ACCUPLACER Advanced Algebra & Functions Quiz: Factoring Trinomials
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Factoring TrinomialsQuestion 1 of 20

Which of the following is the correct factorization of 10x2−11x−610x^2 - 11x - 6?

(5x+2)(2x−3)(5x+2)(2x-3)
(5x−2)(2x+3)(5x-2)(2x+3)
(10x+3)(x−2)(10x+3)(x-2)
(10x−3)(x+2)(10x-3)(x+2)
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ACCUPLACER Advanced Algebra & Functions Quiz

ACCUPLACER Advanced Algebra & Functions Quiz: Factoring Trinomials

Practice Factoring Trinomials in ACCUPLACER Advanced Algebra & Functions with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Factoring Trinomials, giving you a quick way to practice the rules, question types, and explanations that matter most for ACCUPLACER Advanced Algebra & Functions.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Which of the following is the correct factorization of 10x2−11x−610x^2 - 11x - 6?

  1. (5x+2)(2x−3)(5x+2)(2x-3) (correct answer)
  2. (5x−2)(2x+3)(5x-2)(2x+3)
  3. (10x+3)(x−2)(10x+3)(x-2)
  4. (10x−3)(x+2)(10x-3)(x+2)
Explanation: To factor 10x2−11x−610x^2 - 11x - 6, we look for two numbers that multiply to 10×−6=−6010 \times -6 = -60 and add to -11. These numbers are -15 and 4. We can rewrite the middle term: 10x2−15x+4x−610x^2 - 15x + 4x - 6. Factor by grouping: 5x(2x−3)+2(2x−3)=(5x+2)(2x−3)5x(2x - 3) + 2(2x - 3) = (5x+2)(2x-3).

Question 2

The area of a rectangular garden is represented by the expression 3x2+10x−83x^2 + 10x - 8. If the width of the garden is represented by x+4x+4, what expression represents the length?

  1. 3x−23x-2 (correct answer)
  2. 3x+23x+2
  3. x−2x-2
  4. x+2x+2
Explanation: The area of a rectangle is length times width. To find the length, we must factor the area expression 3x2+10x−83x^2 + 10x - 8. Since we know one factor is x+4x+4, the other factor must be of the form (3x+k)(3x+k) to get the 3x23x^2 term. The product of the constant terms must be -8, so 4k=−84k = -8, which gives k=−2k = -2. Thus, the other factor (the length) is 3x−23x-2. Checking the expansion: (x+4)(3x−2)=3x2−2x+12x−8=3x2+10x−8(x+4)(3x-2) = 3x^2 - 2x + 12x - 8 = 3x^2 + 10x - 8.

Question 3

For the trinomial ax2+bx+cax^2 + bx + c, if a=3a=3 and c=−4c=-4, which of the following values of bb allows the trinomial to be factored over the integers?

  1. -13
  2. -11 (correct answer)
  3. 8
  4. 10
Explanation: The trinomial is 3x2+bx−43x^2 + bx - 4. For it to be factorable, bb must be the sum of the inner and outer products of its binomial factors. The factors of a=3a=3 are (1, 3). The factors of c=−4c=-4 are (1, -4), (-1, 4), (2, -2). We test the combinations: (x+1)(3x−4)  ⟹  b=−4+3=−1(x+1)(3x-4) \implies b = -4+3 = -1. (x−4)(3x+1)  ⟹  b=1−12=−11(x-4)(3x+1) \implies b = 1-12 = -11. (x−1)(3x+4)  ⟹  b=4−3=1(x-1)(3x+4) \implies b = 4-3 = 1. (x+4)(3x−1)  ⟹  b=−1+12=11(x+4)(3x-1) \implies b = -1+12 = 11. (x+2)(3x−2)  ⟹  b=−2+6=4(x+2)(3x-2) \implies b = -2+6 = 4. (x−2)(3x+2)  ⟹  b=2−6=−4(x-2)(3x+2) \implies b = 2-6 = -4. Of the choices provided, -11 is a possible value for bb.

Question 4

If 4x2+12x+9=(px+q)24x^2 + 12x + 9 = (px + q)^2 for all values of xx, what is p+qp + q?

  1. 4
  2. 5 (correct answer)
  3. 6
  4. 7
Explanation: Since 4x2+12x+94x^2 + 12x + 9 is a perfect square trinomial, we recognize it as (2x+3)2(2x + 3)^2. We can verify: (2x+3)2=4x2+12x+9(2x + 3)^2 = 4x^2 + 12x + 9. Comparing with (px+q)2(px + q)^2, we have p=2p = 2 and q=3q = 3, so p+q=5p + q = 5.

Question 5

The trinomial x2−6x+5x^2 - 6x + 5 and the trinomial 2x2−7x−152x^2 - 7x - 15 share a common binomial factor. What is that common factor?

  1. x−1x-1
  2. x−5x-5 (correct answer)
  3. x+3x+3
  4. 2x+32x+3
Explanation: First, factor the simpler trinomial, x2−6x+5x^2 - 6x + 5. This factors into (x−1)(x−5)(x-1)(x-5). The common factor must be either x−1x-1 or x−5x-5. Now, factor the second trinomial, 2x2−7x−152x^2 - 7x - 15. This factors into (2x+3)(x−5)(2x+3)(x-5). Comparing the factors of both trinomials, the common factor is x−5x-5.

Question 6

If x+4x+4 is a factor of the trinomial x2+kx−24x^2 + kx - 24, what is the other factor?

  1. x−6x-6 (correct answer)
  2. x+6x+6
  3. x−2x-2
  4. x+2x+2
Explanation: If x+4x+4 is a factor, then the factorization must be of the form (x+4)(x+n)(x+4)(x+n). Expanding this gives x2+(4+n)x+4nx^2 + (4+n)x + 4n. Comparing this to x2+kx−24x^2 + kx - 24, we can see that 4n=−244n = -24, which means n=−6n = -6. Therefore, the other factor is x−6x-6. The full factorization is (x+4)(x−6)=x2−2x−24(x+4)(x-6) = x^2 - 2x - 24, so k=−2k = -2.

Question 7

Which of the following is equivalent to the expression 2x2−x−15x2−9\frac{2x^2 - x - 15}{x^2 - 9} for all x≠±3x \neq \pm3?

  1. 2x+5x−3\frac{2x+5}{x-3}
  2. 2x−5x−3\frac{2x-5}{x-3}
  3. 2x+5x+3\frac{2x+5}{x+3} (correct answer)
  4. 2x−5x+3\frac{2x-5}{x+3}
Explanation: To simplify the expression, factor the numerator and the denominator. The numerator, 2x2−x−152x^2 - x - 15, factors into (2x+5)(x−3)(2x+5)(x-3). The denominator, x2−9x^2 - 9, is a difference of squares and factors into (x−3)(x+3)(x-3)(x+3). The expression becomes (2x+5)(x−3)(x−3)(x+3)\frac{(2x+5)(x-3)}{(x-3)(x+3)}. The (x−3)(x-3) terms cancel, leaving 2x+5x+3\frac{2x+5}{x+3}.

Question 8

If 16x2−kx+916x^2 - kx + 9 is a perfect square trinomial, which of the following could be the value of kk?

  1. 12
  2. 24 (correct answer)
  3. 36
  4. 72
Explanation: A perfect square trinomial has the form a2−2ab+b2=(a−b)2a^2 - 2ab + b^2 = (a-b)^2 or a2+2ab+b2=(a+b)2a^2 + 2ab + b^2 = (a+b)^2. In 16x2−kx+916x^2 - kx + 9, we have a2=16x2a^2 = 16x^2, so a=4xa=4x, and b2=9b^2 = 9, so b=3b=3. The middle term, −kx-kx, must be equal to −2ab-2ab. Substituting the values for aa and bb, we get −2(4x)(3)=−24x-2(4x)(3) = -24x. Therefore, −kx=−24x-kx = -24x, which means k=24k=24.

Question 9

Which of the following is a factor of the expression x2−3xy−10y2x^2 - 3xy - 10y^2?

  1. x−2yx - 2y
  2. x+5yx + 5y
  3. x−5yx - 5y (correct answer)
  4. x−10yx - 10y
Explanation: To factor the trinomial x2−3xy−10y2x^2 - 3xy - 10y^2, we look for two numbers that multiply to -10 and add to -3. These numbers are -5 and 2. The factorization is (x−5y)(x+2y)(x - 5y)(x + 2y). Therefore, x−5yx - 5y is one of the factors.

Question 10

Three of the following trinomials can be factored over the integers. Which one CANNOT?

  1. 2x2+7x+32x^2 + 7x + 3
  2. x2−5x−14x^2 - 5x - 14
  3. 3x2−5x+23x^2 - 5x + 2
  4. 2x2+3x−42x^2 + 3x - 4 (correct answer)
Explanation: We test each option. A) 2x2+7x+3=(2x+1)(x+3)2x^2 + 7x + 3 = (2x+1)(x+3). B) x2−5x−14=(x−7)(x+2)x^2 - 5x - 14 = (x-7)(x+2). C) 3x2−5x+2=(3x−2)(x−1)3x^2 - 5x + 2 = (3x-2)(x-1). D) For 2x2+3x−42x^2 + 3x - 4, we need two integers that multiply to 2×−4=−82 \times -4 = -8 and sum to 3. The integer factor pairs of -8 are (1, -8), (-1, 8), (2, -4), and (-2, 4). None of these pairs sum to 3, so the trinomial cannot be factored over the integers.

Question 11

The trinomial x2−10x+kx^2 - 10x + k can be factored into the product of two distinct binomials with integer coefficients. Which of the following could be a value for kk?

  1. 20
  2. 21 (correct answer)
  3. 25
  4. 30
Explanation: If x2−10x+kx^2 - 10x + k factors into (x−a)(x−b)(x-a)(x-b), then a+b=10a+b=10 and ab=kab=k. Since the binomials must be distinct, a≠ba \neq b. We need to find pairs of distinct integers (a,b)(a, b) that sum to 10 and then find their product kk. The pairs are (1, 9), (2, 8), (3, 7), and (4, 6). Their products are 1×9=91\times9=9, 2×8=162\times8=16, 3×7=213\times7=21, and 4×6=244\times6=24. Of the choices, only 21 is a possible value for kk. The value 25 corresponds to the pair (5, 5), which would result in identical factors (x−5)(x−5)(x-5)(x-5), not distinct ones.

Question 12

A rectangular prism has a volume of 3x3+17x2+10x3x^3 + 17x^2 + 10x. If the height is xx, which of the following could represent the length and width of the prism's base?

  1. 3x+53x+5 and x+2x+2
  2. 3x+23x+2 and x+5x+5 (correct answer)
  3. 3x+103x+10 and x+1x+1
  4. 3x+13x+1 and x+10x+10
Explanation: The volume of a rectangular prism is V=length×width×heightV = \text{length} \times \text{width} \times \text{height}. We are given V=3x3+17x2+10xV = 3x^3 + 17x^2 + 10x and height =x= x. The area of the base is V/h=(3x3+17x2+10x)/x=3x2+17x+10V/h = (3x^3 + 17x^2 + 10x)/x = 3x^2 + 17x + 10. To find the length and width, we must factor this trinomial. We need two numbers that multiply to 3×10=303 \times 10 = 30 and add to 17. These numbers are 15 and 2. Rewriting the trinomial: 3x2+15x+2x+10=3x(x+5)+2(x+5)=(3x+2)(x+5)3x^2 + 15x + 2x + 10 = 3x(x+5) + 2(x+5) = (3x+2)(x+5). Thus, the length and width could be 3x+23x+2 and x+5x+5.

Question 13

If x−2x-2 is a factor of both x2+ax+10x^2+ax+10 and x2+bx−2x^2+bx-2, what is the value of a−ba-b?

  1. -10
  2. -8
  3. -6 (correct answer)
  4. 8
Explanation: If x−2x-2 is a factor of a polynomial, then x=2x=2 is a root. Substitute x=2x=2 into the first expression and set it to zero: (2)2+a(2)+10=0  ⟹  4+2a+10=0  ⟹  2a=−14  ⟹  a=−7(2)^2+a(2)+10 = 0 \implies 4+2a+10 = 0 \implies 2a = -14 \implies a = -7. Substitute x=2x=2 into the second expression and set it to zero: (2)2+b(2)−2=0  ⟹  4+2b−2=0  ⟹  2b=−2  ⟹  b=−1(2)^2+b(2)-2 = 0 \implies 4+2b-2 = 0 \implies 2b = -2 \implies b = -1. Therefore, a−b=−7−(−1)=−7+1=−6a-b = -7 - (-1) = -7+1 = -6.

Question 14

The trinomial 2x2+9x−52x^2 + 9x - 5 is factored into the form (ax+b)(cx+d)(ax+b)(cx+d). If a>0a > 0 and c>0c > 0, what is the value of the sum of the four coefficients, a+b+c+da+b+c+d?

  1. 5
  2. 7 (correct answer)
  3. 9
  4. 15
Explanation: First, factor the trinomial 2x2+9x−52x^2 + 9x - 5. We need two numbers that multiply to 2×−5=−102 \times -5 = -10 and add to 9. These numbers are 10 and -1. Rewriting the trinomial and factoring by grouping: 2x2+10x−x−5=2x(x+5)−1(x+5)=(2x−1)(x+5)2x^2 + 10x - x - 5 = 2x(x+5) - 1(x+5) = (2x-1)(x+5). The factors are (2x−1)(2x-1) and (x+5)(x+5). So, we can set a=2,b=−1,c=1,d=5a=2, b=-1, c=1, d=5. The sum of the coefficients is a+b+c+d=2+(−1)+1+5=7a+b+c+d = 2 + (-1) + 1 + 5 = 7.

Question 15

The trinomial x2+bx−18x^2 + bx - 18 can be factored into two binomials with integer coefficients. Which of the following is NOT a possible value for bb?

  1. -17
  2. -7
  3. 3
  4. 12 (correct answer)
Explanation: For the trinomial to be factorable, bb must be the sum of a pair of integer factors of -18. The factor pairs of -18 are (1, -18), (-1, 18), (2, -9), (-2, 9), (3, -6), and (-3, 6). The possible sums (values for bb) are -17, 17, -7, 7, -3, and 3. Of the given choices, -17, -7, and 3 are all possible values for bb. The value 12 is not a possible sum of any integer factor pair of -18.

Question 16

What is the sum of the solutions to the equation 2x2−8x=102x^2 - 8x = 10?

  1. -5
  2. -4
  3. 4 (correct answer)
  4. 5
Explanation: First, set the equation to zero: 2x2−8x−10=02x^2 - 8x - 10 = 0. Then, factor out the greatest common factor, which is 2: 2(x2−4x−5)=02(x^2 - 4x - 5) = 0. Next, factor the trinomial: 2(x−5)(x+1)=02(x-5)(x+1) = 0. The solutions are the values of xx that make the factors zero, which are x=5x=5 and x=−1x=-1. The sum of the solutions is 5+(−1)=45 + (-1) = 4.

Question 17

The expression (5−x)(x+2)(5-x)(x+2) is the factored form of which of the following trinomials?

  1. x2−3x−10x^2 - 3x - 10
  2. x2+3x−10x^2 + 3x - 10
  3. −x2−3x+10-x^2 - 3x + 10
  4. −x2+3x+10-x^2 + 3x + 10 (correct answer)
Explanation: To find the trinomial, expand the factored expression using the FOIL method: (5−x)(x+2)=(5)(x)+(5)(2)+(−x)(x)+(−x)(2)=5x+10−x2−2x(5-x)(x+2) = (5)(x) + (5)(2) + (-x)(x) + (-x)(2) = 5x + 10 - x^2 - 2x. Combining like terms gives −x2+(5x−2x)+10=−x2+3x+10-x^2 + (5x - 2x) + 10 = -x^2 + 3x + 10.

Question 18

The expression 3(x+2)2−5(x+2)−23(x+2)^2 - 5(x+2) - 2 can be factored into the form (ax+b)(cx+d)(ax+b)(cx+d). What is the value of a+ca+c?

  1. 3
  2. 4 (correct answer)
  3. 5
  4. 6
Explanation: Let u=x+2u = x+2. The expression becomes 3u2−5u−23u^2 - 5u - 2. We can factor this trinomial. We need two numbers that multiply to 3×−2=−63 \times -2 = -6 and add to -5. These numbers are -6 and 1. So, 3u2−6u+u−2=3u(u−2)+1(u−2)=(3u+1)(u−2)3u^2 - 6u + u - 2 = 3u(u-2) + 1(u-2) = (3u+1)(u-2). Now substitute back u=x+2u=x+2: (3(x+2)+1)((x+2)−2)=(3x+6+1)(x)=(3x+7)(x)(3(x+2)+1)((x+2)-2) = (3x+6+1)(x) = (3x+7)(x). The factors are xx and 3x+73x+7. We can write xx as 1x+01x+0. So the factored form is (3x+7)(1x+0)(3x+7)(1x+0). We have a=3,c=1a=3, c=1 (or vice versa). The sum a+c=3+1=4a+c = 3+1 = 4.

Question 19

When the expression 4x3+10x2−6x4x^3 + 10x^2 - 6x is factored completely, which of the following is one of its factors?

  1. x−3x-3
  2. 2x+12x+1
  3. 2x−12x-1 (correct answer)
  4. x−1x-1
Explanation: First, factor out the greatest common factor (GCF), which is 2x2x. This gives 2x(2x2+5x−3)2x(2x^2 + 5x - 3). Next, factor the trinomial 2x2+5x−32x^2 + 5x - 3. We look for two numbers that multiply to 2×−3=−62 \times -3 = -6 and add to 5. These numbers are 6 and -1. This leads to the factorization (2x−1)(x+3)(2x-1)(x+3). The complete factorization is 2x(2x−1)(x+3)2x(2x-1)(x+3). The factors are 2x2x, 2x−12x-1, and x+3x+3. Of the choices provided, 2x−12x-1 is a factor.

Question 20

If 2x2−8x+k2x^2 - 8x + k can be factored as (2x−a)(x−b)(2x - a)(x - b) where aa and bb are positive integers, what is the value of kk?

  1. 6 (correct answer)
  2. 8
  3. 12
  4. 16
Explanation: Expanding (2x−a)(x−b)=2x2−2bx−ax+ab=2x2−(2b+a)x+ab(2x - a)(x - b) = 2x^2 - 2bx - ax + ab = 2x^2 - (2b + a)x + ab. Comparing with 2x2−8x+k2x^2 - 8x + k, we need 2b+a=82b + a = 8 and k=abk = ab. Since aa and bb are positive integers, possible pairs are: (a,b)=(2,3),(4,2),(6,1)(a,b) = (2,3), (4,2), (6,1). Only (2,3)(2,3) gives 2(3)+2=82(3) + 2 = 8, so k=2⋅3=6k = 2 \cdot 3 = 6.