Exponents and Polynomials — Free Algebra 1 Review Games.

When multiplying same bases, add exponents: \(3+4=7\).

Power of a power: multiply exponents: \(2 \cdot 3=6\).

Any nonzero number raised to the \(0\) power equals \(1\).

Combine like terms: 3+5 = 8, so 8x^2.

The degree is the highest exponent, which is 3.

When dividing same bases, subtract exponents: \(5-2=3\).

Cube both: \(2^3 \cdot x^3 = 8x^3\).

Combine like terms: \(4x^2 - 2x + 4\).

Distribute \(x\) to each term.

A negative exponent means reciprocal: \(x^{-2} = 1/x^2\).

FOIL: \(x^2 + 5x + 3x + 15 = x^2 + 8x + 15\).

6x^2 + 8x - 3x - 4 = 6x^2 + 5x - 4.

Square each factor: 3^2 * x^(2*2) * y^2 = 9x^4y^2.

\((x-4)^2 = x^2 - 2(4)x + 16 = x^2 - 8x + 16\).

\(x^{3-5} \cdot y^{2-1} = x^{-2} \cdot y = y/x^2\).

Multiply the coefficients: \(2 \cdot 3 = 6\). Then add the exponents using the Product Rule: \(x^2 \cdot x^3 = x^{2+3} = x^5\). The result is \(6x^5\). A common mistake is multiplying the exponents instead of adding them, which incorrectly gives \(6x^6\).

The degree of a monomial with multiple variables is the sum of all the variable exponents. Here, 2 + 3 = 5, so the degree is 5. Choosing 3 (the largest single exponent) is a common error — you must add all variable exponents, not just pick the highest one.

A binomial has exactly two terms. \(4x - 7\) has two terms: \(4x\) and \(-7\). \(5x^3\) is a monomial (one term), while \(3x^2 - 2x + 1\) and \(x^2 + x - 3\) are both trinomials (three terms each).

When raising a power to a power, multiply the exponents: (y^3)^4 = y^(3 * 4) = y^12. A very common mistake is adding the exponents instead of multiplying, which gives y^7 — that rule applies to multiplication like y^3 * y^4, not to a power raised to a power.

Any nonzero constant can be written as 8x^0, so its degree is 0. The number 8 is the coefficient, not the degree. The degree is not undefined — only the zero polynomial is considered to have no defined degree.

\(8x^4\) and \(3x^4\) are like terms because they share the same variable and exponent. Subtract only the coefficients: \(8 - 3 = 5\). The variable part \(x^4\) stays unchanged. The result is \(5x^4\). Do not subtract the exponents or add the coefficients.

A monomial is a single-term expression consisting of a number, a variable, or a product of numbers and variables with non-negative integer exponents. \(4x^3\) is a monomial. The other choices all contain addition or subtraction, making them polynomials with two or more terms.

Apply the exponent to both the coefficient and the variable separately: \(2^3 = 8\) and \((x^3)^3 = x^{3 \cdot 3} = x^9\). The result is \(8x^9\). A common mistake is multiplying \(2\) by \(3\) instead of cubing it (giving \(6x^9\)), or forgetting to apply the Power Rule correctly to the exponent (giving \(8x^6\)).

Distribute the negative sign to every term in the second polynomial: \((4x^2 - 3x + 2) - x^2 - 2x + 5\). Combine like terms: \((4-1)x^2 = 3x^2\), \((-3-2)x = -5x\), and \(2+5 = 7\). The result is \(3x^2 - 5x + 7\). The most common error is forgetting to distribute the negative to \(-5\), which incorrectly gives \(-3\) as the constant instead of \(+7\).

Distribute \(3x\) to every term inside the parentheses: \(3x \cdot x^2 = 3x^3\), \(3x \cdot (-2x) = -6x^2\), and \(3x \cdot 4 = 12x\). The result is \(3x^3 - 6x^2 + 12x\). A frequent error is multiplying \(3x\) by \(4\) as just \(4\) (forgetting the variable), which gives \(+4\) in the last term instead of \(+12x\).

Add the exponents using the Product Rule: \(x^2 \cdot x^{-5} = x^{2 + (-5)} = x^{-3}\). Convert the negative exponent to a fraction: \(x^{-3} = 1/x^3\). Subtracting instead of adding gives \(x^{2-(-5)} = x^7\), which is incorrect — always add exponents when multiplying.

Divide the coefficients: \(12 / 4 = 3\). Apply the Quotient Rule for exponents: \(x^5 / x^2 = x^{5-2} = x^3\). The result is \(3x^3\). Subtracting the coefficients (\(12 - 4 = 8\)) instead of dividing is a common error, as is adding the exponents instead of subtracting them.

Work from the inside out. First apply the Product Rule inside the parentheses: \(x^3 \cdot x^2 = x^5\). Then apply the outer exponent using the Power Rule: \((x^5)^2 = x^{10}\). A common mistake is adding all three exponents together (\(3 + 2 + 2 = 7\)) rather than following the correct order of operations.

A negative exponent means take the reciprocal of the base raised to the positive exponent: 2^(-3) = 1 / 2^3 = 1/8. The result is positive, not negative — negative exponents do not create negative values. 1/6 is wrong because 2^3 = 8, not 6.

Apply the Power Rule to each factor by multiplying exponents: \((x^4)^3 = x^{4 \cdot 3} = x^{12}\) and \((y^2)^3 = y^{2 \cdot 3} = y^6\). The result is \(x^{12} y^6\). Adding exponents instead of multiplying them gives \(x^7 y^5\), which confuses the Product Rule with the Power Rule.

Group and combine like terms: \(x^3\) terms: \(3x^3 + (-x^3) = 2x^3\). \(x\) terms: \(-x + 4x = 3x\). Constants: \(2 + (-5) = -3\). The result is \(2x^3 + 3x - 3\). A sign error on the \(x\) terms — treating \(-x + 4x\) as \(-3x\) instead of \(+3x\) — is a frequent mistake.

Multiply each exponent by 4 using the Power Rule: (a^2)^4 = a^(2*4) = a^8 and (b^3)^4 = b^(3*4) = b^12. The result is a^8 b^12. Adding exponents instead of multiplying (2+4=6 and 3+4=7) gives a^6 b^7, which is the Product Rule incorrectly applied here.

The negative exponent means take the reciprocal: \((2x)^{-2} = 1/(2x)^2\). Now square the entire expression in the denominator: \((2x)^2 = 2^2 \cdot x^2 = 4x^2\). So the result is \(1/(4x^2)\). Forgetting to square the coefficient \(2\) gives \(1/(2x^2)\). The expression is positive, not negative.

Distribute: \(x(x^2 - 3x + 9) = x^3 - 3x^2 + 9x\), and \(3(x^2 - 3x + 9) = 3x^2 - 9x + 27\). Combine: \(x^3 + (-3x^2 + 3x^2) + (9x - 9x) + 27 = x^3 + 27\). The middle terms cancel completely. This is the Sum of Cubes pattern: \((a + b)(a^2 - ab + b^2) = a^3 + b^3\) with \(a = x\) and \(b = 3\).

Use the Perfect Square Trinomial pattern: (a + b)^2 = a^2 + 2ab + b^2 with a = 2x and b = 3. Compute each part: a^2 = 4x^2, 2ab = 2(2x)(3) = 12x, b^2 = 9. The result is 4x^2 + 12x + 9. The most common error is writing just 4x^2 + 9, completely omitting the middle term 2ab.

Simplify the numerator first: \((3x^2 y^3)^2 = 9x^4 y^6\). Now divide: \(\frac{9x^4 y^6}{9xy^4}\). The 9s cancel to give 1. Apply the Quotient Rule: \(x^{4-1} = x^3\) and \(y^{6-4} = y^2\). The result is \(x^3 y^2\). Failing to cancel the 9s would give \(9x^3 y^2\), and adding exponents instead of subtracting gives \(x^3 y^{10}\).

First multiply using FOIL: \((x - 2)(x + 5) = x^2 + 5x - 2x - 10 = x^2 + 3x - 10\). Then subtract \(x^2\): \((x^2 + 3x - 10) - x^2 = 3x - 10\). Forgetting to subtract \(x^2\) at the end leaves the answer as \(x^2 + 3x - 10\), which is choice C.

First apply the exponent to \((2x^2 y)^2\): square each factor to get \(2^2 = 4\), \((x^2)^2 = x^4\), and \(y^2\). This gives \(4x^4 y^2\). Now multiply by \(xy^3\): coefficients \(4\cdot1 = 4\), exponents \(x^{4+1} = x^5\) and \(y^{2+3} = y^5\). The result is \(4x^5 y^5\). Forgetting to square the coefficient 2 gives \(2x^5 y^5\).

Multiply the first two factors using the Difference of Squares shortcut: \((x + 2)(x - 2) = x^2 - 4\). Then multiply \((x^2 - 4)(x + 1)\): \(x^2 \cdot x = x^3\), \(x^2 \cdot 1 = x^2\), \(-4 \cdot x = -4x\), \(-4 \cdot 1 = -4\). The result is \(x^3 + x^2 - 4x - 4\). Skipping the strategic first grouping and distributing incorrectly often leads to a missing \(x^2\) term.

Distribute \(3x\): \(3x(2x^2) = 6x^3\), \(3x(x) = 3x^2\), \(3x(-4) = -12x\). Distribute \(-1\): \(-1(2x^2) = -2x^2\), \(-1(x) = -x\), \(-1(-4) = +4\). Combine like terms: \(6x^3\), \((3x^2 - 2x^2) = x^2\), \((-12x - x) = -13x\), and \(+4\). The result is \(6x^3 + x^2 - 13x + 4\). Sign errors — especially on the constant term where \(-1\) times \(-4\) gives \(+4\), not \(-4\) — are the most common mistakes.

Any nonzero base raised to the power of \(0\) equals \(1\). This is the zero exponent rule. A common mistake is choosing \(0\), confusing the value of the exponent with the result of the expression.

A monomial has exactly one term, such as 5x^2 or -3y. A binomial has two terms, a trinomial has three terms. 'Quadratic' describes the degree of a polynomial, not the number of terms.

The degree of a monomial is the exponent on its variable. Since x is raised to the power of 4, the degree is 4. The coefficient 7 does not affect the degree.

3^0 = 1 by the zero exponent rule, and 4^1 = 4 because any base raised to the first power equals itself. So 1 + 4 = 5. Choosing 4 is a common error from forgetting that 3^0 = 1, not 0.

A binomial has exactly two terms. '\(x + 3\)' has two terms: \(x\) and \(3\). The expression \(5x^2\) is a monomial (one term), \(x^2 + x + 1\) is a trinomial (three terms), and \(4\) is a monomial constant.

The leading coefficient is the coefficient of the term with the highest degree. The term with the highest degree is -3x^5, so the leading coefficient is -3. The exponent 5 and the constant -7 are not the leading coefficient.

By the zero exponent rule, \(x^0 = 1\) for any nonzero \(x\). So \(x^3 \cdot x^0 = x^3 \cdot 1 = x^3\). Alternatively, the product rule gives \(x^{3+0} = x^3\). Choosing \(0\) is wrong because multiplying by \(x^0\) multiplies by \(1\), not \(0\).

The degree of a polynomial is the highest exponent among all its terms. The terms have degrees 2, 5, 1, and 0 respectively. The highest is 5, from the term -4x^5, so the degree of the polynomial is 5.

The power of a power rule states \((x^a)^b = x^{a \cdot b}\). So \((x^3)^4 = x^{3 \cdot 4} = x^{12}\). A common error is adding instead of multiplying the exponents to get \(x^7\). Multiplying the base \(x\) by the exponents to get \(x^{81}\) is also incorrect.

Distribute the negative sign across the second polynomial: \((5x^2 - 3x + 1) + (-2x^2 - x + 4)\). Now combine like terms: \((5-2)x^2 + (-3-1)x + (1+4) = 3x^2 - 4x + 5\). A common error is forgetting to flip the sign of \(-4\) to \(+4\), which gives \(3x^2 - 4x - 3\) instead.

The quotient rule for exponents states \(\frac{x^a}{x^b} = x^{a-b}\). So \(\frac{x^7}{x^3} = x^{7-3} = x^4\). Adding instead of subtracting gives the wrong answer \(x^{10}\). Dividing the exponents is not a valid rule.

Distribute \(2x\) to each term: \(2x \cdot 3x^2 = 6x^3\), \(2x \cdot (-x) = -2x^2\), and \(2x \cdot 4 = 8x\). The result is \(6x^3 - 2x^2 + 8x\). Choice B is wrong because \(2x \cdot (-x) = -2x^2\), not \(-2x\), and \(2x \cdot 4 = 8x\), not \(8\).

Multiply the coefficients: \(3 \cdot 4 = 12\). Apply the product rule to the variable parts: \(x^2 \cdot x^3 = x^{2+3} = x^5\). The result is \(12x^5\). Choice B is wrong because the exponents are added (\(2+3=5\)), not multiplied (\(2\cdot3=6\)).

A negative exponent means take the reciprocal of the base raised to the positive exponent: 2^(-4) = 1/(2^4) = 1/16. Choice B (1/8) would equal 2^(-3). Negative exponents do not make the result negative, so -8 and -16 are incorrect.

First simplify the numerator using the product rule: y^5 * y^(-2) = y^(5+(-2)) = y^3. Then since y^0 = 1, dividing by y^0 is dividing by 1, which leaves y^3 unchanged. Choice A incorrectly adds all three exponents as 5 + (-2) + 0 = 3 but misidentifies the sign.

Add like terms: \((4+1)x^2 = 5x^2\), \((-1+2)x = x\), and \((3-5) = -2\). The sum is \(5x^2 + x - 2\). Choice C incorrectly adds \(-1\) and \(2\) as \(+3\) instead of \(+1\). Choice D incorrectly subtracts \(-1\) and \(2\) as \(-1\) instead of \(+1\).

Using the quotient rule: \(\frac{x^2}{x^{-1}} = x^{2-(-1)} = x^{2+1} = x^3\). Subtracting a negative exponent is equivalent to adding, so the exponents add to give \(x^3\). Choice A (\(x\)) would result from \(\frac{x^2}{x^1}\), not \(x^{-1}\).

First find \((x+1)^2 = x^2 + 2x + 1\). Then multiply by \((x+1)\): \((x+1)(x^2+2x+1) = x^3 + 2x^2 + x + x^2 + 2x + 1 = x^3 + 3x^2 + 3x + 1\). Choice C (\(x^3 + 1\)) is a common error from only cubing each term individually without applying the distributive property.

Distribute each term of \((2x+1)\): \(2x(x^2 - x + 3) = 2x^3 - 2x^2 + 6x\) and \(1(x^2 - x + 3) = x^2 - x + 3\). Combining: \(2x^3 + (-2+1)x^2 + (6-1)x + 3 = 2x^3 - x^2 + 5x + 3\). Choice A incorrectly combines \(-2x^2 + x^2\) as \(+x^2\) instead of \(-x^2\).

First expand the numerator: \((3x^2 y)^3 = 3^3 \cdot x^{2\cdot3} \cdot y^3 = 27x^6 y^3\). Then divide: \(27/9 = 3\), \(x^6/x^3 = x^{6-3} = x^3\), and \(y^3/y^2 = y^{3-2} = y\). The result is \(3x^3 y\). Choice A is wrong because \(27/9 = 3\), not \(27\).

This is the difference of squares pattern: \((a + b)(a - b) = a^2 - b^2\). With \(a = x^2\) and \(b = 3\): \((x^2)^2 - 3^2 = x^4 - 9\). Choice A (\(x^4 - 6x^2 + 9\)) is the expansion of \((x^2 - 3)^2\), which is a different expression.

First apply the power rule: \((x^{-2} y^3)^2 = x^{-4} y^6\). Then multiply: \(x^{-4} y^6 \cdot x^4 y^{-1} = x^{-4+4} y^{6-1} = x^0 y^5 = 1 \cdot y^5 = y^5\). The \(x\) terms cancel completely since \(-4 + 4 = 0\).

Area = length * width = (3x + 2)(x - 1). Using FOIL: 3x*x + 3x*(-1) + 2*x + 2*(-1) = 3x^2 - 3x + 2x - 2 = 3x^2 - x - 2. Choice A (3x^2 + 5x - 2) comes from incorrectly computing -3x + 2x as +5x instead of -x.

Apply the power of a power rule: (x^a)^3 = x^(3a). Then use the product rule: x^(3a) * x^(-2a) = x^(3a + (-2a)) = x^a. Choice A is wrong because 3a + (-2a) = a, not 5a. Exponents cannot become squared (like -6a^2) through basic exponent rules.

Step 1: Expand \((2x^2)^3 = 8x^6\). Step 2: Multiply by \(x^{-4}\): \(8x^6 \cdot x^{-4} = 8x^{6-4} = 8x^2\). Step 3: Divide by \(4x\): \(\frac{8x^2}{4x} = 2x^{2-1} = 2x\). Choice A is wrong because \(8/4 = 2\) and \(x^2/x = x^1\), not \(x^3\).

By the zero exponent rule, any nonzero base raised to the power of \(0\) equals \(1\). So \(x^0 = 1\) for any nonzero \(x\). A common mistake is thinking \(x^0 = 0\) because the exponent is zero, but that confuses multiplying by zero with raising to the zero power.

A trinomial is a polynomial with exactly three terms. \(x^2 + 2x - 1\) has three terms: \(x^2\), \(2x\), and \(-1\). Choice A is a monomial (one term), Choice B is a binomial (two terms), and Choice D is a polynomial with four terms.

The degree of a polynomial is the highest exponent of the variable. The terms are 4x^3 (degree 3), -2x (degree 1), and 7 (degree 0). The highest is 3, so the polynomial has degree 3. A common error is counting the number of terms instead of finding the highest exponent.

When raising a power to a power, multiply the exponents: \((x^3)^2 = x^{3 \times 2} = x^6\). A common mistake is adding the exponents (getting \(x^5\)) instead of multiplying them. The rule is \((x^m)^n = x^{m \times n}\).

The leading coefficient is the coefficient of the term with the highest degree. The term with the highest degree is 5x^4, so the leading coefficient is 5. Choice A (-2) is the constant term, Choice B (1) is the coefficient of x, and Choice C (4) is the degree of the polynomial itself.

A monomial is a single algebraic term — one number, one variable, or their product with no addition or subtraction. \(3x^2\) is a single term, making it a monomial. Choice A is a binomial (two terms), Choice C is a trinomial (three terms), and Choice D is a binomial with two different variables.

When multiplying powers with the same base, add the exponents: a^4 * a^3 = a^(4+3) = a^7. Choice B (a^12) results from multiplying the exponents (4 times 3 = 12) instead of adding them. Choice C incorrectly introduces a coefficient of 2.

Multiply the coefficients and add the exponents of like bases: \((2)(3) = 6\), \(x^1 \cdot x^2 = x^{1+2} = x^3\), and \(y^2 \cdot y^1 = y^{2+1} = y^3\). The result is \(6x^3y^3\). Choices A and D incorrectly add the coefficients \((2+3=5)\) instead of multiplying them.

Combine like terms: \((3x^2 + x^2) + (2x - 5x) + (-4 + 1) = 4x^2 - 3x - 3\). Choice B incorrectly adds the x coefficients \((2+5=7)\) instead of recognizing the subtraction. Choice C subtracts the \(x^2\) terms instead of adding them.

\((x + 5)^2 = (x + 5)(x + 5) = x^2 + 5x + 5x + 25 = x^2 + 10x + 25\). Choice A is the most common error — forgetting the middle term entirely. The formula \((a + b)^2 = a^2 + 2ab + b^2\) gives \(2ab = 2(x)(5) = 10x\).

Use FOIL: First: \(x \cdot x = x^2\), Outer: \(x \cdot (-4) = -4x\), Inner: \(3 \cdot x = 3x\), Last: \(3 \cdot (-4) = -12\). Combining the middle terms: \(-4x + 3x = -x\). Result: \(x^2 - x - 12\). Choice D (\(x^2 + 7x - 12\)) results from adding 3 and 4 for the middle term instead of combining \(-4x\) and \(+3x\).

Apply the power rule to each factor inside the parentheses: (a^3)^2 = a^(3 times 2) = a^6 and (b^2)^2 = b^(2 times 2) = b^4. The result is a^6b^4. Choice A adds exponents (3+2=5) instead of multiplying. Choice D multiplies 3 by 3 for the a exponent rather than 3 by 2.

Distribute the negative sign to every term in the second polynomial: \(5x^2 - 3x + 2 - 2x^2 - x + 4\). Then combine like terms: \((5-2)x^2 + (-3-1)x + (2+4) = 3x^2 - 4x + 6\). Choice A gets \(-2\) for the constant term, which is the error of treating \(-(-4)\) as \(-4\) instead of \(+4\).

A negative exponent indicates a reciprocal: \(x^{-n} = \frac{1}{x^n}\). So \(5x^{-3} = 5 \cdot \frac{1}{x^3} = \frac{5}{x^3}\). The positive coefficient 5 remains in the numerator. Choice A incorrectly negates the coefficient, and Choice D incorrectly moves the coefficient 5 to the denominator.

Use FOIL: First: 2x * x = 2x^2, Outer: 2x * 5 = 10x, Inner: -3 * x = -3x, Last: -3 * 5 = -15. Combine middle terms: 10x - 3x = 7x. Result: 2x^2 + 7x - 15. Choice B (2x^2 - 7x - 15) results from subtracting 10x - 3x incorrectly by using the wrong signs.

Apply the exponent to each factor: \(4^2 = 16\) and \((x^2)^2 = x^{2 \times 2} = x^4\). The result is \(16x^4\). Choice A (\(8x^4\)) doubles the coefficient instead of squaring it. Choice D (\(16x^2\)) squares the coefficient correctly but fails to apply the outer exponent to \(x^2\).

When multiplying powers with the same base, add the exponents: b^(-2) * b^6 = b^(-2+6) = b^4. Choice A results from multiplying the exponents (-2 times 6 = -12) instead of adding. Choice D adds the absolute values (2+6=8) without accounting for the negative sign on -2.

\(-3x^2 + x - 5\) has degree 2 (the highest exponent is 2) and its leading coefficient is \(-3\). Choice A has degree 3, though \(-3\) appears as a coefficient of \(x\). Choice C has degree 2 but a leading coefficient of 1. Choice D has degree 1.

First expand the numerator: \((2x^2y^{-1})^3 = 2^3 \cdot x^{2 \times 3} \cdot y^{-1 \times 3} = 8x^6y^{-3}\). Then divide: \(\frac{8x^6y^{-3}}{4x^3y^2} = \frac{8}{4} \cdot x^{6-3} \cdot y^{-3-2} = 2x^3y^{-5} = \frac{2x^3}{y^5}\). Choice D (\(\frac{8x^3}{y^5}\)) forgets to divide the coefficients \(\frac{8}{4}\).

Distribute each term: \(2x(3x^2 + 2x - 4) - 1(3x^2 + 2x - 4) = 6x^3 + 4x^2 - 8x - 3x^2 - 2x + 4\). Combine like terms: \(6x^3 + (4-3)x^2 + (-8-2)x + 4 = 6x^3 + x^2 - 10x + 4\). Choice D makes a sign error on the constant: \(-1 \times -4 = +4\), not \(-4\).

Area = \((x+3)^2 = x^2 + 6x + 9\). Perimeter = \(4(x+3) = 4x + 12\). Area minus Perimeter = \((x^2 + 6x + 9) - (4x + 12) = x^2 + (6-4)x + (9-12) = x^2 + 2x - 3\). Choice D (\(x^2 + 2x + 3\)) makes an arithmetic error on the constant: \(9 - 12 = -3\), not \(+3\).

Add exponents in the numerator: \(x^{n+2} \cdot x^{n-1} = x^{(n+2)+(n-1)} = x^{2n+1}\). Then subtract exponents when dividing: \(\frac{x^{2n+1}}{x^{2n-1}} = x^{(2n+1)-(2n-1)} = x^{2n+1-2n+1} = x^2\). The variable \(n\) cancels completely, so the result is always \(x^2\) regardless of \(n\).

Step 1: \((3x^{-1}y^2)^2 = 9x^{-2}y^4\). Step 2: Multiply by \(x^3y^{-2}\): \(9 \cdot x^{-2+3} \cdot y^{4-2} = 9xy^2\). Step 3: Divide by \(9y^2\): \(\frac{9xy^2}{9y^2} = x\). The coefficient 9 and the \(y^2\) cancel, leaving just \(x\). Choice D (\(xy^2\)) forgets to cancel the \(y^2\) in the denominator.

Distribute each term of \((x^2 + 3x)\) across \((2x - 1)\): \(x^2(2x - 1) + 3x(2x - 1) = 2x^3 - x^2 + 6x^2 - 3x\). Combine like terms: \(2x^3 + (-1+6)x^2 - 3x = 2x^3 + 5x^2 - 3x\). Choice C (\(2x^3 + 7x^2 - 3x\)) adds \(x^2 + 6x^2 = 7x^2\) instead of correctly computing \(-x^2 + 6x^2 = 5x^2\).

Using (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 with a = 2x and b = y: (2x)^3 = 8x^3, 3(2x)^2(y) = 3(4x^2)(y) = 12x^2y, 3(2x)(y^2) = 6xy^2, y^3 = y^3. Result: 8x^3 + 12x^2y + 6xy^2 + y^3. Choice A omits all middle terms. Choice D writes 3y^3 for the last term, but (b)^3 = y^3, not 3y^3.

Any nonzero number raised to the zero power equals 1. This is the Zero Exponent Rule: a^0 = 1 for any a that does not equal 0. Choosing 0 is a common mistake that confuses the exponent with multiplying by zero. Choosing 7 would be correct for 7^1, not 7^0.

The Product of Powers Rule states that when multiplying expressions with the same base, you add the exponents: \(x^4 \cdot x^3 = x^{4+3} = x^7\). Choice A (\(x^{12}\)) results from multiplying the exponents instead of adding them — that applies the Power of a Power Rule, not the Product Rule.

The degree of a monomial with multiple variables is the sum of all the exponents. The exponents are 3 (for x) and 2 (for y), so the degree is 3 + 2 = 5. Choosing 3 ignores the y exponent entirely, and choosing 6 confuses the coefficient with the degree.

Combine like terms: 3x + 2x = 5x, and 5 + (-1) = 4. The sum is 5x + 4. Choice B (5x + 6) results from adding 5 + 1 = 6, treating the minus sign as a plus sign and incorrectly making the constant larger instead of smaller.

A negative exponent means take the reciprocal: \(x^{-n} = \frac{1}{x^n}\). Therefore \(x^{-4} = \frac{1}{x^4}\). Choice A (\(-x^4\)) is a common error that confuses a negative exponent with a negative coefficient — the negative exponent only signals a reciprocal and does not make the expression negative.

The Quotient of Powers Rule states that when dividing expressions with the same base, you subtract the exponents: y^8 / y^3 = y^(8-3) = y^5. Choice A (y^11) results from adding the exponents instead of subtracting them, which is the Product Rule for multiplication, not the Quotient Rule for division.

The degree of a polynomial is the highest exponent among all its terms. The term degrees are 3, 5, 1, and 0 respectively. The greatest is 5, so the degree of the polynomial is 5. Choice A (3) is wrong because -2x^5 has a higher degree than 4x^3, even though 4x^3 is written first in the expression.

The Power of a Product Rule states \((ab)^n = a^n \cdot b^n\). Therefore \((xy)^4 = x^4 \cdot y^4\). Both variables must be raised to the fourth power. Choices A and B each apply the exponent to only one of the two variables, which is incorrect — every factor inside the parentheses must be raised to the given power.

This is the Difference of Squares pattern: \((a + b)(a - b) = a^2 - b^2\). With \(a = x\) and \(b = 6\): \((x + 6)(x - 6) = x^2 - 36\). The middle terms cancel because \(x(-6) + 6(x) = -6x + 6x = 0\). Choice B (\(x^2 - 12x + 36\)) is actually \((x - 6)^2\), a completely different product.

Apply the Product of Powers Rule by adding exponents: \(x^3 \cdot x^{-7} = x^{3 + (-7)} = x^{-4}\). Then apply the Negative Exponent Rule: \(x^{-4} = \frac{1}{x^4}\). Choice B (\(x^4\)) results from computing \(7 - 3 = 4\) and treating the result as positive, ignoring that the exponent is negative.

Combine like terms: \(2x^2 + x^2 = 3x^2\); \(-5x + 3x = -2x\); \(1 + (-6) = -5\). The result is \(3x^2 - 2x - 5\). Choice A (\(3x^2 + 8x - 5\)) results from treating \(-5x\) as \(+5x\) when combining the x-terms, giving \(5x + 3x = 8x\) — a sign error in reading the first polynomial.

Distribute \(3x^2\) to each term: \(3x^2 \cdot 2x = 6x^3\) (since \(x^2 \cdot x = x^3\) by adding exponents) and \(3x^2 \cdot (-5) = -15x^2\). The result is \(6x^3 - 15x^2\). Choice A (\(6x^2 - 15x\)) fails to apply the Product of Powers Rule — \(3x^2 \cdot 2x\) should give \(x^{2+1} = x^3\), not \(x^2\).

Apply the Power of a Product Rule — raise each factor to the third power: \(3^3 = 27\), \((x^1)^3 = x^3\), and \((y^2)^3 = y^{2 \times 3} = y^6\). The result is \(27x^3y^6\). Choice B (\(9x^3y^6\)) uses \(3^2 = 9\) instead of \(3^3 = 27\), and Choice C adds the y exponents \((2 + 3 = 5)\) instead of multiplying them.

Apply the Quotient of Powers Rule: \(\frac{x^4}{x^7} = x^{4-7} = x^{-3}\). Rewriting with a positive exponent gives \(\frac{1}{x^3}\). Choice A (\(x^3\)) subtracts in the wrong order \((7 - 4 = 3)\) and ignores the negative sign. Choice B (\(x^{11}\)) adds instead of subtracts, confusing division with multiplication.

Rewrite as \((x + 5)(x + 5)\) and apply FOIL: \(x \cdot x + x \cdot 5 + 5 \cdot x + 5 \cdot 5 = x^2 + 5x + 5x + 25 = x^2 + 10x + 25\). Choice A (\(x^2 + 25\)) is the most common error — squaring only the first and last terms while omitting the middle term, which equals \(2(x)(5) = 10x\).

Standard form requires terms arranged from highest to lowest degree. Choice D has degrees 3, 2, 1, 0 in descending order, which is correct standard form. Choice C starts with -2x^2 (degree 2) before 5x^3 (degree 3), violating the descending order requirement.

Apply the Power of a Product Rule: raise each factor to the fourth power. \(2^4 = 16\) and \((x^3)^4 = x^{3 \times 4} = x^{12}\). The result is \(16x^{12}\). Choice B (\(8x^{12}\)) uses \(2^3 = 8\) instead of \(2^4 = 16\). Choice C (\(16x^7\)) adds the exponents \(3 + 4 = 7\) instead of multiplying them, which is the Product Rule error.

Distribute each term of \((x - 2)\) across the trinomial: \(x(x^2 + 2x + 4) = x^3 + 2x^2 + 4x\); \((-2)(x^2 + 2x + 4) = -2x^2 - 4x - 8\). Combine: \(x^3 + (2x^2 - 2x^2) + (4x - 4x) - 8 = x^3 - 8\). This is the Difference of Cubes pattern: \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\). Choice B incorrectly distributes the \(-2\) with wrong signs across each term.

Step 1 — Expand the numerator: \((3x^2y^{-3})^2 = 9x^4y^{-6}\). Step 2 — Divide: \(9x^4y^{-6} / (9x^{-1}y^2) = x^{4 - (-1)} \cdot y^{(-6) - 2} = x^5 \cdot y^{-8} = x^5/y^8\). Choice A (\(x^3/y^8\)) computes \(x^{4 - 1} = x^3\) instead of \(x^{4 - (-1)} = x^5\), missing that subtracting a negative exponent increases the result.

Area = length times width = (3x + 2)(2x - 1). Using FOIL: 3x * 2x = 6x^2; 3x * (-1) = -3x; 2 * 2x = 4x; 2 * (-1) = -2. Combine: 6x^2 + (-3x + 4x) - 2 = 6x^2 + x - 2. Choice A (6x^2 - x - 2) incorrectly combines -3x + 4x as -x instead of +x. Choice C (5x + 1) is a perimeter-style linear expression, not an area calculation.

Distribute each term: \(3x(x^2 - x + 4) = 3x^3 - 3x^2 + 12x\); \((-2)(x^2 - x + 4) = -2x^2 + 2x - 8\). Combine like terms: \(3x^3 + (-3x^2 - 2x^2) + (12x + 2x) - 8 = 3x^3 - 5x^2 + 14x - 8\). Choice D has \(+8\) instead of \(-8\), which results from computing \((-2)(4) = +8\) — a sign error when distributing the negative.

Add the exponents in the numerator: 3n + (2n - 1) = 5n - 1. Then subtract the denominator exponent: (5n - 1) - (n + 2) = 5n - 1 - n - 2 = 4n - 3. The result is x^(4n-3). Choice B (x^(5n-3)) forgets to subtract the denominator's exponent entirely. Choice C (x^(4n-1)) makes an arithmetic error treating -(n+2) as -n+2 instead of -n-2.

Substitute \((x + 2)\) for every \(x\) in \(P(x)\): \(P(x+2) = (x+2)^2 - 4(x+2) + 3\). Expand: \((x+2)^2 = x^2 + 4x + 4\); \(-4(x+2) = -4x - 8\). Combine all terms: \(x^2 + 4x + 4 - 4x - 8 + 3 = x^2 + (4x - 4x) + (4 - 8 + 3) = x^2 - 1\). Choice A (\(x^2 + 4x - 1\)) results from not recognizing that the \(+4x\) and \(-4x\) cancel completely.

Multiply the numerator: \(4x^3y^2 \cdot 2xy^{-3} = 8 \cdot x^{3+1} \cdot y^{2+(-3)} = 8x^4y^{-1}\). Then divide: \(8x^4y^{-1} / (8x^2y^{-1}) = x^{4-2} \cdot y^{(-1)-(-1)} = x^2 \cdot y^0 = x^2 \cdot 1 = x^2\). Choice B (\(2x^2\)) forgets that the \(8\)s in the numerator and denominator cancel to \(1\), not to \(2\). Choice A incorrectly computes \(y^{-1} / y^{-1} = y^{-2}\) rather than \(y^0 = 1\).

Step 1 — Expand the power: \((2x^2)^3 = 2^3 \cdot x^6 = 8x^6\). Step 2 — Multiply by \(x^{-4}\): \(8x^6 \cdot x^{-4} = 8x^{6-4} = 8x^2\). Step 3 — Divide by \(4x^2\): \(8x^2 / (4x^2) = (8/4) \cdot x^{2-2} = 2 \cdot x^0 = 2 \cdot 1 = 2\). Choice A (\(2x^2\)) correctly divides by \(4\) but forgets to cancel the \(x^2\) terms. Choice B (\(8x^2\)) forgets to divide by \(4\) entirely.

The degree of a polynomial is the highest exponent of the variable. The terms are 4x^3 (degree 3), -2x (degree 1), and 7 (degree 0). The highest is 3, so the degree is 3. A common error is confusing the leading coefficient (4) with the degree.

Any nonzero base raised to the power of \(0\) equals \(1\). This follows from the quotient rule: \(x^n / x^n = x^{n-n} = x^0\), and any nonzero number divided by itself equals \(1\). The restriction \(x\) not equal to \(0\) is necessary because \(0^0\) is undefined.

When multiplying powers with the same base, add the exponents: \(x^3 \cdot x^5 = x^{3+5} = x^8\). A common mistake is multiplying the exponents to get \(x^{15}\) — that rule applies to a power raised to a power, such as \((x^3)^5\), not to products of the same base.

A negative exponent means take the reciprocal: \(x^{-n} = 1/x^n\). Therefore \(x^{-2} = 1/x^2\). The negative exponent does not make the expression negative — it indicates a reciprocal. Choice A (\(-x^2\)) is a common misreading of the negative sign.

The leading coefficient is the coefficient of the term with the highest degree. The highest-degree term is -5x^4 (degree 4), so the leading coefficient is -5. Choice A (9) is the constant term, and choice D (-1) is the coefficient of the -x term.

A binomial is a polynomial with exactly two terms. A monomial has one term, a trinomial has three terms, and 'quadratic' describes the degree (degree 2), not the number of terms. So 3x^2 - 7 is correctly called a binomial.

When raising a power to a power, multiply the exponents: \((x^4)^3 = x^{4 \cdot 3} = x^{12}\). Choice A (\(x^7\)) results from adding the exponents instead of multiplying — a very common error. Choice D incorrectly moves the outer exponent to the coefficient position.

Distribute the negative sign to every term in the second polynomial: \(4x^2 - 3x + 5 - 2x^2 - x + 3\). Combine like terms: \((4-2)x^2 + (-3-1)x + (5+3) = 2x^2 - 4x + 8\). A key error is forgetting to flip the sign of \(-3\), which becomes \(+3\) after distribution, giving a constant of \(8\) not \(2\).

This is a difference of squares: \((a + b)(a - b) = a^2 - b^2\). With \(a = x\) and \(b = 3\), the result is \(x^2 - 9\). The outer and inner terms (\(+3x\) and \(-3x\)) cancel, leaving no middle term. Choice D (\(x^2 - 6x + 9\)) is \((x - 3)^2\), a different expression entirely.

When dividing powers with the same base, subtract the exponents: y^8 / y^3 = y^(8-3) = y^5. Choice B (y^11) results from adding instead of subtracting. Choice C (y^24) results from multiplying the exponents, which is the rule for (y^8)^3, not for division.

Apply the exponent to each factor: \((2x^2 y)^3 = 2^3 \cdot (x^2)^3 \cdot y^3 = 8 \cdot x^6 \cdot y^3 = 8x^6 y^3\). Choice A (\(6x^6 y^3\)) is the error of multiplying \(2\) by \(3\) instead of raising \(2\) to the \(3\)rd power. Choice C (\(8x^5 y^3\)) results from computing \((x^2)^3 = x^{2+3} = x^5\) by adding instead of multiplying the exponents.

Use FOIL: First \(= x \cdot x = x^2\), Outer \(= x \cdot 2 = 2x\), Inner \(= 4 \cdot x = 4x\), Last \(= 4 \cdot 2 = 8\). Combine: \(x^2 + 2x + 4x + 8 = x^2 + 6x + 8\). Choice B swaps the middle and constant terms. Choice D incorrectly uses \(4+4=8\) as the middle term coefficient instead of \(2+4=6\).

Distribute -2x to each term: (-2x)(3x^2) = -6x^3, (-2x)(-4x) = +8x^2, (-2x)(1) = -2x. Result: -6x^3 + 8x^2 - 2x. Choice B results from not properly distributing the negative to the -4x term. Choice D incorrectly makes the last term positive by losing the negative sign from -2x*1.

Multiply coefficients and add exponents: \((5)(2) = 10\) and \(x^3 \cdot x^{-2} = x^{3+(-2)} = x^1 = x\). Result: \(10x\). Choice B (\(10x^5\)) is the error of treating \(x^{-2}\) as \(x^2\) and adding \(3+2=5\). Choice C (\(7x\)) adds the coefficients \(5+2=7\) instead of multiplying them.

Apply the outer exponent to each factor: (a^2)^4 = a^(2*4) = a^8 and b^4. So (a^2 b)^4 = a^8 b^4. Choice B (a^6 b^4) results from computing (a^2)^4 = a^(2+4) = a^6 by adding instead of multiplying exponents. Choice D treats the exponent as a multiplier on the coefficient.

Use FOIL: First = 2x*x = 2x^2, Outer = 2x*3 = 6x, Inner = -5*x = -5x, Last = -5*3 = -15. Combine the middle terms: 6x + (-5x) = x. Result: 2x^2 + x - 15. Choice B (2x^2 - x - 15) reverses the sign on the middle term, a common error when students compute 5x - 6x instead of 6x - 5x.

When multiplying two polynomials, the degree of the product is the sum of their degrees. Degree 2 + degree 3 = degree 5. Choice C (6) likely results from multiplying the degrees rather than adding them. Choice D assumes the degree of the product matches the larger of the two degrees.

Apply the product rule by adding exponents: 3^2 * 3^(-4) = 3^(2 + (-4)) = 3^(-2), which equals 1/9. Choice B (3^8) results from multiplying exponents 2 and 4 instead of adding them. Choice D (3^(-6)) incorrectly subtracts in the wrong direction: 2 - 4 = -2, not -6.

Expand the numerator: \((2x^2 y^{-1})^3 = 8x^6 y^{-3}\). Divide: \(8x^6 y^{-3} / (4x^{-1} y^2) = (8/4) \cdot x^{6-(-1)} \cdot y^{-3-2} = 2 \cdot x^7 \cdot y^{-5} = 2x^7/y^5\). Choice B has \(x^5\) because \(6 - 1 = 5\) — forgetting that subtracting a negative exponent means adding. Choice D places \(y\) in the numerator by mishandling the negative exponent.

Expand step by step: \((x+2)^3 = (x+2)(x+2)^2 = (x+2)(x^2+4x+4)\). Distribute: \(x^3+4x^2+4x+2x^2+8x+8 = x^3+6x^2+12x+8\). Choice A (\(x^3+8\)) is the classic error of cubing only the first and last terms. Choice C uses \(4\) and \(8\) as middle coefficients, which come from the intermediate step \((x+2)^2 = x^2+4x+4\) but are not the final coefficients.

Area = side^2 = (2x+3)^2. Use the perfect square pattern (a+b)^2 = a^2 + 2ab + b^2 with a = 2x and b = 3: (2x)^2 + 2(2x)(3) + 3^2 = 4x^2 + 12x + 9. Choice A forgets the middle term entirely. Choice B uses 2(1)(3) = 6 as the middle coefficient, incorrectly treating 2x as having coefficient 1 in the 2ab term.

Substitute \((x-1)\) for every \(x\): \(f(x-1) = 2(x-1)^2 - 3(x-1) + 1\). Expand each part: \(2(x^2-2x+1) = 2x^2-4x+2\) and \(-3(x-1) = -3x+3\). Sum all parts: \(2x^2-4x+2-3x+3+1 = 2x^2-7x+6\). Choice B (constant \(4\)) results from not fully expanding \((x-1)^2\) and instead treating it as \(x^2-1\).

Subtract exponents using the quotient rule: x^(2n+3) / x^(n-1) = x^((2n+3)-(n-1)). Distribute the negative sign carefully: (2n+3) - (n-1) = 2n+3-n+1 = n+4. Result: x^(n+4). Choice B (x^(n+2)) is the error of computing (2n+3)-(n+1)=n+2, treating -(n-1) as -(n+1) and losing the sign flip on -1.

Simplify the numerator first: (3x^2)^2 = 9x^4, so 9x^4 * 2x^3 = 18x^7. Then divide: 18x^7 / (6x^5) = (18/6) * x^(7-5) = 3x^2. Choice B (6x^2) results from dividing 18 by 3 instead of 18 by 6. Choice C (3x^4) results from computing the exponent as 7-3=4 instead of 7-5=2.

Volume \(= x \cdot (x+1) \cdot (x-1)\). Recognize \((x+1)(x-1)\) as a difference of squares: \(x^2-1\). Then multiply: \(x(x^2-1) = x^3-x\). Choice B (\(x^3-1\)) results from forgetting to multiply by the factor of \(x\) at the final step. Choice A may arise from incorrectly expanding one of the binomial products and introducing an extra \(x^2\) term through a sign error.

Any nonzero number raised to the power of \(0\) equals \(1\). This is called the zero exponent rule. A common error is choosing \(0\), which confuses \(x^0\) with \(0^x\).

The degree of a polynomial is the highest exponent of its variable. The term 5x^4 has the greatest exponent of 4, so the degree is 4. The value 5 is the leading coefficient, not the degree.

The product rule for exponents states that when multiplying powers with the same base, add the exponents: \(x^4 \cdot x^6 = x^{4+6} = x^{10}\). Multiplying the exponents (\(4 \cdot 6 = 24\)) is a common error.

The leading coefficient is the coefficient of the term with the highest degree. The highest-degree term is -4x^3, so the leading coefficient is -4. The negative sign is part of the coefficient, and 6 is the constant term, not the leading coefficient.

A binomial is a polynomial with exactly two terms. The expression \(3x - 7\) has two terms: \(3x\) and \(-7\). The expression \(5x^2\) is a monomial (one term), \(x^2 + 2x - 1\) is a trinomial (three terms), and \(4\) is a constant monomial (one term).

The power of a power rule states \((x^m)^n = x^{m \cdot n}\). So \((x^3)^4 = x^{3 \cdot 4} = x^{12}\). A common error is adding the exponents instead of multiplying, which gives \(3 + 4 = 7\).

Combine like terms: (3x + 2x) + (5 + (-8)) = 5x + (-3) = 5x - 3. The constants 5 and -8 add to give -3, not +3.

A negative exponent indicates a reciprocal: \(x^{-1} = 1/x^1 = 1/x\). This is the negative exponent rule. The expression \(x^{-1}\) is positive and equals \(1/x\), not \(-x\) or \(-1/x\).

Combine like terms: \((3x^2 + 2x^2) + (-x + 5x) + (4 + (-3)) = 5x^2 + 4x + 1\). Note that \(-x + 5x = 4x\), not \(-4x\). The constants \(4\) and \(-3\) add to give \(+1\), not \(+7\).

Distribute the negative sign to every term of the second polynomial: \(5x^2 + 3x - 2 - 2x^2 + x - 4\). Combine: \((5-2)x^2 + (3+1)x + (-2-4) = 3x^2 + 4x - 6\). A common error is leaving the sign on \(-x\) unchanged, which gives \(3x^2 + 2x - 6\).

Using FOIL: First = \(x^2\), Outer = \(-6x\), Inner = \(-4x\), Last = \((-4)(-6) = +24\). Combining the middle terms: \(x^2 - 10x + 24\). The last term is positive because a negative times a negative equals a positive: \((-4)(-6) = +24\), not \(-24\).

The quotient rule states that when dividing powers with the same base, subtract the exponents: y^9 / y^3 = y^(9-3) = y^6. Dividing the exponents (9 / 3 = 3) is a common error that produces the incorrect answer y^3.

Using FOIL: First = \(x^2\), Outer = \(-7x\), Inner = \(+7x\), Last = \(-49\). The outer and inner terms cancel: \(x^2 + 0x - 49 = x^2 - 49\). This is the difference of squares pattern: \((a+b)(a-b) = a^2 - b^2\). The middle terms do not cancel if the signs are the same, so choices A and B are incorrect.

(-3)^4 = (-3)(-3)(-3)(-3). Multiplying in pairs: (-3)(-3) = 9, then 9 * 9 = 81. When a negative number is raised to an even power, the result is always positive. Only odd powers of negative numbers produce negative results.

Apply the product rule by adding the exponents: a^5 * a^(-2) = a^(5 + (-2)) = a^3. The key is to add, not multiply, the exponents when the bases are the same. The value 5 * (-2) = -10 is a multiplication error.

First apply the power of a power rule: \((x^2)^3 = x^{2 \cdot 3} = x^6\). Then apply the product rule: \(x^6 \cdot x^5 = x^{6+5} = x^{11}\). A common error is computing \((x^2)^3 = x^5\) by adding instead of multiplying, which leads to \(x^5 \cdot x^5 = x^{10}\).

Distribute -3x^2 to each term: (-3x^2)(2x) = -6x^3 and (-3x^2)(-5) = +15x^2. The second product is positive because a negative times a negative equals a positive. Result: -6x^3 + 15x^2. Choice B is incorrect because (-3x^2)(-5) = +15x^2, not -15x^2.

Distribute each term of \((x+2)\) across \((x^2-3x+4)\): \(x(x^2-3x+4) + 2(x^2-3x+4) = x^3-3x^2+4x + 2x^2-6x+8\). Combine like terms: \(x^3 + (-3+2)x^2 + (4-6)x + 8 = x^3 - x^2 - 2x + 8\). A sign error on the \(x^2\) terms gives Choice B.

Apply the pattern (a - b)^2 = a^2 - 2ab + b^2 with a = 4x and b = 1: (4x)^2 - 2(4x)(1) + 1^2 = 16x^2 - 8x + 1. A common error is treating it like a difference of squares and writing 16x^2 - 1, which omits the middle term entirely.

Distribute each term: \(2x(x^2-x+3) + 1(x^2-x+3) = 2x^3-2x^2+6x + x^2-x+3\). Combine: \(2x^3 + (-2+1)x^2 + (6-1)x + 3 = 2x^3 - x^2 + 5x + 3\). Watch the \(x^2\) term: \(-2x^2 + x^2 = -x^2\), not \(+x^2\), which rules out Choice A.

Expand the numerator: \((3x^4y^2)^2 = 3^2 \cdot x^{4 \cdot 2} \cdot y^{2 \cdot 2} = 9x^8y^4\). Then divide: \((9x^8y^4) / (9x^5y) = (9/9) \cdot x^{8-5} \cdot y^{4-1} = x^3y^3\). The 9s cancel completely. Choice B results from forgetting the 9 in the denominator cancels.

Distribute \((x+4)\) across each term: \(x^2(x+4) + 2x(x+4) + (-3)(x+4) = x^3+4x^2 + 2x^2+8x - 3x-12\). Combine: \(x^3 + 6x^2 + 5x - 12\). Choice B is a common error from collecting only \(4x^2\) and missing the \(2x^2\) contribution.

Apply the product rule by adding the exponents: x^(n+3) * x^(2n-1) = x^((n+3)+(2n-1)). Combine like terms inside the exponent: (n + 2n) + (3 + (-1)) = 3n + 2. Choice A results from multiplying the two binomial exponents together as polynomials instead of adding them.

Expand the numerator: \((2x^{-1}y^2)^3 = 2^3 \cdot x^{-3} \cdot y^6 = 8x^{-3}y^6\). Divide by \(8y^3\): \((8x^{-3}y^6) / (8y^3) = x^{-3} \cdot y^{6-3} = x^{-3}y^3\). Since \(x^{-3} = 1/x^3\), the result is \(y^3/x^3\). The 8s cancel completely. Choice C results from failing to cancel the 8 in the denominator.

Expand each square: \((x+3)^2 = x^2+6x+9\) and \((x-3)^2 = x^2-6x+9\). Subtract: \((x^2+6x+9) - (x^2-6x+9) = 12x\). The \(x^2\) terms and the constants cancel, leaving only \(12x\). Choice C results from adding the two expressions instead of subtracting them.

Any nonzero base raised to the power of 0 equals 1. This is the zero exponent rule: \(x^0 = 1\) as long as \(x\) is not equal to 0. The answer 0 is a common misconception — raising to the zero power does not produce zero. The 'undefined' case applies only when \(x = 0\).

When multiplying powers with the same base, add the exponents: 2^3 · 2^2 = 2^(3+2) = 2^5 = 32. The distractor 2^6 results from multiplying the exponents (3 × 2 = 6), which is the power-to-a-power rule and does not apply here.

The degree of a polynomial is the highest exponent of the variable. The largest exponent is 3 in the term 5x^3, so the degree is 3. The number of terms (4) and the constant term (-7) do not determine the degree.

The leading coefficient is the coefficient of the term with the highest degree. In standard form, the highest-degree term is \(-4x^3\), so the leading coefficient is \(-4\). The constant \(6\) and the coefficient \(2\) of \(x^2\) are common distractors.

When raising a power to a power, multiply the exponents: \((x^4)^2 = x^{4 \times 2} = x^8\). A common error is adding instead of multiplying: \(4 + 2 = 6\), giving \(x^6\). Adding exponents applies to \(x^4 \cdot x^2\) (multiplying two separate powers), not to a power raised to a power.

The exponent applies to every factor inside the parentheses: \((3x)^2 = 3^2 \cdot x^2 = 9x^2\). A common mistake is squaring only the variable and leaving the coefficient unchanged, giving \(3x^2\). Another error is doubling instead of squaring the coefficient, giving \(6x^2\).

Combine like terms: \(2x^2 + x^2 = 3x^2\), \(3x + (-x) = 2x\), and \(-5 + 2 = -3\). Result: \(3x^2 + 2x - 3\). The distractor \(3x^2 + 4x - 3\) incorrectly adds \(3x + x = 4x\) instead of subtracting. The distractor \(3x^2 + 2x - 7\) incorrectly computes \(-5 - 2 = -7\).

Distribute the minus sign to every term in the second polynomial: \((4x^2 - 3x + 2) - x^2 - 5x + 1\). Then combine like terms: \(3x^2 - 8x + 3\). A common error is forgetting to distribute the negative to the last term, changing \(-(-1)\) to \(-1\) instead of \(+1\), giving \(3x^2 - 8x + 1\).

Distribute \(3x\) to each term: \(3x \cdot 2x^2 = 6x^3\), \(3x \cdot (-4x) = -12x^2\), and \(3x \cdot 1 = 3x\). Result: \(6x^3 - 12x^2 + 3x\). The distractor \(6x^3 - 12x + 3\) incorrectly treats the last multiplication as \(3 \cdot 1 = 3\) instead of \(3x \cdot 1 = 3x\).

Add exponents for matching bases: \(x^3 \cdot x^2 = x^{3+2} = x^5\) and \(y^2 \cdot y^4 = y^{2+4} = y^6\). Result: \(x^5 y^6\). The distractor \(x^6 y^8\) incorrectly multiplies the exponents (\(3 \times 2\) and \(2 \times 4\)) instead of adding them.

Use FOIL: First = \(x \cdot x = x^2\), Outer = \(x \cdot 2 = 2x\), Inner = \(5 \cdot x = 5x\), Last = \(5 \cdot 2 = 10\). Combine the middle terms: \(2x + 5x = 7x\). Result: \(x^2 + 7x + 10\). The distractor \(x^2 + 7x + 7\) incorrectly computes the last term as \(5 + 2 = 7\) instead of \(5 \times 2 = 10\).

Raise each factor to the third power: \(2^3 = 8\) and \((x^3)^3 = x^{3 \times 3} = x^9\). Result: \(8x^9\). The distractor \(6x^9\) incorrectly multiplies \(2 \times 3 = 6\) instead of computing \(2^3 = 8\). The distractor \(8x^6\) incorrectly adds the exponents (\(3 + 3 = 6\)) instead of multiplying them (\(3 \times 3 = 9\)).

Divide the coefficients and subtract the exponents of matching bases: \(15 / 5 = 3\) and \(x^6 / x^2 = x^{6-2} = x^4\). Result: \(3x^4\). The distractor \(3x^3\) mistakenly subtracts \(6 - 3\) instead of \(6 - 2\), confusing the value of the coefficient divisor (5) with the exponent.

Use FOIL: \(x \cdot x = x^2\), \(x \cdot 8 = 8x\), \(-3 \cdot x = -3x\), \(-3 \cdot 8 = -24\). Combine the middle terms: \(8x + (-3x) = 5x\). Result: \(x^2 + 5x - 24\). The distractor \(x^2 - 5x - 24\) incorrectly computes the middle as \(3 - 8 = -5x\) instead of \(8x - 3x = +5x\).

Apply the power-to-a-power rule to each factor: (a^3)^4 = a^(3 × 4) = a^12 and (b^2)^4 = b^(2 × 4) = b^8. Result: a^12 b^8. The distractor a^7 b^6 incorrectly adds the outer exponent to each inner exponent (3+4 = 7 and 2+4 = 6) instead of multiplying.

Any nonzero expression raised to the power of 0 equals 1. The entire expression \((4x^2 y)\) is the base, and since \(x\) and \(y\) are nonzero, the result is simply \(1\). The distractor \(4\) incorrectly applies the zero exponent only to \(x^2 y\) while leaving the coefficient unchanged.

A negative exponent means take the reciprocal of the base raised to the positive exponent: 4^(-2) = 1 / 4^2 = 1/16. A negative exponent does NOT produce a negative result. The distractor -16 confuses a negative exponent with a negative sign, computing -(4^2) = -16 instead.

Standard form requires terms to be written in descending order of degree. \(x^2 - 3x + 4\) correctly lists: degree 2 first, then degree 1, then degree 0 (constant). Choice A starts with the constant. Choice C begins with the degree-1 term. Choice D places the constant before the degree-1 term.

Distribute each term of the first binomial across the trinomial. \(2x(3x^2 + x - 2) = 6x^3 + 2x^2 - 4x\). Then \(-3(3x^2 + x - 2) = -9x^2 - 3x + 6\). Combine like terms: \(6x^3 + (2 - 9)x^2 + (-4 - 3)x + 6 = 6x^3 - 7x^2 - 7x + 6\). The distractor ending in \(-6\) forgets that \(-3\) times \(-2\) is positive \(6\).

First apply the outer exponent to the numerator: \((x^2 y^{-3})^2 = x^4 y^{-6}\). Then divide by subtracting exponents: \(x^{4-3} y^{(-6)-(-4)} = x^1 y^{-2} = x / y^2\). The distractor \(x y^2\) incorrectly treats \(y^{-2}\) as positive. The distractor \(x^5 / y^2\) adds instead of subtracts the \(x\)-exponents.

Apply the binomial cube formula \((a + b)^3 = a^3 + 3a^2 b + 3ab^2 + b^3\) with \(a = x\) and \(b = 1\): \(x^3 + 3x^2(1) + 3x(1)^2 + 1^3 = x^3 + 3x^2 + 3x + 1\). A common error is to cube each term individually (\(x^3 + 1^3\)), which ignores the three middle cross-product terms.

Expand each part: \((x + 3)^2 = x^2 + 6x + 9\), and \((x - 1)(x + 1)\) is a difference of squares equal to \(x^2 - 1\). Adding the results: \((x^2 + 6x + 9) + (x^2 - 1) = 2x^2 + 6x + 8\). The distractor \(2x^2 + 6x + 10\) incorrectly computes \((x - 1)(x + 1)\) as \(x^2 + 1\) instead of \(x^2 - 1\).

Distribute each term of the trinomial: \(x^2(x^2 + 2) = x^4 + 2x^2\), \(-x(x^2 + 2) = -x^3 - 2x\), and \(3(x^2 + 2) = 3x^2 + 6\). Combine like terms: \(x^4 - x^3 + (2 + 3)x^2 - 2x + 6 = x^4 - x^3 + 5x^2 - 2x + 6\). The distractor with \(+2x\) results from forgetting the negative sign in \(-x(x^2 + 2)\).

First simplify inside the parentheses: \((4/2) \cdot x^{3-(-1)} \cdot y^{1-2} = 2x^4 y^{-1}\). Then square the entire result: \((2x^4 y^{-1})^2 = 2^2 x^8 y^{-2} = 4x^8 / y^2\). The distractor \(16x^8 / y^2\) incorrectly squares the already-simplified coefficient \(4\) instead of the intermediate result \(2\), computing \(4^2 = 16\).

First expand \((x - 2)^2 = x^2 - 4x + 4\). Then multiply by \((x + 2)\): \(x^2(x + 2) = x^3 + 2x^2\), \(-4x(x + 2) = -4x^2 - 8x\), \(4(x + 2) = 4x + 8\). Combine: \(x^3 + (2 - 4)x^2 + (-8 + 4)x + 8 = x^3 - 2x^2 - 4x + 8\). The distractor \(x^3 - 4x + 8\) skips squaring \((x - 2)\) first and simply multiplies \((x - 2)(x - 2)(x + 2)\) as if the first two factors cancel.

When multiplying two powers with the same base, add the exponents: \(x^4 \cdot x^6 = x^{4+6} = x^{10}\). A common mistake is multiplying the exponents (getting \(x^{24}\)), which is only correct when raising a power to a power, not when multiplying.

The degree of a polynomial is the highest exponent on the variable. The terms have degrees 3, 1, and 0 respectively, so the degree of the polynomial is 3. The number 5 is the leading coefficient, not the degree.

Combine like terms: \((3x^2 + x^2) + (2x - 5x) + (-1 + 4) = 4x^2 - 3x + 3\). Choice B is wrong because it adds \(2x\) and \(5x\) instead of subtracting. Remember to keep the sign attached to each term when removing parentheses.

Distribute the negative sign to every term in the second polynomial: 5x^2 - 3x + 2 - 2x^2 - x + 6. Then combine like terms: 3x^2 - 4x + 8. Choice B is a common error from forgetting to flip the sign on -6, treating it as -2 - 6 = -4 rather than +2 + 6 = +8.

Multiply the coefficients and apply the product rule to the exponents: \(3 \cdot 4 = 12\) and \(x^2 \cdot x^3 = x^{2+3} = x^5\), giving \(12x^5\). Choice C (\(12x^6\)) results from multiplying the exponents instead of adding them. Choice A adds coefficients instead of multiplying.

Use FOIL: First: 3x * x = 3x^2. Outer: 3x * 4 = 12x. Inner: -2 * x = -2x. Last: -2 * 4 = -8. Combining the middle terms: 12x - 2x = 10x. Result: 3x^2 + 10x - 8. Choice C is wrong because it omits the inner product (-2x), keeping only the outer term.

Apply the product rule by adding exponents: \(x^{-3} \cdot x^5 = x^{-3 + 5} = x^2\). Choice B (\(x^{-2}\)) results from subtracting instead of adding the exponents. Choice A (\(x^{-15}\)) incorrectly multiplies the exponents.

First expand each factor using the power rule: \((2x^2 y^3)^2 = 4x^4 y^6\) and \((x y^{-1})^3 = x^3 y^{-3}\). Then multiply: \(4 \cdot x^{4+3} \cdot y^{6+(-3)} = 4x^7 y^3\). Choice B forgets the \(y\) term. Choice C uses \(8\) because it adds \(2+3\) instead of multiplying coefficients \(4 \cdot 1\).

\((2x + 1)^2 = 4x^2 + 4x + 1\). \((x - 3)(x + 3)\) is a difference of squares \(= x^2 - 9\). Subtracting: \(4x^2 + 4x + 1 - (x^2 - 9) = 4x^2 + 4x + 1 - x^2 + 9 = 3x^2 + 4x + 10\). Choice A is wrong because it subtracts \(9\) rather than adding it (forgetting to distribute the negative sign). Choice C does not subtract the \(x^2\) term.

Distribute \((x - 1)\) across each term: \(x^2(x-1) + 2x(x-1) - 3(x-1) = x^3 - x^2 + 2x^2 - 2x - 3x + 3\). Combine like terms: \(x^3 + (-1+2)x^2 + (-2-3)x + 3 = x^3 + x^2 - 5x + 3\). Choice A has \(-x^2\) instead of \(+x^2\), an error from forgetting to combine the \(-x^2\) and \(+2x^2\) terms correctly.