 - x + 1) = x^2 - x + 1\); subtract to get \(2x - 2\). Since \(2x - 2\) has degree less than the divisor, it is the remainder. Therefore \(Q(x) = 3x + 1\) and \(R(x) = 2x - 2\). Choice B gives \(Q(x) = 3x - 1\), an error in the second quotient term where \(+1\) should be added, not subtracted.

Since the polynomial has real coefficients and \(1 + 3i\) is a zero, its conjugate \(1 - 3i\) is also a zero. The quadratic from this pair: \((x-(1+3i))(x-(1-3i)) = (x-1)^2 + 9 = x^2 - 2x + 10\). The zero \(x = -2\) gives factor \((x + 2)\). The double zero at \(x = 0\) gives \(x^2\). Combined: \(f(x) = x^2(x + 2)(x^2 - 2x + 10)\), which is degree 5. Choice B uses \((x - 2)\), placing the zero at \(x = +2\) instead of \(x = -2\). Choice C gives \(x^2 + 2x + 10\), corresponding to zeros \(-1 \pm 3i\), not \(1 \pm 3i\).

Possible rational zeros are plus or minus (factors of 6) over (factors of 2): plus or minus 1, 2, 3, 6, 1/2, 3/2. Test x = 3: f(3) = 54 - 27 - 33 + 6 = 0. Synthetic division by (x - 3) yields quotient 2x^2 + 3x - 2, which factors as (2x - 1)(x + 2). Setting each factor to zero: 2x - 1 = 0 gives x = 1/2, and x + 2 = 0 gives x = -2. All three rational zeros are x = 3, x = 1/2, and x = -2. Choice B uses x = -1/2, but solving 2x - 1 = 0 gives x = +1/2, not -1/2.

Since the polynomial has real coefficients, 2i being a zero forces its conjugate -2i to also be a zero — that accounts for 2 non-real complex zeros. The zero x = -1 (multiplicity 3) accounts for 3 real zeros. So far: 3 + 2 = 5 zeros. A degree-6 polynomial requires exactly 6 zeros counting multiplicity, so one more is needed. Adding another non-real complex zero would require its conjugate as well (two more), bringing the total to 7 and exceeding degree 6. Therefore the sixth zero must be real. Final count: 4 real zeros and exactly 2 non-real complex zeros. Choice B incorrectly assumes no additional real zero is required.

The Fundamental Theorem of Algebra guarantees that every degree-n polynomial has exactly n zeros in the complex number system, counting multiplicity. A degree-4 polynomial therefore has exactly 4 zeros. Choosing 8 is a common error that comes from doubling the degree rather than using it directly.

End behavior is controlled solely by the leading term, 4x^6. The degree is even and the leading coefficient is positive, so both ends of the graph rise toward positive infinity. A negative leading coefficient would make both ends fall (choice A). Choices C and D describe the behavior of odd-degree polynomials.

The Remainder Theorem states that the remainder when f(x) is divided by (x - c) equals f(c), the value of the polynomial evaluated at c. Choice A gives the y-intercept, not the remainder. Choice D confuses this theorem with the Factor Theorem, which applies only when the remainder happens to be zero.

A zero of even multiplicity causes the graph to touch but not cross the x-axis, because the factor appears an even number of times and the function does not change sign there. A zero of odd multiplicity (like multiplicity 1) produces a crossing as in choice A. A flattened S-shape through the axis (choice D) is characteristic of an odd multiplicity of 3 or higher.

The degree of a product of polynomials equals the sum of their individual degrees. Multiplying a degree-3 polynomial by a degree-2 polynomial yields a degree-5 result, since the leading term \(x^3\) times \(x^2\) produces \(x^5\). Choice C (degree 6) is the common error of multiplying the degrees rather than adding them.

A degree-n polynomial can have at most n - 1 turning points. For a degree-6 polynomial, the maximum is 6 - 1 = 5. The polynomial may have fewer turning points but never more. A common error is confusing n with n - 1 and choosing 6.

Setting \(x^3 + 27 = 0\) gives \(x^3 = -27\), so \(x = -3\). Confirming: \((-3)^3 + 27 = -27 + 27 = 0\). Choice A is incorrect because \((3)^3 + 27 = 54\), not zero. This polynomial is a sum of cubes and factors as \((x + 3)(x^2 - 3x + 9)\), confirming \(x = -3\) is the only real zero.

The Complex Conjugate Zeros Theorem states that if a polynomial has real coefficients and a + bi is a zero (with b not equal to 0), then its complex conjugate a - bi must also be a zero. The conjugate of 3 - 2i is 3 + 2i. Choices B and D change the sign of the real part, and choice C swaps the real and imaginary parts, neither of which produces the correct conjugate.

By the Remainder Theorem, the remainder equals \(f(3)\). Evaluating: \(f(3) = (3)^3 - 4(3)^2 + 2(3) + 5 = 27 - 36 + 6 + 5 = 2\). A common mistake is evaluating at \(x = -3\) instead of \(x = 3\), since the divisor is \((x - 3)\) and its root is \(x = 3\), not \(x = -3\).

Setting up synthetic division with \(c = 2\) and coefficients \(1, 2, -5, -6\): bring down \(1\); \(1\) times \(2 = 2\), and \(2 + 2 = 4\); \(4\) times \(2 = 8\), and \(-5 + 8 = 3\); \(3\) times \(2 = 6\), and \(-6 + 6 = 0\). The quotient is \(x^2 + 4x + 3\) with remainder \(0\). This factors as \((x + 1)(x + 3)\), confirming that \(x = -1\) and \(x = -3\) are also zeros of \(f(x)\).

The Rational Zero Theorem states that possible rational zeros have the form (factor of constant term) divided by (factor of leading coefficient). The constant is \(2\), with factors \(1\) and \(2\). The leading coefficient is \(2\), also with factors \(1\) and \(2\). This gives \(\pm \frac{1}{1}, \pm \frac{2}{1}, \pm \frac{1}{2}\), and \(\pm \frac{2}{2}\), which simplifies to \(\pm 1, \pm 2, \pm \frac{1}{2}\). Choice B incorrectly includes \(\pm 4\), and choices C and D are incomplete.

Touching without crossing indicates an even multiplicity at x = -2, so (x + 2)^2 must be a factor. Crossing at x = 1 indicates odd multiplicity, so (x - 1) appears to an odd power. Choice B, with (x + 2)^2(x - 1), satisfies both conditions. Choice C reverses the behavior: it would produce a touch at x = 1 and a crossing at x = -2. Choice D has odd multiplicity 3 at x = -2, which would also produce a crossing.

End behavior depends only on the leading term, -2x^5. The degree is odd and the leading coefficient is negative. For an odd-degree polynomial with a negative leading coefficient, the left end rises (toward positive infinity) and the right end falls (toward negative infinity). Choice A describes an odd-degree polynomial with a positive leading coefficient, while choices B and D describe even-degree polynomials.

By the Factor Theorem, \((x + 1)\) is a factor if and only if \(f(-1) = 0\). Substituting: \(f(-1) = (-1)^3 + k(-1)^2 - 4(-1) - 4 = -1 + k + 4 - 4 = k - 1 = 0\), giving \(k = 1\). With \(k = 1\) the polynomial becomes \(x^3 + x^2 - 4x - 4\), which factors as \((x + 1)(x^2 - 4) = (x + 1)(x - 2)(x + 2)\). Choosing \(k = -1\) gives \(f(-1) = -2\), not zero, so \((x + 1)\) would not be a factor.

The Fundamental Theorem of Algebra guarantees a degree-5 polynomial has exactly 5 zeros counting multiplicity, regardless of what those zeros are. Additionally, since the polynomial has real coefficients and 2 + 3i is a zero, its conjugate 2 - 3i must also be a zero. That accounts for four of the five zeros; the fifth must be a real zero (possibly a repeat of x = 4 or x = -1). The total is always 5.

Rewrite \(x^4 - 16\) using placeholder zeros for missing terms: coefficients \(1, 0, 0, 0, -16\). Dividing by \(c = 2\) yields \(1, 2, 4, 8\) with remainder \(0\), giving the quotient \(x^3 + 2x^2 + 4x + 8\). This result makes sense because \(x^4 - 16 = (x - 2)(x + 2)(x^2 + 4)\), so dividing by \((x - 2)\) leaves \((x + 2)(x^2 + 4) = x^3 + 2x^2 + 4x + 8\).

With zeros \(x = -1\) (multiplicity 2) and \(x = 3\) (multiplicity 1) and leading coefficient \(1\), the polynomial is \(f(x) = (x + 1)^2(x - 3)\). Expanding: \((x + 1)^2 = x^2 + 2x + 1\), then multiplying by \((x - 3)\): \(x^3 - 3x^2 + 2x^2 - 6x + x - 3 = x^3 - x^2 - 5x - 3\). A frequent error is writing \((x - 1)\) instead of \((x + 1)\) for the zero at \(x = -1\), which would produce choice C.

Substituting \(u = x^2\) transforms the equation into \(u^2 - 13u + 36 = 0\), which factors as \((u - 4)(u - 9) = 0\). This gives \(u = 4\) and \(u = 9\), so \(x^2 = 4\) yields \(x = \pm 2\) and \(x^2 = 9\) yields \(x = \pm 3\). A common error is stopping at \(u = 4\) and \(u = 9\) and reporting those as the answers, forgetting to take square roots of each solution.

Using synthetic division with \(c = 1\) and coefficients \(2, -1, -7, 6\): the quotient is \(2x^2 + x - 6\). Factoring: \(2x^2 + x - 6 = (2x - 3)(x + 2) = 0\) gives \(x = \frac{3}{2}\) and \(x = -2\). The three zeros are \(x = 1\), \(x = \frac{3}{2}\), and \(x = -2\). Choice B is wrong because dividing by \(x = -1\) rather than \(x = 1\) would leave a non-zero remainder, as \(f(-1) = -2 - 1 + 7 + 6 = 10\) does not equal zero.

Because the polynomial has real coefficients, the conjugate \(1 - i\) must also be a zero, giving four zeros: \(1 + i\), \(1 - i\), \(2\), \(-2\). The factor for the complex pair is \((x - (1 + i))(x - (1 - i)) = (x - 1)^2 + 1 = x^2 - 2x + 2\). Multiplying by \((x - 2)(x + 2) = x^2 - 4\): \((x^2 - 2x + 2)(x^2 - 4) = x^4 - 4x^2 - 2x^3 + 8x + 2x^2 - 8 = x^4 - 2x^3 - 2x^2 + 8x - 8\).

The Rational Zero Theorem gives possible rational zeros including +/-{1, 2, 4, 8, 1/3, 2/3, 4/3, 8/3}. Testing x = 4: f(4) = 3(64) - 7(16) - 22(4) + 8 = 192 - 112 - 88 + 8 = 0. Synthetic division by (x - 4) yields the quotient 3x^2 + 5x - 2, which factors as (3x - 1)(x + 2), giving x = 1/3 and x = -2. The three zeros are x = 4, 1/3, and -2. Testing x = -4 gives f(-4) = -192 - 112 + 88 + 8 = -208, which is not zero, ruling out choice B.

Since \(2i\) is a zero of a real-coefficient polynomial, its conjugate \(-2i\) must also be a zero. The five zeros are \(x = -3\) (mult. 2), \(x = 1\), \(x = 2i\), and \(x = -2i\). The complex conjugate pair produces \((x - 2i)(x + 2i) = x^2 + 4\). The factor for \(x = -3\) (mult. 2) is \((x + 3)^2\), not \((x - 3)^2\) as in choice B. The factor for \(x = 1\) is \((x - 1)\), not \((x + 1)\) as in choice C. Choice D uses \(x^2 - 4 = (x - 2)(x + 2)\), which would add real zeros at \(\pm 2\) instead of complex zeros.

Synthetic division by \((x - 1)\) gives the quotient \(x^3 - 2x^2 + x - 2\). Factoring by grouping: \(x^2(x - 2) + 1(x - 2) = (x - 2)(x^2 + 1)\). Setting each factor to zero: \(x = 2\) from \((x - 2)\), and \(x = \pm i\) from \(x^2 + 1 = 0\). All four zeros are \(x = 1, 2, i, -i\). Choice C incorrectly uses \(x^2 - 1 = 0\) (which gives \(\pm 1\)) instead of \(x^2 + 1 = 0\).

Counting sign changes in \(f(x) = x^5 - 2x^4 + x^3 - x^2 + 2x - 1\): the signs are \(+, -, +, -, +, -\), giving 5 sign changes. By Descartes' Rule, the number of positive real zeros equals \(5\) or decreases by any even integer, so the possible counts are \(5, 3\), or \(1\). For \(f(-x) = -x^5 - 2x^4 - x^3 - x^2 - 2x - 1\), all coefficients are negative, yielding 0 sign changes and hence no negative real zeros. Choice B would apply to a polynomial with 4 sign changes.

The coefficients \(1, -4, 6, -4, 1\) match the binomial expansion of \((x - 1)^4\), since the binomial theorem gives \((x - 1)^4 = x^4 - 4x^3 + 6x^2 - 4x + 1\) using the row-4 Pascal's triangle coefficients \(1, 4, 6, 4, 1\). Therefore \(x = 1\) is the only zero, with multiplicity 4. The graph touches the x-axis at \(x = 1\) without crossing, consistent with even multiplicity. Choice B is wrong because \(f(-1) = 1 + 4 + 6 + 4 + 1 = 16\), confirming \(x = -1\) is not a zero.

End behavior is determined solely by the leading term, 2x^6. Because the degree is even and the leading coefficient is positive (2 > 0), both ends of the graph rise to +∞. Choice A is wrong because it describes an odd-degree polynomial with a positive leading coefficient, where the two ends go in opposite directions.

The Remainder Theorem states that the remainder of f(x) divided by (x - c) is f(c). You substitute the value c (not -c) directly into the polynomial. Choice B, f(-c), would be correct only if dividing by (x + c), which can be rewritten as (x - (-c)).

The Fundamental Theorem of Algebra guarantees that every polynomial of degree \(n\) has exactly \(n\) zeros in the complex number system, counting multiplicity. Because the degree of \(f(x)\) is \(5\), there are exactly \(5\) complex zeros. The number of terms (3) and the constant term do not determine the zero count.

Setting \(f(x) = 0\) gives \(x^3 = 27\), so \(x = \sqrt[3]{27} = 3\). Checking: \(3^3 - 27 = 27 - 27 = 0\). Choice A, \(x = -3\), gives \((-3)^3 - 27 = -27 - 27 = -54 \neq 0\). The other choices do not satisfy the equation either.

When a zero has odd multiplicity (1, 3, 5, …), the factor changes sign on either side of x = a, so the graph crosses the x-axis. Choice A describes even-multiplicity zeros, where the factor does not change sign and the graph bounces back without crossing.

The degree of a polynomial is the highest exponent of the variable. The term with the greatest exponent here is 7x^4, so the degree is 4. Choice B, 11, is the absolute value of the constant term, not the degree.

The Factor Theorem is a corollary of the Remainder Theorem. When f(x) is divided by (x - k), the remainder is f(k). For (x - k) to be a factor, the remainder must be zero, so f(k) = 0. Choice D, f(-k) = 0, would apply to the factor (x + k), not (x - k).

By the Remainder Theorem, the remainder equals f(2). Evaluating: f(2) = 3(8) - 2(4) + 2 - 5 = 24 - 8 + 2 - 5 = 13. Choice D, -5, is the constant term of f(x) and equals f(0), not f(2).

Writing \(f(x)\) with a placeholder: \(x^3 + 3x^2 + 0x - 4\), divide by \((x - (-2))\). Synthetic division with \(-2\): bring down \(1\); \(-2(1) = -2\), add to \(3 \to 1\); \(-2(1) = -2\), add to \(0 \to -2\); \(-2(-2) = 4\), add to \(-4 \to 0\). The coefficients \(1, 1, -2\) give quotient \(x^2 + x - 2\) with remainder \(0\). Choice B has the wrong sign on the x-term because the middle step produces \(+1\), not \(-1\).

To count negative real zeros, evaluate \(f(-x) = -x^5 + 2x^4 + x^3 - 3x^2 - x - 1\). The sign sequence is \(-, +, +, -, -, -\). The sign changes are: \(-\) to \(+\) (one) and \(+\) to \(-\) (one) \(= 2\) changes. By Descartes' Rule, there are \(2\) or \(0\) negative real zeros. Choice B (\(3\) or \(1\)) comes from incorrectly counting sign changes in the original \(f(x)\) rather than \(f(-x)\).

When a polynomial has real coefficients, complex zeros must appear in conjugate pairs. So if x = 3i is a zero, then x = -3i must also be a zero. Combined with the two real zeros x = 2 and x = -1, the minimum total number of zeros is 4, meaning the minimum degree is 4. Choice A, 3, is wrong because it would require omitting the conjugate -3i.

The degree is 3 + 2 = 5 and the leading coefficient is positive (1 × 1 = 1). An odd degree with a positive leading coefficient gives opposite end behaviors: f(x)→-∞ as x→-∞ and f(x)→+∞ as x→+∞. Choice C describes an even-degree polynomial where both ends go to +∞.

A zero at x = -2 requires the factor (x + 2), a zero at x = 0 requires the factor x, and a zero at x = 5 requires the factor (x - 5). Only choice A contains all three correct factors. Choice B shifts the signs incorrectly, producing zeros at x = 2, x = 0, and x = -5.

By the Factor Theorem, if \((x - 3)\) is a factor then \(f(3) = 0\). Substituting: \(27 + 9b - 21 + 3 = 0 \to 9 + 9b = 0 \to b = -1\). Choice B, \(b = 1\), gives \(f(3) = 27 + 9 - 21 + 3 = 18 \neq 0\) and can be ruled out.

The three distinct zero values are x = -3, x = 1, and x = 2, producing 3 distinct x-intercepts. At x = -3 and x = 1 the graph crosses (odd multiplicity), while at x = 2 it touches and turns around (even multiplicity). The total count of zeros including multiplicity is 1 + 1 + 2 = 4, matching the degree, but that is the zero count, not the x-intercept count.

Substituting \(u = x^2\) gives \(u^2 - 5u + 4 = (u - 1)(u - 4) = 0\), so \(u = 1\) or \(u = 4\). Solving back: \(x^2 = 1 \to x = \pm 1\) and \(x^2 = 4 \to x = \pm 2\). All four zeros are distinct and have multiplicity 1, so the graph crosses the x-axis at each one, giving 4 crossings. Choice A, \(2\), is wrong because it counts only positive zeros.

Evaluate \(f\) at the endpoints: \(f(2) = 8 - 4 - 5 = -1 < 0\) and \(f(3) = 27 - 6 - 5 = 16 > 0\). Since \(f\) is continuous and changes sign on \((2, 3)\), the IVT guarantees a real zero there. For choices A and B: \(f(0) = -5 < 0\), \(f(1) = -6 < 0\), \(f(2) = -1 < 0\), so no sign change occurs on \((0, 1)\) or \((1, 2)\).

By the Fundamental Theorem of Algebra, a degree-6 polynomial has exactly 6 zeros counting multiplicity. With 2 real zeros accounted for, the remaining 4 zeros are non-real. Since non-real complex zeros of a real-coefficient polynomial always come in conjugate pairs, having 4 non-real zeros (2 conjugate pairs) is consistent. Choice A, 2, would mean only 1 conjugate pair, leaving the total at 4 — not enough for degree 6.

Testing \(x = 1\): \(1 - 3 + 6 - 12 + 8 = 0\). Synthetic division by \((x - 1)\) gives quotient \(x^3 - 2x^2 + 4x - 8\). Factor by grouping: \(x^2(x - 2) + 4(x - 2) = (x^2 + 4)(x - 2)\), so \(x = 2\) is a real zero and \(x^2 + 4 = 0\) gives \(x = \pm 2i\). Choice C is wrong because \(x^2 + 4 = 0\) yields \(x = \pm 2i\), not \(\pm 4i\).

Because coefficients are real, \(x = 2 - i\) is also a zero. Build the factors: \((x + 1)^2 = x^2 + 2x + 1\) and \((x - (2+i))(x - (2-i)) = (x-2)^2 + 1 = x^2 - 4x + 5\). Multiplying: \((x^2 + 2x + 1)(x^2 - 4x + 5) = x^4 - 4x^3 + 5x^2 + 2x^3 - 8x^2 + 10x + x^2 - 4x + 5 = x^4 - 2x^3 - 2x^2 + 6x + 5\). Choice A incorrectly has \(+2x^2\) instead of \(-2x^2\).

Possible rational zeros are \(\pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}\). Testing \(x = 2\): \(2(8) + 4 - 26 + 6 = 0\). Synthetic division by \(2\) yields \(2x^2 + 5x - 3\), which factors as \((2x - 1)(x + 3)\), giving \(x = \frac{1}{2}\) and \(x = -3\). The three real zeros are \(2, \frac{1}{2}\), and \(-3\). Choice D incorrectly lists \(x = 3\) instead of \(x = -3\).

Divide by \((x - 1)\) via synthetic division to get \(x^3 + 3x^2 - 4x - 12\), then divide by \((x - 2)\) to get \(x^2 + 5x + 6\). Factoring: \(x^2 + 5x + 6 = (x + 2)(x + 3)\), giving \(x = -2\) and \(x = -3\). Choice C mixes the signs; with a positive x-coefficient of 5, both remaining zeros must be negative by Vieta's formulas.

Counting sign changes in \(f(x) = 2x^4 - x^3 + 3x^2 - x + 2\): the signs are \(+, -, +, -, +\) with 4 changes, meaning 4, 2, or 0 positive real zeros. For \(f(-x) = 2x^4 + x^3 + 3x^2 + x + 2\), all coefficients are positive, giving 0 sign changes and therefore 0 negative real zeros. Choice B incorrectly allows for negative real zeros when the sign pattern of \(f(-x)\) precludes them.

\(x^6 - 1 = (x^3 - 1)(x^3 + 1) = (x - 1)(x^2 + x + 1)(x + 1)(x^2 - x + 1)\). Each quadratic factor is irreducible over the reals: the discriminants are \(1 - 4 = -3 < 0\) for both. Choice B is incomplete because \(x^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x + 1)\) can be factored further, and choices C and D have not been fully reduced.

Because coefficients are real, \(x = 1 - 2i\) is also a zero. The conjugate pair contributes \((x - (1+2i))(x - (1-2i)) = (x-1)^2 + 4 = x^2 - 2x + 5\). The zero \(x = -3\) contributes \((x + 3)\), and \(x = 0\) (multiplicity 2) contributes \(x^2\). Including the leading coefficient: \(f(x) = -2x^2(x + 3)(x^2 - 2x + 5)\). Choice A incorrectly uses \(x^2 + 2x + 5\), which would correspond to the conjugate pair centered at \(-1\), not \(+1\).

The degree of a polynomial is the highest power of \(x\) with a nonzero coefficient. Here the highest power is \(x^5\), so the degree is 5. The constant term \(-7\) does not affect the degree.

The leading coefficient belongs to the term with the highest degree. Written in standard form, f(x) = 2x^5 - 4x^3 - x + 9, so the leading coefficient is 2. A common mistake is using the coefficient of the first written term (-4) before reordering.

The Fundamental Theorem of Algebra states that every polynomial of degree n has exactly n zeros in the complex number system, counted with multiplicity. A degree-4 polynomial therefore has exactly 4 zeros.

Setting \(f(x) = 0\) gives \(x^3 = 8\), so \(x = 2\) is the real zero. You can verify: \(2^3 - 8 = 8 - 8 = 0\). The value \(x = -2\) gives \((-2)^3 - 8 = -16\), not zero.

By the Remainder Theorem, the remainder equals \(f(2) = (2)^3 - 4(2) + 1 = 8 - 8 + 1 = 1\). A common error is computing \(f(-2)\) instead of \(f(2)\).

A zero at x = 5 corresponds to the factor (x - 5). Multiplicity 2 means this factor appears twice, giving (x - 5)^2. The factor (x + 5) would correspond to a zero at x = -5.

For \(f(x) = x^4\), the leading term has an even degree and a positive leading coefficient. Both ends of the graph rise, so as \(x\) approaches positive infinity, \(f(x)\) approaches positive infinity.

A degree-3 polynomial has exactly 3 zeros in the complex number system. All 3 could be real (e.g., f(x) = x(x-1)(x+1)), so the maximum number of real zeros is 3. Complex zeros always come in conjugate pairs, so a cubic must have at least one real zero.

In synthetic division, you list the coefficients of f(x) in order: 2, -5, 4, -1. The first step is to bring down the leading coefficient, which is 2. The divisor root is 1 (from x - 1 = 0).

The leading term is -3x^5: odd degree and negative leading coefficient. For odd-degree polynomials with negative leading coefficients, the left end rises (approaches positive infinity) and the right end falls (approaches negative infinity).

Dividing: \(x^3 / x = x^2\). Then \(x^2(x+3) = x^3 + 3x^2\). Subtracting gives \(-x^2 - 5x - 6\). Next, \(-x^2 / x = -x\). Then \(-x(x+3) = -x^2 - 3x\). Subtracting gives \(-2x - 6\). Finally, \(-2x / x = -2\), and \(-2(x+3) = -2x - 6\) with remainder 0. Quotient is \(x^2 - x - 2\).

Complex zeros of polynomials with real coefficients come in conjugate pairs. x = i requires x = -i as a partner (2 complex zeros), and x = 2 + 3i requires x = 2 - 3i as a partner (2 more complex zeros). That accounts for 4 of the 5 zeros. The remaining zero must be real. So the minimum number of real zeros is 1.

First, \(x^4 - 16 = (x^2 - 4)(x^2 + 4)\). Then \(x^2 - 4\) factors further as \((x - 2)(x + 2)\). The factor \(x^2 + 4\) is irreducible over the reals (it has no real zeros). So the complete real factorization is \((x - 2)(x + 2)(x^2 + 4)\). Choice D includes complex factors, which goes beyond real factorization.

The Rational Zero Theorem states that any rational zero p/q must have p as a factor of the constant term (6) and q as a factor of the leading coefficient (3). So possible rational zeros are formed by (factors of 6)/(factors of 3). Choice D reverses this relationship incorrectly.

When a zero has even multiplicity, the graph touches the x-axis and turns around (bounces) at that point rather than crossing through it. Since x = 3 has multiplicity 2 (even), the graph touches but does not cross at x = 3.

Test x = 3: f(3) = (3+1)(3-2)(3-4) = (4)(1)(-1) = -4 < 0. On the interval (2, 4), there are no zeros of f, so the sign does not change. The function is always negative on (2, 4). A common mistake is not checking the sign of each factor separately.

By the Factor Theorem, if f(c) = 0 then (x - c) is a factor of f(x). When synthetic division yields remainder 0, it confirms f(c) = 0, meaning x = c is a zero and (x - c) is a factor. The factor would be (x - c), not (x + c).

To find negative real zeros, evaluate \(f(-x) = -x^5 - x^3 - x\). There are no sign changes in the coefficients (\(-1, -1, -1\)), so by Descartes' Rule of Signs, \(f(x)\) has 0 negative real zeros.

Since \(x = 1\) is a zero, divide by \((x - 1)\) using synthetic division: coefficients \(1, -1, -4, 4\) with root \(1\) give quotient \(x^2 - 4\). Setting \(x^2 - 4 = 0\) yields \(x = 2\) and \(x = -2\). So all real zeros are \(x = 1\), \(x = 2\), and \(x = -2\).

Complex zeros come in conjugate pairs, so \(x = 1 - 2i\) is also a zero. The factor from the complex pair: \((x - (1+2i))(x - (1-2i)) = ((x-1) - 2i)((x-1) + 2i) = (x-1)^2 + 4 = x^2 - 2x + 5\). The factor from \(x = 3\) (multiplicity 2, since we need degree 4): \((x-3)^2 = x^2 - 6x + 9\). Then \(f(x) = (x^2 - 2x + 5)(x^2 - 6x + 9)\). Expanding: \(x^4 - 6x^3 + 9x^2 - 2x^3 + 12x^2 - 18x + 5x^2 - 30x + 45 = x^4 - 8x^3 + 26x^2 - 48x + 45\). Rechecking the cross terms: \(-18x - 30x = -48x\). Answer is \(x^4 - 8x^3 + 26x^2 - 48x + 45\). The closest match is choice A.

Test \(x = 3\): \(f(3) = 81 - 135 + 45 + 15 - 6 = 0\). So \((x - 3)\) is a factor. Testing \(x = -2\): \(f(-2) = 16 + 40 + 20 - 10 - 6 = 60\), not zero. Testing \(x = 6\): \(f(6) = 1296 - 1080 + 180 + 30 - 6 = 420\), not zero.

A degree-6 polynomial has exactly 6 zeros (counting multiplicity). With 4 distinct real zeros (each multiplicity 1), 2 zeros remain. Non-real complex zeros must come in conjugate pairs, so the remaining 2 zeros form exactly one conjugate pair. Therefore the polynomial has exactly 2 non-real complex zeros.

Possible rational zeros: factors of 6 over factors of 2 = ±1, ±2, ±3, ±6, ±1/2, ±3/2. Test x = 2: 2(8) + 3(4) - 11(2) - 6 = 16 + 12 - 22 - 6 = 0. So (x - 2) is a factor. Synthetic division by 2 gives 2x^2 + 7x + 3. Factor 2x^2 + 7x + 3 = (2x + 1)(x + 3). So f(x) = (x - 2)(2x + 1)(x + 3).

Factor out \(x\): \(f(x) = x(x^4 - x^3 - 7x^2 + x + 6)\). One zero is \(x = 0\). For the quartic, test \(x = -1\): \(1 + 1 - 7 - 1 + 6 = 0\). Divide by \((x+1)\): quotient is \(x^3 - 2x^2 - 5x + 6\). Test \(x = 1\): \(1 - 2 - 5 + 6 = 0\). Divide by \((x-1)\): quotient is \(x^2 - x - 6 = (x-3)(x+2)\). So all zeros are \(x = 0, -1, 1, 3, -2\).

Since \(f(x) = x^4 + ax^2 + b\) contains only even powers, if \(r\) is a zero then \(-r\) is also a zero. Given \(x = 2\) and \(x = -3\), their negatives \(x = -2\) and \(x = 3\) must also be zeros. So \(f(x) = (x-2)(x+2)(x-3)(x+3) = (x^2 - 4)(x^2 - 9) = x^4 - 13x^2 + 36\). Therefore \(a = -13\) and \(b = 36\).

The degree of a polynomial is the highest exponent of x among all terms. The terms have degrees 5, 3, 1, and 0 respectively. The highest is 5, so the degree is 5. A common error is confusing the degree with the number of terms or with the constant term.

The Fundamental Theorem of Algebra guarantees that every polynomial of degree \(n\) has exactly \(n\) zeros in the complex number system, counting multiplicity. Since \(f(x)\) has degree 7, it has exactly 7 complex zeros. The number of terms (4) or the highest non-leading exponent (4) should not be confused with the degree.

End behavior is determined by the leading term, -2x^4. The degree is even (4), so both ends behave the same way. The leading coefficient is negative (-2), so both ends point downward toward -infinity. Choice C would be correct for a positive leading coefficient with even degree.

The Factor Theorem states that (x - c) is a factor of f(x) if and only if f(c) = 0. Here c = 3, so (x - 3) is a factor if and only if f(3) = 0. Choice D confuses the sign: (x - 3) relates to c = 3, not c = -3.

The leading term comes from multiplying the highest-degree terms of each factor: \(x^2\) times \(x^3 = x^5\). All other products (\(x^2\) times \(-2x = -2x^3\), and \(1\) times \(x^3 = x^3\)) produce lower-degree terms. The degree of the product equals the sum of the degrees: \(2 + 3 = 5\).

Factor \(f(x) = (x - 2)(x - 3)\). Setting each factor equal to zero gives \(x = 2\) and \(x = 3\). Of the choices, only \(x = 3\) is a zero. Verification: \(f(3) = 9 - 15 + 6 = 0\). The values \(-2\) and \(-3\) are not zeros: \(f(-2) = 4 + 10 + 6 = 20\).

A polynomial of degree n can have at most n - 1 turning points (local maxima or minima). For degree 6, the maximum is 6 - 1 = 5 turning points. A polynomial cannot have more turning points than one less than its degree, so 6 or 7 are impossible for a degree-6 polynomial.

Set up synthetic division with \(c = 2\) and coefficients \(1, -3, 0, 4\) (note the 0 placeholder for the missing \(x\) term). Bring down 1; multiply 1 times 2 to get 2, add to \(-3\) for \(-1\); multiply \(-1\) times 2 to get \(-2\), add to 0 for \(-2\); multiply \(-2\) times 2 to get \(-4\), add to 4 for remainder 0. The quotient is \(x^2 - x - 2\).

Count sign changes in the coefficients of \(f(x)\): \((+), (-), (+), (-), (+)\) — that is 4 sign changes. Descartes' Rule says the number of positive real zeros equals the number of sign changes or is less by an even number. So possible counts are 4, 2, or 0. The rule does not guarantee exactly 4; unaccounted-for zeros are complex.

Factor by grouping: \(x^2(x + 2) - 1(x + 2) = (x + 2)(x^2 - 1) = (x + 2)(x - 1)(x + 1)\). Setting each factor to zero gives \(x = -2\), \(x = 1\), and \(x = -1\). Verify: \(f(-2) = -8 + 8 + 2 - 2 = 0\), \(f(1) = 1 + 2 - 1 - 2 = 0\), \(f(-1) = -1 + 2 + 1 - 2 = 0\).

A zero at x = 1 with multiplicity 2 contributes the factor (x - 1)^2. A zero at x = -3 contributes (x + 3), and a zero at x = 2 contributes (x - 2). The polynomial is (x - 1)^2(x + 3)(x - 2). Choice D incorrectly uses (x - 3) and (x + 2), which would produce zeros at x = 3 and x = -2, not x = -3 and x = 2.

Factoring gives \(f(x) = x^2(x - 1)(x + 1)\), so \(x = 0\) is a zero of multiplicity 2, which is even. Zeros of even multiplicity cause the graph to touch the x-axis and bounce back without crossing, resembling a parabola at that point. Zeros of odd multiplicity cross the x-axis, like a linear factor.

The Complex Conjugate Root Theorem states that for polynomials with real coefficients, complex zeros always appear in conjugate pairs. The conjugate of 3 + 2i is 3 - 2i (negate only the imaginary part). Choice A negates the real part only, and choice C negates both parts, neither of which is the conjugate.

Synthetic division with c = 3 and coefficients 2, -3, -11, 6: bring down 2; 2 times 3 = 6, add to -3 for 3; 3 times 3 = 9, add to -11 for -2; -2 times 3 = -6, add to 6 for remainder 0. The quotient is 2x^2 + 3x - 2. Note: 2x^2 + 3x - 2 = (2x - 1)(x + 2), so the full factorization is (x - 3)(2x - 1)(x + 2).

The leading term is \(-x^5\). The degree is odd (5) and the leading coefficient is negative (\(-1\)). For odd-degree polynomials with negative leading coefficients: as \(x\) approaches \(+\infty\), \(f(x)\) approaches \(-\infty\), and as \(x\) approaches \(-\infty\), \(f(x)\) approaches \(+\infty\). Choice B describes an odd-degree polynomial with a positive leading coefficient.

The Remainder Theorem states that when f(x) is divided by (x - c), the remainder equals f(c). Rewrite the divisor: x + 4 = x - (-4), so c = -4. The remainder equals f(-4). A common error is using c = 4 (ignoring the sign change), which would apply to the divisor (x - 4), not (x + 4).

Treat as a quadratic in \(x^2\): \(x^4 - 5x^2 + 4 = (x^2 - 1)(x^2 - 4)\). Each factor is a difference of squares: \(x^2 - 1 = (x - 1)(x + 1)\) and \(x^2 - 4 = (x - 2)(x + 2)\). The fully factored form is \((x - 1)(x + 1)(x - 2)(x + 2)\). Choice A is only partially factored; choices B and C each leave one quadratic unfactored over the reals.

Since \(x = 2\) has multiplicity 2, divide \(f(x)\) by \((x - 2)^2 = x^2 - 4x + 4\). Polynomial long division yields quotient \(x^2 - 2x + 1 = (x - 1)^2\). So \(f(x) = (x - 2)^2(x - 1)^2\), giving zeros \(x = 2\) (multiplicity 2) and \(x = 1\) (multiplicity 2) — four zeros total, consistent with degree 4. Choice D would give 5 total zeros, which is impossible for a degree-4 polynomial.

By the Complex Conjugate Root Theorem, since 1 + i is a zero of a real-coefficient polynomial, its conjugate 1 - i must also be a zero. The zero x = -2 with multiplicity 2 contributes two copies. That gives four zeros: 1 + i, 1 - i, -2, -2, which matches the degree. Choice A incorrectly replaces -2 (multiplicity 2) with -2 and 2, which are different values.

Since \(x = 2\) is a double zero, \((x - 2)^2 = x^2 - 4x + 4\) is a factor. Dividing \(f(x)\) by \(x^2 - 4x + 4\) via long division yields quotient \(x^2 + 2x + 1 = (x + 1)^2\). Therefore \(f(x) = (x - 2)^2(x + 1)^2\). Verify by expanding: \((x^2 - 4x + 4)(x^2 + 2x + 1) = x^4 - 2x^3 - 3x^2 + 4x + 4\). Choice C would give zeros at \(x = 2\) and \(x = 1\), but \(f(1) = 1 - 2 - 3 + 4 + 4 = 4\), not 0.

By the Complex Conjugate Root Theorem, since the polynomial has real coefficients and x = i is a zero, x = -i must also be a zero. The full list is x = 3 (mult 2), x = -1, x = i, x = -i, giving 2 + 1 + 1 + 1 = 5 zeros, consistent with degree 5. Choice A incorrectly pairs i with 1 instead of -i.

The Rational Zero Theorem gives possible zeros \(\pm 1, \pm 2, \pm 3, \pm 6\). Test \(x = 1\): \(f(1) = 1 - 7 + 6 = 0\). Synthetic division by \((x - 1)\) with coefficients \(1, 0, -7, 6\) yields quotient \(x^2 + x - 6 = (x + 3)(x - 2)\). All zeros: \(x = 1\), \(x = 2\), \(x = -3\). Verify: \(f(2) = 8 - 14 + 6 = 0\) and \(f(-3) = -27 + 21 + 6 = 0\).

Synthetic division by \((x - 1)\) with coefficients \(2, -3, -9, 15, -5\) gives quotient \(2x^3 - x^2 - 10x + 5\). Factor by grouping: \(x^2(2x - 1) - 5(2x - 1) = (2x - 1)(x^2 - 5)\). Setting \(2x - 1 = 0\) gives \(x = 1/2\); setting \(x^2 - 5 = 0\) gives \(x = \sqrt{5}\) and \(x = -\sqrt{5}\). The remaining zeros are \(1/2\), \(\sqrt{5}\), and \(-\sqrt{5}\).

The Complex Conjugate Root Theorem applies to polynomials with real coefficients: since \(2 + 3i\) is a zero, its conjugate \(2 - 3i\) must also be a zero. The values \(\sqrt{5}\) and \(-1\) are real, so they do not generate complex conjugates. Note that \(-\sqrt{5}\) would only be required if the polynomial had rational (not merely real) coefficients, by the Conjugate Irrational Root Theorem — that is a stricter condition.

Since \(x = 1\) and \(x = -3\) are zeros, \((x - 1)(x + 3) = x^2 + 2x - 3\) is a factor. Dividing \(f(x)\) by \(x^2 + 2x - 3\): the first step gives \(x^2\) as the leading quotient term; subtracting \(x^2(x^2 + 2x - 3) = x^4 + 2x^3 - 3x^2\) leaves \(-4x^2 - 8x + 12 = -4(x^2 + 2x - 3)\), so the quotient is \(x^2 - 4 = (x - 2)(x + 2)\). Complete factorization: \((x - 1)(x + 3)(x - 2)(x + 2)\).