Factoring — Free Algebra 1 Review Games.

GCF is 6: 6x + 12 = 6(x + 2).

Find two numbers that multiply to 6 and add to 5: 2 and 3.

Difference of squares: \(a^2 - b^2 = (a+b)(a-b)\). Here \(a=x\), \(b=3\).

GCF of 8 and 12 is 4; smallest power of \(x\) is \(x^2\). GCF = \(4x^2\).

Factor out 3: \(3(x^2-1) = 3(x+1)(x-1)\).

Two numbers that multiply to 12 and add to \(-7\): \(-3\) and \(-4\).

Two numbers multiplying to \(-12\) and adding to 1: 4 and \(-3\).

Find factors: 2x^2 + 6x + x + 3 = 2x(x+3) + 1(x+3) = (2x+1)(x+3).

Difference of squares: (2x)^2 - 5^2 = (2x+5)(2x-5).

Factor \(x\): \(x(x^2-4)\), then difference of squares: \(x(x+2)(x-2)\).

ac method: 6*(-3)=-18. Factors of -18 that sum to -7: 2 and -9. Split: 6x^2+2x-9x-3 = 2x(3x+1)-3(3x+1) = (3x+1)(2x-3).

Perfect square trinomial: \(x^2 - 2(5)x + 5^2 = (x-5)^2\).

Group: \(x^2(x+2) + 3(x+2) = (x^2+3)(x+2)\).

Factor out 3: \(3(x^2 - 4x + 4) = 3(x-2)^2\).

Difference of squares twice: \((x^2+4)(x^2-4) = (x^2+4)(x+2)(x-2)\).

The GCF of 4x and 20 is 4. Dividing each term: 4x ÷ 4 = x and 20 ÷ 4 = 5, giving 4(x + 5). Choice B is technically equivalent but not fully factored since 2(2x + 10) still has a common factor of 2 inside the parentheses. Choice C incorrectly leaves the 20 inside without dividing. Always factor out the greatest common factor, not just any common factor.

To factor x² + 8x + 15, find two numbers that multiply to 15 and add to 8. Those numbers are 3 and 5, since 3 × 5 = 15 and 3 + 5 = 8. This gives (x + 3)(x + 5). Choice C gives x² + 8x + 16 (off by 1 in the constant). Choice D gives x² + 8x + 12 (wrong constant). Choice A multiplies correctly to 15 but adds to 16, not 8.

x² − 49 is a difference of squares: a² − b² = (a − b)(a + b), where a = x and b = 7. So x² − 49 = (x − 7)(x + 7). Choice A expands to x² − 14x + 49, which has a middle term — a key sign that it is not the difference of squares pattern. Choice B expands to x² + 14x + 49. The difference of squares always produces two binomials with opposite signs.

To find the GCF, take the GCF of the coefficients and the lowest power of x. GCF(12, 18) = 6. The lowest power of x between x⁴ and x² is x². Therefore the GCF is 6x². Choice A uses 3 instead of 6 as the coefficient GCF. Choice B uses the higher power x⁴, but the GCF must use the lower power so it divides evenly into both terms.

This is a perfect square trinomial. The pattern is a² + 2ab + b² = (a + b)². Here a = x and b = 5, since 5² = 25 and 2(x)(5) = 10x. So x² + 10x + 25 = (x + 5)². Choice B gives x² − 10x + 25 (negative middle term). Choice C gives x² − 25 (difference of squares, no middle term). The positive middle term confirms b is positive.

The GCF of 7x³ and 14x² is 7x² (GCF of coefficients 7 and 14 is 7; lowest power of x is x²). Factoring out 7x²: 7x³ ÷ 7x² = x and 14x² ÷ 7x² = 2, giving 7x²(x + 2). Choice A only pulls out 7x, leaving x² + 2x which still has a common factor of x — not fully factored. Choice C only pulls out the constant 7, ignoring the x² that can also be factored out.

(x + 4)(x − 4) is the product of a sum and difference, fitting the pattern (a + b)(a − b) = a² − b². The middle terms cancel: x · (−4) + 4 · x = −4x + 4x = 0. Only the difference of the squares remains: x² − 16. Choices A and B incorrectly include a middle term of ±8x. Choice D incorrectly adds the squares instead of subtracting them.

Find two numbers that multiply to +15 and add to −8. Both numbers must be negative: −3 and −5, since (−3)(−5) = 15 and (−3) + (−5) = −8. This gives (x − 3)(x − 5). Choice A uses −5 and +3, which multiply to −15 (wrong sign). Choice B uses +3 and +5, which add to +8 (wrong sign of middle term). When both the product and sum are negative, both factors must be negative.

Find two numbers that multiply to −15 and add to +2. Those are +5 and −3, since 5 × (−3) = −15 and 5 + (−3) = 2. So x² + 2x − 15 = (x + 5)(x − 3). Choice B reverses the signs: (x − 5)(x + 3) gives x² − 2x − 15, not +2x. Choice D has both factors negative and would produce a positive constant term, which does not match the −15.

Using the AC method: multiply a × c = 3 × 4 = 12. Find two numbers that multiply to 12 and add to 8: those are 6 and 2. Rewrite: 3x² + 6x + 2x + 4 = 3x(x + 2) + 2(x + 2) = (3x + 2)(x + 2). Choice A: (3x + 4)(x + 1) = 3x² + 3x + 4x + 4 = 3x² + 7x + 4 — the middle term is 7x, not 8x. Choice D produces a negative constant when expanded.

16x² − 9 is a difference of squares: (4x)² − 3² = (4x − 3)(4x + 3). Choice A expands to 16x² − 24x + 9, which is a perfect square trinomial — it has a middle term, so it cannot be the difference of squares. Choice B gives (2x − 3)(2x + 3) = 4x² − 9, not 16x² − 9. The coefficient of x must square to 16, so it is 4x, not 2x.

First, factor out the GCF: 3x³ − 27x = 3x(x² − 9). Then recognize x² − 9 as a difference of squares: x² − 9 = (x − 3)(x + 3). The complete factorization is 3x(x − 3)(x + 3). Choice C is partially correct but stops too early — x² − 9 can still be factored. Choice B incorrectly treats −27 as 3 × (−9) applied to only one x, giving 3x(x − 9), which expands to 3x² − 27x, not 3x³ − 27x.

Group the terms: (x³ + 4x²) + (2x + 8). Factor each group: x²(x + 4) + 2(x + 4). Both groups share (x + 4), so factor it out: (x² + 2)(x + 4). Choice B gives (x + 2)(x² + 4) = x³ + 4x + 2x² + 8 = x³ + 2x² + 4x + 8, which has different middle terms. Choice C gives (x + 4)(x² − 2) = x³ − 2x + 4x² − 8, with a wrong sign on both the linear and constant terms.

Using the AC method: a × c = 2 × (−3) = −6. Find factors of −6 that add to +5: those are +6 and −1. Rewrite: 2x² + 6x − x − 3 = 2x(x + 3) − 1(x + 3) = (2x − 1)(x + 3). Choice A: (2x + 1)(x − 3) = 2x² − 6x + x − 3 = 2x² − 5x − 3 — wrong sign on the middle term. Choice C: (2x + 3)(x − 1) = 2x² − 2x + 3x − 3 = 2x² + x − 3 — middle term is +x, not +5x.

25x² − 1 fits the difference of squares pattern: (5x)² − 1² = (5x − 1)(5x + 1). Choice B: (5x − 1)² = 25x² − 10x + 1 — this is a perfect square trinomial with a middle term. Choice D: (5x + 1)² = 25x² + 10x + 1 — again a trinomial. Difference of squares requires no middle term and a subtraction sign between two perfect squares.

First factor out the GCF: 5x² − 20 = 5(x² − 4). Then factor x² − 4 as a difference of squares: (x − 2)(x + 2). The complete factorization is 5(x − 2)(x + 2). Choice B is only partially factored — x² − 4 can still be factored further. Choice A: 5x(x − 4) = 5x² − 20x, which introduces an x-term that is not in the original expression. Always check if a binomial inside parentheses is a difference of squares.

Find two numbers that multiply to −24 and add to −2: those are −6 and +4, since (−6)(4) = −24 and (−6) + 4 = −2. This gives (x − 6)(x + 4). Choice B reverses the signs: (x + 6)(x − 4) = x² + 2x − 24, where the middle term is +2x instead of −2x. When the constant is negative, one factor is positive and one is negative — the larger absolute value takes the sign of the middle term.

Factor x² + 4x − 21 by finding two numbers that multiply to −21 and add to +4: those are +7 and −3, since 7 × (−3) = −21 and 7 + (−3) = 4. So x² + 4x − 21 = (x + 7)(x − 3). Therefore (x + 7) is a factor. Choice A uses (x − 7), but the correct factor has +7 not −7. Choice D uses (x + 3), but the correct factor is (x − 3) — note the sign.

Using the AC method: a × c = 6 × 4 = 24. Both signs are negative in the original, so find two negative numbers that multiply to +24 and add to −11: those are −8 and −3. Rewrite: 6x² − 8x − 3x + 4 = 2x(3x − 4) − 1(3x − 4) = (2x − 1)(3x − 4). Choice A: (6x − 1)(x − 4) = 6x² − 24x − x + 4 = 6x² − 25x + 4 — wrong middle coefficient. Choice B: (2x + 1)(3x − 4) = 6x² − 8x + 3x − 4 = 6x² − 5x − 4 — wrong signs on both middle and constant.

Check for a perfect square trinomial: a² − 2ab + b² = (a − b)². Here 4x² = (2x)², 25 = 5², and the middle term −20x = −2(2x)(5). All conditions are satisfied, so 4x² − 20x + 25 = (2x − 5)². Choice B: (2x − 5)(2x + 5) = 4x² − 25, which is the difference of squares with no middle term. Choice C gives +20x (positive middle term). The negative middle term indicates (a − b)², not (a + b)².

Group the terms: (2x³ − 6x²) + (x − 3). Factor each group: 2x²(x − 3) + 1(x − 3). Since both groups share (x − 3), factor it out: (2x² + 1)(x − 3). Choice B: (2x² − 1)(x + 3) = 2x³ + 6x² − x − 3, which has different signs on the x² and x terms. Choice D shows the intermediate step before finishing the factoring — always complete the process by pulling out the shared binomial.

First apply difference of squares: x⁴ − 81 = (x²)² − 9² = (x² − 9)(x² + 9). Then check if either factor can be factored further: x² − 9 = (x − 3)(x + 3) (another difference of squares). However, x² + 9 cannot be factored over the real numbers. The complete factorization is (x − 3)(x + 3)(x² + 9). Choice A stops too early — x² − 9 is still factorable. Choice C applies difference of squares as if x⁴ = x² and 81 = 9, which is incorrect since √(x⁴) = x² and √81 = 9.

Check for a perfect square trinomial: (3x)² = 9x², 4² = 16, and −2(3x)(4) = −24x. All three conditions match, so 9x² − 24x + 16 = (3x − 4)². Choice A: (3x − 4)(3x + 4) = 9x² − 16, the difference of squares — there is no middle term, which rules it out immediately. Choice B gives +24x (wrong sign). For a perfect square trinomial with a negative middle term, the binomial must have a minus sign.

First factor out the GCF: 2x³ + 4x² − 6x = 2x(x² + 2x − 3). Then factor the trinomial: find two numbers that multiply to −3 and add to +2, which are +3 and −1. So x² + 2x − 3 = (x + 3)(x − 1). The complete factorization is 2x(x + 3)(x − 1). Choice A: 2x(x − 3)(x + 1) = 2x(x² − 2x − 3) = 2x³ − 4x² − 6x — the sign on x² is wrong. Choice B: x(2x + 6)(x − 1) is equivalent but not fully factored since 2x + 6 = 2(x + 3) still has a common factor.

Using the AC method: a × c = 8 × (−15) = −120. Find two numbers that multiply to −120 and add to +2: those are +12 and −10. Rewrite: 8x² + 12x − 10x − 15 = 4x(2x + 3) − 5(2x + 3) = (4x − 5)(2x + 3). Choice B: (4x + 5)(2x − 3) = 8x² − 12x + 10x − 15 = 8x² − 2x − 15 — the middle term has the wrong sign. Choice C: (8x − 5)(x + 3) = 8x² + 24x − 5x − 15 = 8x² + 19x − 15 — middle coefficient is far too large.

The GCF of the coefficients 15 and 10 is 5. The GCF of x² and x is x, since x is the lowest power present. So the GCF is 5x. Choice A (5x²) is wrong because x² does not divide evenly into 10x — you would get 10/x, which is not a polynomial term.

x² − 25 is a difference of squares using the pattern a² − b² = (a + b)(a − b), where a = x and b = 5. So x² − 25 = (x + 5)(x − 5). Choices A and D both represent (x − 5)², which expands to x² − 10x + 25 — that has a middle term, making it a trinomial, not x² − 25.

Find two positive numbers that multiply to 12 and add to 7: 3 and 4. So x² + 7x + 12 = (x + 3)(x + 4). Choice A gives a middle term of 8x (2 + 6 = 8), and Choice B gives 13x (1 + 12 = 13). Choice D uses negative signs, producing x² − 7x + 12 instead.

The GCF of the coefficients 6, 12, and 18 is 6. The GCF of x³, x², and x is x (take the lowest exponent). So the GCF is 6x. Choice A (6x²) is wrong because x² does not divide evenly into 18x. Choice B (3x) is a common factor but not the greatest one.

4x² − 9 = (2x)² − 3², which fits the pattern a² − b² exactly. Choice A fails because 8 is not a perfect square. Choice B is a sum of squares, not a difference, and cannot be factored over real numbers. Choice D is a perfect square trinomial — it equals (x − 1)² — not a binomial difference.

Find two numbers that multiply to 6 and add to 5: 2 and 3. So x² + 5x + 6 = (x + 2)(x + 3). Choice A gives a middle term of 7x (1 + 6 = 7). Choice C gives 6x (5 + 1 = 6). Choice D equals (x + 2)² = x² + 4x + 4, which has a different constant term.

The GCF of 8x³ and 12x² is 4x². Dividing each term: 8x³ ÷ 4x² = 2x and 12x² ÷ 4x² = 3. So the factored form is 4x²(2x + 3). Choice A uses only 4x as the GCF, leaving an extra x inside. Choice D uses 2x² but leaves 4x + 6 inside, which is not fully simplified since 4x + 6 = 2(2x + 3).

5 × (−3) = −15 and 5 + (−3) = 2, so the pair is 5 and −3. Choice A gives a sum of −2 (not 2). Choice B gives a sum of 14. Choice D gives a sum of −14. Finding such pairs is the key step when factoring trinomials of the form x² + bx + c.

Find two numbers that multiply to −30 and add to −1: −6 and 5. So x² − x − 30 = (x − 6)(x + 5). Choice A uses +6 and −5, which add to +1, giving a middle term of +x instead of −x. Choices B and C produce middle terms of 7x and −7x respectively.

Group the terms: (x³ + 3x²) + (5x + 15) = x²(x + 3) + 5(x + 3). Factor out the common binomial (x + 3): (x² + 5)(x + 3). Choice A incorrectly replaces the x² factor with just x, treating it as a simple linear factoring. Choices C and D have sign errors in one of the factors.

First factor out the GCF of 3: 3(x² + 2x − 8). Then factor the trinomial by finding numbers that multiply to −8 and add to 2: 4 and −2. Result: 3(x + 4)(x − 2). Choice A is not fully factored because (3x − 6) still contains a GCF of 3. Choice D omits the GCF of 3 entirely.

Use the AC method: AC = 2 × 3 = 6. Find factors of 6 that add to 7: 6 and 1. Rewrite: 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3). Check: (2x + 1)(x + 3) = 2x² + 6x + x + 3 = 2x² + 7x + 3 ✓. Choice A gives (2x+3)(x+1) = 2x² + 5x + 3, not 7x.

First factor out the GCF, 4x: 4x(x² − 4). Then recognize x² − 4 as a difference of squares and factor it: (x − 2)(x + 2). Complete factorization: 4x(x − 2)(x + 2). Choice A stops too early — x² − 4 must be factored further. Choices C and D have not pulled out the full GCF.

Group: (2x³ + x²) + (6x + 3) = x²(2x + 1) + 3(2x + 1). Factor out the common binomial (2x + 1): (x² + 3)(2x + 1). Choice B is wrong because the first factor should be x², not x — grouping the first two terms gives x², not x. Choice A results from incorrect grouping of terms.

AC method: AC = 6 × (−3) = −18. Find factors of −18 that add to 7: 9 and −2. Rewrite: 6x² + 9x − 2x − 3 = 3x(2x + 3) − 1(2x + 3) = (3x − 1)(2x + 3). Check: (2x + 3)(3x − 1) = 6x² − 2x + 9x − 3 = 6x² + 7x − 3 ✓. Choice A gives (3x+1)(2x−3) = 6x²−9x+2x−3 = 6x²−7x−3, flipping the sign of the middle term.

Find two numbers that multiply to −12 and add to −1: −4 and 3. So x² − x − 12 = (x + 3)(x − 4). Check: (x + 3)(x − 4) = x² − 4x + 3x − 12 = x² − x − 12 ✓. Choice A gives (x − 3)(x + 4) = x² + 4x − 3x − 12 = x² + x − 12, which has the wrong sign on the middle term.

Find two numbers that multiply to +28 and add to −11: −7 and −4. Both must be negative since the product is positive and the sum is negative. So x² − 11x + 28 = (x − 7)(x − 4). Choice B uses (x − 7)(x + 4) = x² − 3x − 28, where the product is negative. Choice A gives a product of −28, not +28.

AC method: AC = 12 × (−10) = −120. Find factors of −120 that add to −7: −15 and 8. Rewrite: 12x² − 15x + 8x − 10 = 3x(4x − 5) + 2(4x − 5) = (3x + 2)(4x − 5). Choice D gives (3x − 2)(4x + 5) = 12x² + 15x − 8x − 10 = 12x² + 7x − 10, where the sign of the middle term is flipped.

First factor out the GCF, 3x²: 3x²(x² − 25). Then factor x² − 25 as a difference of squares: (x − 5)(x + 5). Complete factorization: 3x²(x − 5)(x + 5). Choice A pulls out only 3x, missing one factor of x. Choice B stops before applying difference of squares to x² − 25. Choice D incorrectly factors the expression.

AC method: AC = 6 × (−5) = −30. Find factors of −30 that add to 13: 15 and −2. Rewrite: 6x² + 15x − 2x − 5 = 3x(2x + 5) − 1(2x + 5) = (3x − 1)(2x + 5). Check: (3x − 1)(2x + 5) = 6x² + 15x − 2x − 5 = 6x² + 13x − 5 ✓. Choice B gives (3x+5)(2x−1) = 6x² − 3x + 10x − 5 = 6x² + 7x − 5, not 13x.

Factor out the GCF, 2x: 2x(x² − 4x − 5). Factor the trinomial by finding numbers that multiply to −5 and add to −4: −5 and 1. So x² − 4x − 5 = (x − 5)(x + 1). Complete factorization: 2x(x − 5)(x + 1). Choice A gives 2x(x+5)(x−1) = 2x(x²+4x−5), which expands to 2x³ + 8x² − 10x, not the original expression.

Group: (3x³ − 9x²) + (4x − 12) = 3x²(x − 3) + 4(x − 3). Factor out the common binomial (x − 3): (3x² + 4)(x − 3). Choice C uses (x + 3) instead of (x − 3), which is a sign error — the grouped terms produce (x − 3), not (x + 3). Choice B rearranges incorrectly.

AC method: AC = 10 × (−1) = −10. Find factors of −10 that add to −3: −5 and 2. Rewrite: 10x² − 5x + 2x − 1 = 5x(2x − 1) + 1(2x − 1) = (5x + 1)(2x − 1). Check: (5x + 1)(2x − 1) = 10x² − 5x + 2x − 1 = 10x² − 3x − 1 ✓. Choice A gives (5x−1)(2x+1) = 10x² + 5x − 2x − 1 = 10x² + 3x − 1, where the sign of the middle term is flipped.

A perfect square trinomial has the form (x + a)² = x² + 2ax + a². For a² = 25, a = 5 or a = −5. The middle coefficient is k = 2a, so k = 10 or k = −10. Check: (x + 5)² = x² + 10x + 25 and (x − 5)² = x² − 10x + 25. Choice C confuses k with a — the coefficient of the middle term is 2a, not a itself.

Apply difference of squares twice. First: x⁴ − 16 = (x²)² − 4² = (x² + 4)(x² − 4). Then factor x² − 4 again as a difference of squares: (x − 2)(x + 2). Note that x² + 4 cannot be factored over the real numbers. Final answer: (x² + 4)(x − 2)(x + 2). Choice D treats the expression as if it were x² − 16 rather than x⁴ − 16, skipping the squared variable.

The GCF of 15 and 25 is 5. The GCF of x³ and x² is x² (the lower power). So the GCF is 5x². Choice A uses x³, which is the higher power and is not shared by both terms — 25x² does not contain x³.

A difference of squares has the form a² − b², where both terms are perfect squares. x² − 81 = x² − 9² fits this pattern perfectly. x² + 16 is a sum, not a difference. x² − 5 fails because 5 is not a perfect square. 4x² + 9 is a sum of squares, which does not factor over real numbers.

A perfect square trinomial has the form a² + 2ab + b². For x² + 8x + 16: the last term is 4² = 16 and the middle term is 2(1)(4) = 8, so it factors as (x + 4)². Choice D has last term 16 = 4², but 2(1)(4) = 8, not 4 — the middle term is wrong. Choice B: the middle term 8 requires the constant to be 16, not 12.

x² − 64 = x² − 8² is a difference of squares, which factors as (x − 8)(x + 8). Choice A: (x − 8)² = x² − 16x + 64, which has a middle term and a positive constant. Choice D: (x − 4)(x + 16) = x² + 12x − 64, which has a middle term and a wrong constant.

Find two numbers that multiply to 6 and add to 5: 2 and 3. So the factored form is (x + 2)(x + 3). Choice A gives a middle term of 1 + 6 = 7, not 5. Choice C gives a middle term of −5x, since (−2) + (−3) = −5.

The GCF of 6 and 9 is 3. The GCF of a² and a is a (lower power). The GCF of b and b² is b (lower power). So the GCF is 3ab. Choice B: 6 does not divide evenly into 9. Choice C: a² appears in 6a²b but not in 9ab², so a² cannot be the GCF.

The GCF of 7x² and 14x is 7x. Factoring out 7x gives 7x(x − 2). Choice A factors out only 7, leaving x² − 2 inside, which is incorrect. Choice B factors out only x, leaving 7 inside the parentheses — not fully factored. Choice D: 7x(x + 2) = 7x² + 14x, which has the wrong sign.

x² − 14x + 49 is a perfect square trinomial: 49 = 7² and the middle term 14 = 2(1)(7). It factors as (x − 7)². Choice C: (x − 7)(x + 7) = x² − 49, which eliminates the middle term and changes the constant. Choice B: (x + 7)² = x² + 14x + 49 has the wrong sign on the middle term.

Find two numbers that multiply to −18 and add to −3: −6 and 3. So (x − 6)(x + 3). Verify: (x − 6)(x + 3) = x² + 3x − 6x − 18 = x² − 3x − 18. Choice B: (x + 6)(x − 3) = x² + 3x − 18, where the middle term is +3x instead of −3x.

First factor out the GCF of 3: 3(x² − 9). Then x² − 9 = (x − 3)(x + 3) is a difference of squares. The fully factored form is 3(x − 3)(x + 3). Choice D is only partially factored — x² − 9 can be factored further. Choice A: 3(x − 3)² = 3x² − 18x + 27, which does not equal 3x² − 27. Choice C is missing the factor of 3.

Group as (x³ + 4x²) + (3x + 12). Factor each group: x²(x + 4) + 3(x + 4). Factor out the common binomial (x + 4): (x + 4)(x² + 3). Choice B: (x + 3)(x² + 4) = x³ + 4x + 3x² + 12, which has the middle terms switched — 3x² and 4x instead of 4x² and 3x.

Factor out GCF of 2: 2(x² − 25). Then x² − 25 = (x − 5)(x + 5) by difference of squares. Complete factoring: 2(x − 5)(x + 5). Choice D is only partially factored — x² − 25 can be factored further. Choice B: while (2x − 10)(x + 5) equals 2x² − 50 numerically, it is not fully factored because 2x − 10 = 2(x − 5).

x² + 10x + 25 is a perfect square trinomial: 25 = 5² and 10 = 2(1)(5). It factors as (x + 5)². Choice C: (x + 5)(x − 5) = x² − 25, losing the middle term and changing the sign of the constant. Choice B: (x − 5)² = x² − 10x + 25 has a negative middle term.

9x² − 16 = (3x)² − 4² is a difference of squares. It factors as (3x − 4)(3x + 4). Choice C: (3x − 4)² = 9x² − 24x + 16, which has a middle term and a positive constant. Choice B: (9x − 4)(x + 4) = 9x² + 36x − 4x − 16 = 9x² + 32x − 16, which includes an unwanted middle term.

Group as (3x³ + 6x²) + (x + 2). Factor each group: 3x²(x + 2) + 1(x + 2). Factor out the common binomial: (x + 2)(3x² + 1). Choice B: (3x + 1)(x² + 2) = 3x³ + 6x + x² + 2, which has x² and 6x instead of 6x² and x. Choice D: (x + 2)(3x + 1) = 3x² + 7x + 2, only a degree-2 polynomial.

4x² − 25 = (2x)² − 5² is a difference of squares. It factors as (2x − 5)(2x + 5). Choice D: (2x − 5)² = 4x² − 20x + 25, which has a middle term. Choice B: (2x + 5)² = 4x² + 20x + 25, also has a middle term and is positive. Choice A: (4x − 5)(x + 5) = 4x² + 20x − 5x − 25 = 4x² + 15x − 25, wrong.

Find two numbers that multiply to 10 and add to −7: −2 and −5. So (x − 2)(x − 5). Verify: (−2)(−5) = 10 and (−2) + (−5) = −7. Choice B: (x + 2)(x − 5) gives a middle term of 2x − 5x = −3x, not −7x. Choice D: (x − 2)(x + 5) gives a middle term of +3x.

(3x − 2)² = (3x)² − 2(3x)(2) + 2² = 9x² − 12x + 4. Choice A: 9x² − 4 = (3x − 2)(3x + 2), which is the difference of squares — not the square of (3x − 2). Choice C: 9x² + 12x + 4 = (3x + 2)², which results from squaring (3x + 2) with a positive sign. Choice D uses 6x² instead of 9x² as the leading term.

Factor GCF 2: 2(x⁴ − 16). Then x⁴ − 16 = (x²)² − 4² = (x² − 4)(x² + 4). Apply difference of squares again to x² − 4: (x − 2)(x + 2). The sum x² + 4 cannot be factored over real numbers. Complete factoring: 2(x − 2)(x + 2)(x² + 4). Choice A stops too early — x² − 4 factors further. Choice B is incorrect because (x − 2)²(x + 2)² = (x² − 4)² ≠ x⁴ − 16.

Factor GCF 3x: 3x(2x² + x − 6). Use the AC method on 2x² + x − 6: multiply 2 × (−6) = −12; find factors of −12 that add to 1: 4 and −3. Rewrite: 2x² + 4x − 3x − 6 = 2x(x + 2) − 3(x + 2) = (2x − 3)(x + 2). Complete factoring: 3x(2x − 3)(x + 2). Choice A: (2x + 3)(x − 2) = 2x² − x − 6, which has the wrong sign on the middle term.

Use the AC method: a × c = 8 × (−15) = −120. Find factors of −120 that add to 2: 12 and −10. Rewrite: 8x² + 12x − 10x − 15 = 4x(2x + 3) − 5(2x + 3) = (4x − 5)(2x + 3). Choice A: (4x + 5)(2x − 3) = 8x² − 12x + 10x − 15 = 8x² − 2x − 15, which has the wrong sign on the middle term.

Group as (6x³ − 4x²) + (9x − 6). Factor each group: 2x²(3x − 2) + 3(3x − 2). Factor out the common binomial: (3x − 2)(2x² + 3). Choice A: (3x + 2)(2x² − 3) = 6x³ − 9x + 4x² − 6, which has different signs on interior terms. Choice C is not fully factored, since 6x − 4 = 2(3x − 2).

Factor GCF 3x: 3x(x² + 2x − 15). Then factor x² + 2x − 15: find two numbers that multiply to −15 and add to 2: 5 and −3. So x² + 2x − 15 = (x + 5)(x − 3). Complete factoring: 3x(x + 5)(x − 3). Choice B: (x − 5)(x + 3) = x² − 2x − 15, which has the wrong sign on the middle term. Choice C is only partially factored.

Use the AC method: 15 × (−8) = −120. Find factors of −120 that add to −14: −20 and 6. Rewrite: 15x² − 20x + 6x − 8 = 5x(3x − 4) + 2(3x − 4) = (5x + 2)(3x − 4). Choice A: (5x − 2)(3x + 4) = 15x² + 20x − 6x − 8 = 15x² + 14x − 8, where the middle term has the wrong sign.

Group as (x³ + 5x²) + (−x − 5) = x²(x + 5) − 1(x + 5) = (x + 5)(x² − 1). Then factor x² − 1 as a difference of squares: (x − 1)(x + 1). Complete factoring: (x + 5)(x − 1)(x + 1). Choice A is only partially factored — x² − 1 can be factored further. Choice C incorrectly uses (x − 5); the grouping produces (x + 5), not (x − 5).

The GCF of 12 and 8 is 4. The GCF of x³ and x² is x² (the lower power). So the GCF is 4x². Choice B is wrong because x³ is not a factor of x²; the GCF uses the smaller exponent, not the larger one.

x² − 9 is a difference of squares: x² − 3² = (x − 3)(x + 3). Choice C is wrong because (x − 3)² = x² − 6x + 9, which introduces a middle term and a different constant. Choice B is wrong because (x − 9)(x + 1) = x² − 8x − 9, not x² − 9.

Find two numbers that multiply to 12 and add to 7: those are 3 and 4. So x² + 7x + 12 = (x + 3)(x + 4). Choice B is wrong because 2 + 6 = 8, not 7, even though 2 × 6 = 12. Choice D gives x² + 8x + 16, which is a different trinomial.

The GCF of 6x² and 9x is 3x. Dividing each term: 6x² ÷ 3x = 2x and 9x ÷ 3x = 3, giving 3x(2x + 3). Choice B forgets to include x in the GCF. Choice C only factors out x, missing the common factor of 3. Choice D is wrong because 3x × 9 = 27x, not 9x.

A difference of squares has the form a² − b² where both terms are perfect squares and they are subtracted. x² − 25 = x² − 5², which fits perfectly. Choice B is a sum, not a difference. Choice C has 5, which is not a perfect square. Choice D has a middle term, making it a trinomial rather than a difference of squares.

The GCF of 5x² and 20x is 5x. Dividing: 5x² ÷ 5x = x and 20x ÷ 5x = 4, giving 5x(x − 4). Choice B fails to factor out x from the GCF. Choice C only factors out x, not the full GCF of 5x. Choice D is wrong: 5x × (−20) = −100x, not −20x.

Find two numbers that multiply to −6 and add to −1: those are −3 and +2. So x² − x − 6 = (x − 3)(x + 2). Choice B gives (x + 3)(x − 2) = x² + x − 6, which has the wrong sign on the middle term. Choice D gives x² − 5x + 6, which is entirely different.

16x² − 1 = (4x)² − 1², a difference of squares, so it factors as (4x − 1)(4x + 1). Choice B is wrong: (8x − 1)(2x + 1) = 16x² + 6x − 1, which has a middle term. Choice C is wrong: (4x − 1)² = 16x² − 8x + 1, not 16x² − 1.

First factor out the GCF of 3: 3(x² + 4x + 4). Then recognize x² + 4x + 4 = (x + 2)², so the answer is 3(x + 2)². Choice B is wrong: (x + 1)(x + 4) = x² + 5x + 4, not x² + 4x + 4. Choice D gives 3x² + 15x + 12, not the original expression.

Group the terms: (2x³ + x²) + (4x + 2). Factor each group: x²(2x + 1) + 2(2x + 1). Factor out the common binomial (2x + 1): (x² + 2)(2x + 1). Choice C is wrong: (2x² + 1)(x + 2) = 2x³ + 4x² + x + 2, which does not match the original polynomial.

x² − 14x + 49 is a perfect square trinomial: (x − 7)² = x² − 2(7)x + 49 = x² − 14x + 49. Choice B is wrong: (x + 7)² = x² + 14x + 49, which has a positive middle term. Choice C is wrong: (x − 7)(x + 7) = x² − 49, the difference of squares with no middle term.

First factor out the GCF: 4x(x² − 4). Then factor x² − 4 as a difference of squares: (x − 2)(x + 2). The complete factorization is 4x(x − 2)(x + 2). Choice B stops early — x² − 4 can still be factored. Choice D is wrong: (x − 4)(x + 4) = x² − 16, but we need x² − 4.

Multiply a × c = 2 × 3 = 6. Find two numbers that multiply to 6 and add to 5: 2 and 3. Rewrite: 2x² + 2x + 3x + 3 = 2x(x + 1) + 3(x + 1) = (2x + 3)(x + 1). Choice B is wrong: (2x + 1)(x + 3) = 2x² + 7x + 3, not 2x² + 5x + 3.

First factor out the GCF of 5: 5(x² − 1). Then factor x² − 1 as a difference of squares: (x − 1)(x + 1). The complete factorization is 5(x − 1)(x + 1). Choice D is only partially factored because x² − 1 can still be factored. Choice B is wrong: 5(x − 1)² = 5x² − 10x + 5, not 5x² − 5.

Group: (4x³ − 8x²) + (3x − 6). Factor each group: 4x²(x − 2) + 3(x − 2). Factor out (x − 2): (4x² + 3)(x − 2). Choice B is wrong: (4x² − 3)(x + 2) = 4x³ + 8x² − 3x − 6, which has different signs. Choice C is wrong: (4x + 3)(x² − 2) = 4x³ − 8x + 3x² − 6, which rearranges the terms incorrectly.

Find two numbers that multiply to −15 and add to +2: those are +5 and −3. So x² + 2x − 15 = (x + 5)(x − 3). Choice B is wrong: (x − 5)(x + 3) = x² − 2x − 15, which has a negative middle term. Choice D is wrong: (x − 3)(x − 5) = x² − 8x + 15, which is a positive constant.

First factor out the GCF of 6: 6(x² − 4). Then factor x² − 4 as a difference of squares: (x − 2)(x + 2). The complete factorization is 6(x − 2)(x + 2). Choice B is wrong: it uses (x − 4)(x + 4) = x² − 16, but we need x² − 4. Choice D is only partially factored — x² − 4 should be factored further.

First factor out the GCF of 2x: 2x(x² − 4x − 21). Then factor the trinomial: find two numbers that multiply to −21 and add to −4, which are −7 and +3. So 2x(x − 7)(x + 3). Choice B is wrong: (x + 7)(x − 3) = x² + 4x − 21, giving the wrong sign on the middle term. Choice C is only partially factored.

First factor out the GCF of 2x: 2x(9x² − 25). Then recognize 9x² − 25 = (3x)² − 5² as a difference of squares: (3x − 5)(3x + 5). The complete factorization is 2x(3x − 5)(3x + 5). Choice B is only partially factored — 9x² − 25 can still be factored. Choice D is wrong: (3x − 5)² = 9x² − 30x + 25, which is a perfect square trinomial, not a difference of squares.

Multiply a × c = 3 × (−10) = −30. Find two numbers that multiply to −30 and add to −1: −6 and +5. Rewrite: 3x² − 6x + 5x − 10 = 3x(x − 2) + 5(x − 2) = (3x + 5)(x − 2). Choice B is wrong: (3x − 5)(x + 2) = 3x² + x − 10, which has the opposite sign on the middle term. Choice C is wrong: (x + 5)(3x − 2) = 3x² + 13x − 10.

x⁴ − 81 = (x²)² − 9² = (x² − 9)(x² + 9). Then x² − 9 = (x − 3)(x + 3), while x² + 9 is a sum of squares and cannot be factored over the real numbers. The full factorization is (x − 3)(x + 3)(x² + 9). Choice B stops too early — x² − 9 must be factored further. Choice D is wrong: (x − 3)²(x + 3)² = (x² − 9)² = x⁴ − 18x² + 81, not x⁴ − 81.

4x² − 1 = (2x)² − 1² is a difference of squares, so it factors as (2x − 1)(2x + 1). The complete factorization is 3(2x − 1)(2x + 1). Choice B is wrong because there is no common factor of 3 inside the parentheses. Choice D is incorrect — 4x² − 1 is a difference of perfect squares and can always be factored.

Multiply a × c = 4 × (−3) = −12. Find two numbers that multiply to −12 and add to +4: −2 and +6. Rewrite: 4x² − 2x + 6x − 3 = 2x(2x − 1) + 3(2x − 1) = (2x + 3)(2x − 1). Choice C is close but wrong: (2x − 3)(2x + 1) = 4x² − 4x − 3, which has the wrong sign on the middle term. Choice B is wrong: (4x − 1)(x + 3) = 4x² + 11x − 3.

Group: (2x³ − 3x²) + (−8x + 12). Factor each group: x²(2x − 3) − 4(2x − 3). Factor out (2x − 3): (x² − 4)(2x − 3). Then factor x² − 4 as a difference of squares: (x − 2)(x + 2). The complete factorization is (x − 2)(x + 2)(2x − 3). Choice B is only partially factored — x² − 4 should be factored further. Choice C has the wrong sign: the grouping gives −4, not +4.

Check for a perfect square trinomial: √(9x²) = 3x, √25 = 5, and 2(3x)(5) = 30x matches the middle term with a negative sign. So 9x² − 30x + 25 = (3x − 5)². Choice B is wrong: (3x + 5)² = 9x² + 30x + 25, which has a positive middle term. Choice C is wrong: (3x − 5)(3x + 5) = 9x² − 25, which removes the middle term entirely and changes the constant.

The GCF of the coefficients 10 and 15 is 5. The GCF of x³ and x² is x² (the lower power). So the GCF is 5x². Choice B uses 10 instead of the true GCF 5, and choice C incorrectly uses x³ instead of x².

A difference of squares has the form a² − b² where both terms are perfect squares. In 9x² − 4, we have (3x)² − 2², which fits perfectly. Choice C fails because 3 is not a perfect square. Choice D can be written as 4(x² − 2), but 2 is not a perfect square so x² − 2 is not a difference of squares.

Since x² − 49 = x² − 7², this is a difference of squares, which factors as (x − 7)(x + 7). Choice B, (x − 7)², expands to x² − 14x + 49, which has an extra middle term. The difference of squares pattern always produces conjugate binomials, never a perfect square.

3 × 5 = 15 and 3 + 5 = 8, so the pair 3 and 5 works, giving (x + 3)(x + 5). Choice B: 1 + 15 = 16, not 8. Choice C: (−3)(−5) = 15, but −3 + (−5) = −8, not +8. Choice D: 4 × 4 = 16, not 15.

The GCF of coefficients 6, 9, and 3 is 3. The GCF of x⁴, x³, and x² is x² (the lowest power present). So the GCF is 3x². Choice B uses 6 instead of 3, and choice C incorrectly uses x³, which cannot divide evenly into x².

The GCF of 8 and 12 is 4, and the GCF of x³ and x² is x², giving a total GCF of 4x². Dividing each term: 4x²(2x + 3). Choice B, 4x(2x² + 3x), still has x as a common factor inside the parentheses, so the GCF is not fully removed. Choice D uses 2 instead of the complete GCF of 4.

A perfect square trinomial fits a² + 2ab + b². Here x² + 10x + 25 = (x + 5)², since 2(x)(5) = 10x and 5² = 25. Choice B has 20 instead of 25. Choice C has 5x for the middle term, but the correct middle term for (x + 5)² would be 10x. Choice D has a negative constant, so it cannot be written as a squared binomial.

First factor out the GCF: 2(x² + 3x − 10). Then find two integers that multiply to −10 and add to 3: 5 and −2. So x² + 3x − 10 = (x + 5)(x − 2), giving 2(x + 5)(x − 2). Choice B gives 2(x² − 3x − 10), which has the wrong sign on the middle term. Choice D gives a positive constant after expansion, not matching the original.

This is a perfect square trinomial: x² − 8x + 16 = (x − 4)², since (−4)² = 16 and 2(x)(−4) = −8x. Choice C, (x − 4)(x + 4), is the difference of squares pattern and expands to x² − 16, losing the middle term entirely. Choice D expands to x² − 10x + 16, which does not match.

Group the terms: (x³ + 2x²) + (3x + 6) = x²(x + 2) + 3(x + 2). The shared binomial factor is (x + 2), so the result is (x² + 3)(x + 2). Choice B, (x² + 2)(x + 3), would expand to x³ + 3x² + 2x + 6, which does not match the original polynomial.

4x² − 9 = (2x)² − 3², which is a difference of squares. It factors as (2x − 3)(2x + 3). Choice C, (2x − 3)², is a perfect square and expands to 4x² − 12x + 9, which includes a middle term not present in the original expression.

Find two integers that multiply to −20 and add to −1: those are −5 and 4. So x² − x − 20 = (x − 5)(x + 4). Choice B, (x + 5)(x − 4), expands to x² + x − 20, which has the wrong sign on the middle term. Choice D gives a positive constant after expansion, not −20.

Group: (x³ − 4x²) + (3x − 12) = x²(x − 4) + 3(x − 4) = (x² + 3)(x − 4). Choice C, (x² + 4)(x − 3), expands to x³ − 3x² + 4x − 12, which does not match. The key is recognizing that both groups share the factor (x − 4), not (x − 3).

Factor out 3: 3(x² − 4x − 5). Then find two integers that multiply to −5 and add to −4: those are −5 and 1. So x² − 4x − 5 = (x − 5)(x + 1), and the complete factorization is 3(x − 5)(x + 1). The student is correct. Choice B reverses the signs incorrectly. Choice C is wrong because the GCF of 3, 12, and 15 is 3, not 6.

Multiply 3 × 6 = 18. Find factors of 18 that add to 11: 2 and 9. Rewrite: 3x² + 2x + 9x + 6 = x(3x + 2) + 3(3x + 2) = (x + 3)(3x + 2). Choice B, (3x + 6)(x + 1), expands to 3x² + 9x + 6. Choice C, (x + 2)(3x + 3), also expands to 3x² + 9x + 6. Both are missing the correct middle coefficient.

Expanding using FOIL: (x + a)(x − a) = x² − ax + ax − a² = x² − a². The two middle terms cancel. Choice D, x² − 2ax + a², is the expansion of (x − a)², not (x + a)(x − a). This identity — the difference of squares — is fundamental because it shows that conjugate pairs always eliminate the middle term.

First factor out the GCF, 2x: 2x(x² − 9). Then factor the difference of squares: x² − 9 = (x − 3)(x + 3). The complete factoring is 2x(x − 3)(x + 3). Choice B stops prematurely — x² − 9 is still factorable. Choice C is missing the x factor that was part of the GCF.

Expand (x − 9)(x + 2) = x² + 2x − 9x − 18 = x² − 7x − 18. So b = −7 and c = −18. The coefficient b equals the sum of the two constants: −9 + 2 = −7. The constant c equals their product: (−9)(2) = −18. A common error is to use the wrong signs for one or both values.

First factor out the GCF, 3x: 3x(x² + 2x − 15). Then find two integers that multiply to −15 and add to 2: 5 and −3. So x² + 2x − 15 = (x + 5)(x − 3), giving 3x(x + 5)(x − 3). Choice B gives a trinomial with a middle term of −2x after the GCF is removed. Choice C omits the x from the GCF, losing one factor entirely.

First factor out 4: 4(x⁴ − 16). Apply the difference of squares: x⁴ − 16 = (x² − 4)(x² + 4). Apply the difference of squares again to x² − 4: (x − 2)(x + 2). Note that x² + 4 has no real factors. The complete factoring is 4(x − 2)(x + 2)(x² + 4). Choice B is only partially factored — x² − 4 must be factored further. Choice C incorrectly treats x⁴ as x², applying the pattern one level too shallow.

Group: (6x³ + 9x²) + (−4x − 6) = 3x²(2x + 3) − 2(2x + 3) = (3x² − 2)(2x + 3). Verify: (3x² − 2)(2x + 3) = 6x³ + 9x² − 4x − 6. Choice B results from a sign error when factoring the second group, treating −4x − 6 as +4x + 6 in disguise. Choice C comes from incorrectly identifying which terms to group together.

Multiply 6 × 6 = 36. Find factors of 36 that add to −13: −4 and −9. Rewrite: 6x² − 4x − 9x + 6 = 2x(3x − 2) − 3(3x − 2) = (2x − 3)(3x − 2). Choice B, (6x − 1)(x − 6), expands to 6x² − 37x + 6. Choice C, (2x − 6)(3x − 1), expands to 6x² − 20x + 6. Both have the wrong middle coefficient.

By FOIL, the product of the Last terms must equal the constant term of the original polynomial. Here (−4)(1) = −4, but the original has a constant of +4. This immediately reveals the error. The correct factoring is (3x − 2)², since (3x − 2)² = 9x² − 12x + 4. Choice B is a true statement but describes what is correct in the student's work, not what is wrong.

Use grouping: x²(x + 5) − 4(x + 5) = (x² − 4)(x + 5). Then apply the difference of squares to x² − 4: (x − 2)(x + 2). The completely factored form is (x − 2)(x + 2)(x + 5). Choice B is only partially factored — x² − 4 can and must be factored further. Choice D has the wrong sign on the last binomial factor.

Check the perfect square trinomial pattern a² − 2ab + b² = (a − b)². With a = 2x and b = 5: (2x)² = 4x², 2(2x)(5) = 20x, and 5² = 25. All three conditions match, so 4x² − 20x + 25 = (2x − 5)². Choice C, (2x − 5)(2x + 5), applies the difference of squares pattern and expands to 4x² − 25, which is missing the middle term entirely.

The GCF of 12 and 8 is 4. The GCF of x³ and x² is x² (use the lower exponent). So the GCF is 4x². Choice A (4x) incorrectly uses x instead of x² as the variable part.

A difference of squares requires two perfect square terms that are subtracted. x² − 25 = x² − 5² fits this pattern exactly. Choices A and D are sums, not differences. Choice B subtracts 7, which is not a perfect square, so it does not match the a² − b² form.

x² − 16 = x² − 4² fits the difference of squares formula a² − b² = (a + b)(a − b), giving (x + 4)(x − 4). Choice A gives (x − 4)² = x² − 8x + 16, which introduces an unwanted middle term. Choice D uses 8 instead of 4, which would represent x² − 64.

The GCF of 6 and 9 is 3. The GCF of x² and x is x (the lower power). Multiplying these gives 3x. Choice A (just 3) correctly finds the numerical GCF but misses the common variable factor x.

3 × 4 = 12 and 3 + 4 = 7, so 3 and 4 is the correct pair. Choice A (2 and 6) gives a product of 12 but a sum of 8, not 7. Choice B gives a product of 12 but a sum of −7, which would be used for x² − 7x + 12 instead.

The GCF of 5x² and 10x is 5x. Dividing each term by 5x gives x and 2, so the factored form is 5x(x + 2). Choice A incorrectly pulls out only 5 instead of 5x. Choice D, x(5x + 10), is only partially factored — the 5 inside the second factor can still be pulled out.

x² − 36 = x² − 6², which factors as (x + 6)(x − 6) using the difference of squares formula. Choice A gives (x − 6)² = x² − 12x + 36, which includes a middle term and a different constant. Choice D gives x² − 5x − 36, a completely different expression.

Find two positive numbers that multiply to 18 and add to 9: 3 and 6. So x² + 9x + 18 = (x + 3)(x + 6). Choice A, (x + 2)(x + 9), gives x² + 11x + 18 — the middle term is wrong. Choice C uses negative signs, producing a positive constant but a negative middle term (−9x), the opposite of what is needed.

Step 1: Factor out the GCF of 6 to get 6(x² − 1). Step 2: Recognize x² − 1 as a difference of squares and factor it as (x + 1)(x − 1). The completely factored form is 6(x + 1)(x − 1). Choice B stops too early — x² − 1 can be factored further. Choice D is wrong because (x − 1)² = x² − 2x + 1, not x² − 1.

Find two numbers that multiply to −18 and add to −3: −6 and 3 (−6 × 3 = −18, −6 + 3 = −3). So x² − 3x − 18 = (x − 6)(x + 3). Choice A reverses the signs: (x + 6)(x − 3) = x² + 3x − 18, which produces a positive middle term instead of a negative one.

Group: (2x³ + 4x²) + (3x + 6) = 2x²(x + 2) + 3(x + 2). Factor out the common binomial (x + 2): (2x² + 3)(x + 2). Choice B is wrong: (x² + 3)(2x + 2) = 2x³ + 2x² + 6x + 6, which has different coefficients. Choice C: (2x + 3)(x² + 2) = 2x³ + 4x + 3x² + 6 = 2x³ + 3x² + 4x + 6, which is not the original polynomial.

Factor x² + 2x − 15 by finding numbers that multiply to −15 and add to 2: those are 5 and −3. So (x + 5)(x − 3) = 0, giving x = −5 or x = 3. Choice A (x = 5, x = −3) has the signs flipped — it would come from (x − 5)(x + 3) = 0, which produces x² − 2x − 15, not x² + 2x − 15.

Use the ac method: a = 2, c = 3, so ac = 6. Find factors of 6 that add to 7: 1 and 6. Rewrite the middle term: 2x² + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3). Choice A is a common error: (2x + 3)(x + 1) = 2x² + 2x + 3x + 3 = 2x² + 5x + 3, which has the wrong middle term.

First, factor out the GCF of 5x: 5x(x² − 4). Then recognize x² − 4 as a difference of squares and factor it as (x + 2)(x − 2). The complete factoring is 5x(x + 2)(x − 2). Choice A is incorrect because applying difference of squares directly to 5x³ − 20x does not work — the expression is not in the form a² − b².

Find two numbers that multiply to 12 and add to −7: both must be negative. −3 and −4 satisfy (−3)(−4) = 12 and (−3) + (−4) = −7. So x² − 7x + 12 = (x − 3)(x − 4). Choice A uses positive signs: (x + 3)(x + 4) = x² + 7x + 12, giving a positive middle term. Choice B: (x − 2)(x − 6) = x² − 8x + 12, which has the wrong middle coefficient.

Group: (3x³ − 2x²) + (6x − 4) = x²(3x − 2) + 2(3x − 2). Factor out the common binomial (3x − 2): (x² + 2)(3x − 2). Choice C shows the intermediate step before the common binomial is factored out — it is not fully factored. Choice D: (3x² + 2)(x − 2) = 3x³ − 6x² + 2x − 4, which differs from the original polynomial.

Expand (x + c)(x − 4) = x² − 4x + cx − 4c = x² + (c − 4)x − 4c. Match the constant: −4c = −12, so c = 3. Then b = c − 4 = 3 − 4 = −1. The factored form is (x + 3)(x − 4). Choice A (b = 1) would require c = 5, which gives a constant of −20, not −12.

Step 1: Factor out the GCF of 2: 2(x⁴ − 16). Step 2: x⁴ − 16 = (x²)² − 4² = (x² + 4)(x² − 4). Step 3: Factor x² − 4 as another difference of squares: (x + 2)(x − 2). Note that x² + 4 is a sum of squares and cannot be factored over the real numbers. The full result is 2(x² + 4)(x + 2)(x − 2). Choice D is missing the GCF of 2.

A perfect square trinomial follows the pattern (x + a)² = x² + 2ax + a² or (x − a)² = x² − 2ax + a². For the constant term to be 9, we need a² = 9, so a = 3 or a = −3. The middle coefficient is 2a, so k = 2(3) = 6 or k = 2(−3) = −6. Both (x + 3)² = x² + 6x + 9 and (x − 3)² = x² − 6x + 9 are valid. Choices A and C incorrectly use a instead of 2a for k.

Step 1: Factor out the GCF of 3: 3(4x² − 25y²). Step 2: Recognize 4x² − 25y² = (2x)² − (5y)² as a difference of squares. Step 3: Apply the formula: (2x + 5y)(2x − 5y). The complete answer is 3(2x + 5y)(2x − 5y). Choice D uses the perfect square pattern, which would give 3(4x² − 20xy + 25y²) — not the original expression.

The difference of squares pattern a² − b² requires subtraction of two perfect squares. x² + 25 is a sum of squares, not a difference, and cannot be factored over the real numbers. The other three are all valid differences of squares: x² − 49 = (x + 7)(x − 7), 9x² − 16 = (3x + 4)(3x − 4), and 25x² − 4 = (5x + 2)(5x − 2).

Step 1: Factor out the GCF of 3x: 3x(x² − 4x + 3). Step 2: Factor x² − 4x + 3 by finding numbers that multiply to 3 and add to −4: −1 and −3. So x² − 4x + 3 = (x − 1)(x − 3). The complete answer is 3x(x − 1)(x − 3). Choice D stops after one step and does not factor the trinomial. Choice B has the wrong sign on one factor: (x − 1)(x + 3) = x² + 2x − 3, not x² − 4x + 3.

Group: (4x³ − 8x²) + (−3x + 6) = 4x²(x − 2) − 3(x − 2). Factor out the common binomial (x − 2): (4x² − 3)(x − 2). Choice A has the wrong sign in the first factor. Choice B shows only the intermediate grouping step — it is not fully factored. Choice C: (4x − 3)(x² − 2) = 4x³ − 8x − 3x² + 6 = 4x³ − 3x² − 8x + 6, which is not the original polynomial.

x² + 6x + 9 = (x + 3)² and x² − 6x + 9 = (x − 3)². Both fit the pattern a² ± 2ab + b². A perfect square trinomial can have a negative middle term because squaring a binomial with subtraction, such as (x − 3)², always produces one. Choices A and D are incorrect because they impose an unnecessary restriction that the middle term must be positive.

Factor 6x² + x − 12 using the ac method: ac = 6 × (−12) = −72. Factors of −72 that add to 1 are 9 and −8. Rewrite: 6x² + 9x − 8x − 12 = 3x(2x + 3) − 4(2x + 3) = (3x − 4)(2x + 3). Since one dimension is (2x + 3), the other is (3x − 4). Choice A, (3x + 4), is a sign error: (3x + 4)(2x + 3) = 6x² + 9x + 8x + 12 = 6x² + 17x + 12, which is not the original expression.

Find the GCF of the coefficients: GCF(12, 18) = 6. Find the GCF of the variable parts: GCF(x³, x²) = x² (take the smallest exponent). So the GCF is 6x². A common mistake is choosing 6x³, but x³ does not divide evenly into x², so x³ cannot be the GCF.

A difference of squares has the form a² − b², where both terms are perfect squares and they are subtracted. x² − 9 = x² − 3², which fits the form. x² + 9 is a sum of squares (not factorable over the integers). The other two are perfect square trinomials, not differences of squares.

The GCF of 15 and 10 is 5, and the GCF of x³ and x² is x². So the GCF is 5x². Dividing each term: 15x³ ÷ 5x² = 3x and 10x² ÷ 5x² = 2, giving 5x²(3x + 2). Choice A uses only 5x, leaving a common factor of x still inside the parentheses, so it is not fully factored.

A perfect square trinomial has the form a² + 2ab + b². For x² + 8x + 16: a = x, b = 4, and 2ab = 2(x)(4) = 8x. It factors as (x + 4)². For x² + 4x + 16, the middle term would need to be 2(x)(4) = 8x, not 4x, so it is not a perfect square trinomial.

Find two numbers that multiply to 20 and add to 9. The pairs are: 1 and 20 (sum 21), 2 and 10 (sum 12), 4 and 5 (sum 9). The pair 4 and 5 works, so the factored form is (x + 4)(x + 5). You can verify: (x + 4)(x + 5) = x² + 5x + 4x + 20 = x² + 9x + 20.

The GCF of 8x² and 4x is 4x. Dividing: 8x² ÷ 4x = 2x and 4x ÷ 4x = 1, giving 4x(2x − 1). Choice A uses GCF of 4 but ignores the common factor of x, leaving 2x² − x, which still has a common factor x and is not fully factored. Choice C uses 2x but leaves (4x − 2), which has a common factor of 2 and is not fully factored.

x² − 25 is a difference of squares: x² − 5². The difference of squares formula is a² − b² = (a + b)(a − b), so x² − 25 = (x + 5)(x − 5). Choices A and D, which are (x − 5)², expand to x² − 10x + 25, which is a perfect square trinomial — not the same as x² − 25.

Use the AC method: multiply a and c to get 3 × 3 = 9. Find two numbers that multiply to 9 and add to 10: 9 and 1. Rewrite: 3x² + 9x + x + 3 = 3x(x + 3) + 1(x + 3) = (3x + 1)(x + 3). Verify: (3x + 1)(x + 3) = 3x² + 9x + x + 3 = 3x² + 10x + 3. Choice B, (3x + 3)(x + 1), gives 3x² + 6x + 3, which is incorrect.

Group the first two and last two terms: (5x³ + 10x²) + (2x + 4). Factor each group: 5x²(x + 2) + 2(x + 2). The common binomial factor is (x + 2), so the result is (5x² + 2)(x + 2). Choice C uses (x − 2), which would require a minus sign when factoring the second group, giving 2(x − 2), but 2(x − 2) = 2x − 4, not 2x + 4.

Expanding (x + 3)(x + c) = x² + (3 + c)x + 3c. Setting 3c = 18 gives c = 6. Then b = 3 + c = 3 + 6 = 9. A common error is saying b = 3 (just the first term) or b = 6 (just the second term), but b is the sum of both constants in the binomial factors.

Find two numbers that multiply to −18 and add to −3. Testing: −6 and 3 multiply to −18 and add to −3. So the factored form is (x − 6)(x + 3). Choice B, (x + 6)(x − 3), gives a middle term of +3x, not −3x. Choice D uses two negative numbers, giving a positive product and a negative sum, which does not match.

Factor the left side: find two numbers that multiply to 6 and add to −5, which are −2 and −3. So x² − 5x + 6 = (x − 2)(x − 3) = 0. Setting each factor equal to zero: x − 2 = 0 gives x = 2, and x − 3 = 0 gives x = 3. Choice B gives x = −2 and x = −3, which would factor as (x + 2)(x + 3) = x² + 5x + 6, a different polynomial.

4x² − 49 is a difference of squares: (2x)² − 7². Applying the formula a² − b² = (a + b)(a − b) gives (2x + 7)(2x − 7). Choice C, (2x − 7)², expands to 4x² − 28x + 49, which has a middle term and a positive constant, so it is not equivalent. Choice B expands to 4x² − 28x + 7x − 49 = 4x² − 21x − 49, which is incorrect.

First factor out the GCF of 2: 2(x² + 3x − 10). Then factor the trinomial: find two numbers that multiply to −10 and add to 3, which are 5 and −2. So x² + 3x − 10 = (x + 5)(x − 2), giving the final answer 2(x + 5)(x − 2). Choice B is not fully factored because (2x + 10) still has a common factor of 2 that was not extracted.

This is a difference of squares pattern: (a − b)(a + b) = a² − b². With a = x and b = 4: (x − 4)(x + 4) = x² − 16. Choice C, x² − 8x + 16, is (x − 4)², where both binomials have the same sign. The key difference is that when the two binomials have opposite signs but the same terms, the middle terms cancel.

Group: (6x³ − 9x²) + (4x − 6). Factor each group: 3x²(2x − 3) + 2(2x − 3). The common binomial factor is (2x − 3), giving (3x² + 2)(2x − 3). Check: (3x² + 2)(2x − 3) = 6x³ − 9x² + 4x − 6. Choice B would require the second group to factor as −2(2x + 3), but 4x − 6 = 2(2x − 3), not −2(2x + 3).

Use the AC method: 6 × (−10) = −60. Find two numbers that multiply to −60 and add to 11: 15 and −4. Rewrite: 6x² + 15x − 4x − 10 = 3x(2x + 5) − 2(2x + 5) = (3x − 2)(2x + 5). Verify: (2x + 5)(3x − 2) = 6x² − 4x + 15x − 10 = 6x² + 11x − 10. Choice D has (2x − 2), which contains a common factor of 2, indicating it is not fully factored.

The area of a square equals (side length)². Factor x² − 10x + 25 as a perfect square trinomial: find two numbers that multiply to 25 and add to −10, which are −5 and −5. So x² − 10x + 25 = (x − 5)². The side length is x − 5. Choice B, x + 5, would give (x + 5)² = x² + 10x + 25, which has a positive middle term.

First factor out the GCF of 4x: 4x(x² − 4). Then recognize x² − 4 as a difference of squares: x² − 2² = (x + 2)(x − 2). The completely factored form is 4x(x + 2)(x − 2). Choice B stops at 4x(x² − 4) and is not complete because x² − 4 can be factored further. Always check whether remaining factors can be factored again.

A perfect square trinomial has the form (x + k)² = x² + 2kx + k². The middle term is 6x, so 2k = 6, giving k = 3. Therefore c = k² = 3² = 9. The result is (x + 3)². Choice D, c = 36, would be correct if the middle term were 12x (since (x + 6)² = x² + 12x + 36), but the given middle term is 6x, not 12x.

First apply difference of squares: x⁴ − 16 = (x²)² − 4² = (x² + 4)(x² − 4). Then factor x² − 4 again as a difference of squares: (x + 2)(x − 2). Since x² + 4 is a sum of squares, it cannot be factored over the real numbers. The completely factored form is (x² + 4)(x + 2)(x − 2). Choice B stops too early and is not fully factored.

Use the discriminant b² − 4ac to check if integer factors exist (the discriminant must be a perfect square). For x² + 3x + 4: 3² − 4(1)(4) = 9 − 16 = −7, which is negative, so this has no real factors and cannot be factored over the integers. For x² + 5x + 6: 25 − 24 = 1 = 1², factors as (x + 2)(x + 3). For x² − 4x − 5: 16 + 20 = 36 = 6², factors as (x − 5)(x + 1). For x² + x − 6: 1 + 24 = 25 = 5², factors as (x + 3)(x − 2).

First factor out the GCF of 2x: 2x(x² + 2x − 3). Then factor the trinomial x² + 2x − 3: find two numbers that multiply to −3 and add to 2, which are 3 and −1. So x² + 2x − 3 = (x + 3)(x − 1). The complete factoring is 2x(x + 3)(x − 1). Choice D factors out only 2, leaving x³ + 2x² − 3x, which still has a common factor of x and is not fully factored.

Group the terms: (6x³ + 3x²) + (−10x − 5). Factor each group: 3x²(2x + 1) − 5(2x + 1). The common binomial factor is (2x + 1), giving (3x² − 5)(2x + 1). Note that factoring −5 from (−10x − 5) gives −5(2x + 1), keeping the sign of the binomial consistent. Choice B would require the second group to produce +5(2x + 1), meaning −10x − 5 = +5(2x + 1) = 10x + 5, which is false.

Divide the area polynomial by the given length: (3x² − 17x + 10) ÷ (3x − 2). Use the AC method to factor 3x² − 17x + 10: multiply 3 × 10 = 30; find factors of 30 that add to −17: −15 and −2. Rewrite: 3x² − 15x − 2x + 10 = 3x(x − 5) − 2(x − 5) = (3x − 2)(x − 5). The width is x − 5. Verify: (3x − 2)(x − 5) = 3x² − 15x − 2x + 10 = 3x² − 17x + 10.

Find the greatest common factor (GCF) of 12x² and 8x. The GCF of 12 and 8 is 4, and the GCF of x² and x is x, so the overall GCF is 4x. Dividing each term: 12x² ÷ 4x = 3x and 8x ÷ 4x = 2, giving 4x(3x + 2). Choice A is incorrect because 4(3x² + 2x) still has a common factor of x inside the parentheses — it is not fully factored. Choice C is incorrect because 2x(6x + 4) is only partially factored; 6x + 4 still has a common factor of 2.

A difference of squares has the form a² − b², where both terms are perfect squares and they are subtracted. In 4x² − 25, we have (2x)² − 5², which fits the pattern perfectly and factors as (2x + 5)(2x − 5). Choice A (4x² + 25) fails because the terms are added, not subtracted — a sum of squares does not factor over the real numbers. Choice B (4x² − 10) fails because 10 is not a perfect square.

Find the GCF of 15x³ and 10x². The GCF of 15 and 10 is 5, and the GCF of x³ and x² is x². So the GCF is 5x². Dividing: 15x³ ÷ 5x² = 3x and 10x² ÷ 5x² = 2, giving 5x²(3x − 2). Choice A uses 5x instead of 5x², leaving an x inside the parentheses that could still be factored out. Choice B has the wrong sign — the original expression subtracts, so the second term inside must be negative.

To factor x² + 7x + 12, find two numbers that multiply to 12 and add to 7. Testing pairs: 1 and 12 add to 13, 2 and 6 add to 8, 3 and 4 add to 7. So the factors are (x + 3)(x + 4). Choice A is wrong because 2 and 6 multiply to 12 but add to 8, not 7. Choice D is wrong because −3 and −4 multiply to +12 but add to −7, which would give x² − 7x + 12.

Find two numbers that multiply to +20 and add to −9. Since the product is positive and the sum is negative, both numbers must be negative. Testing: −4 and −5 multiply to +20 and add to −9. So the factors are (x − 4)(x − 5). Choice A is wrong: −2 and −10 add to −12, not −9. Choice B is wrong: +4 and −5 multiply to −20, not +20, and they add to −1.

Use the AC method: multiply a = 3 and c = 2 to get 6. Find two numbers that multiply to 6 and add to 7: those are 1 and 6. Rewrite the middle term: 3x² + x + 6x + 2, then group: x(3x + 1) + 2(3x + 1) = (x + 2)(3x + 1). Choice A is wrong: (3x + 2)(x + 1) = 3x² + 3x + 2x + 2 = 3x² + 5x + 2, which has the wrong middle term. Choice C is wrong because it ignores the leading coefficient of 3.

Factor the area expression x² − 36 as a difference of squares: x² − 36 = (x + 6)(x − 6). Since Area = length times width, and one side is (x + 6), divide both sides by (x + 6) to get the other side: (x − 6). Choice B is incorrect — (x + 6)(x + 6) = x² + 12x + 36, which does not equal x² − 36. Always check your factoring by expanding to verify.

First, factor out the GCF of 3: 3(x² − 25). Then recognize x² − 25 as a difference of squares: x² − 25 = (x + 5)(x − 5). The completely factored form is 3(x + 5)(x − 5). Choice B is only partially factored — x² − 25 can be factored further. Choice D is incorrect because it drops the GCF of 3, meaning (x + 5)(x − 5) = x² − 25 ≠ 3x² − 75.

Group the first two and last two terms: (2x³ + 5x²) + (6x + 15). Factor each group: x²(2x + 5) + 3(2x + 5). Both groups share the factor (2x + 5), so factor it out: (x² + 3)(2x + 5). Choice A is incorrect: (2x + 3)(x² + 5) = 2x³ + 10x + 3x² + 15 = 2x³ + 3x² + 10x + 15, which has different coefficients on the x² and x terms. Always verify by expanding your answer.

Use the AC method: a times c = 6 times (−3) = −18. Find two numbers that multiply to −18 and add to −7: those are −9 and +2. Rewrite: 6x² − 9x + 2x − 3, then group: 3x(2x − 3) + 1(2x − 3) = (3x + 1)(2x − 3). Choice A is wrong: (3x − 1)(2x + 3) = 6x² + 9x − 2x − 3 = 6x² + 7x − 3, which has a positive 7x, not negative. Choice D is wrong because (3x − 3) can be factored further as 3(x − 1), meaning the expression is not fully factored.