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Quadratic Factoring Practice Questions

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Updated on Oct 3, 2011

To review these concepts, go to Quadratic Factoring Study Guide.

Quadratic Factoring Practice Questions

Problems

Practice 1

Factor each expression.

  1. x2 + 12x + 35
  2. x2 – 6x + 8
  3. x2x – 12
  4. 2x2 + 3x + 1
  5. 3x2 + 8x – 11

Practice 2

Factor each expression.

  1. x2 – 16
  2. 9x2 – 81
  3. 49x4 – 100

Solutions

Practice 1

1. To factor x2 + 12x + 35, begin by listing the positive and negative factors of the first and last terms:
x2: –x, x
35: –35, –7, –5, –1, 1, 5, 7, 35
x2 is the square of either x or –x. Begin by trying x as the first term in each binomial:
(x + __)(x + __)
The coefficients of each x term are 1. The last terms of each binomial must multiply to 35 and add to 12, since 12x is the sum of the products of the Outside and Inside terms. 12x and 35 are positive, so we are looking for two positive numbers that multiply to 35 and add to 12.
(1)(35) = 35, but 1 + 35 = 36
(5)(7) = 35, and 5 + 7 = 12
The constant of one binomial is 5 and the constant of the other binomial is 7:
(x + 5)(x + 7)
Check the answer using FOIL:
First: (x)(x) = x2
Outside: (x)(7) = 7x
Inside: (5)(x) = 5x
Last: (5)(7) = 35
x2 + 7x + 5x + 35 = x2 + 12x + 35
2. To factor x2 – 6x + 8, begin by listing the positive and negative factors of the first and last terms:
x2: –x, x
8: –8, –4, –2, –1, 1, 2, 4, 8
x2 is the square of either x or –x. Begin by trying x as the first term in each binomial:
(x + __)(x + __)
The coefficients of each x term are 1. The last terms of each binomial must multiply to 8 and add to –6, since –6x is the sum of the products of the Outside and Inside terms. –6x is negative and 8 is positive, so we are looking for two negative numbers that multiply to 8 and add to –6.
(–1)(–8) = 8, but –1 + (–8) = –9
(–2)(–4) = 8, and –2 + (–4) = –6
The constant of one binomial is –2 and the constant of the other binomial is –4:
(x – 2)(x – 4)
Check the answer using FOIL:
First: (x)(x) = x2
Outside: (x)(–4) = –4x
Inside: (–2)(x) = –2x
Last: (–2)(–4) = 8
x2 – 4x – 2x + 8 = x2 – 6x + 8
3. To factor x2x – 12, begin by listing the positive and negative factors of the first and last terms:
x2: –x, x
12: –12, –6, –4, –3, –2, –1, 1, 2, 3, 4, 6, 12
x2 is the square of either x or –x. Begin by trying x as the first term in each binomial:
(x + __)(x + __)
The coefficients of each x term are 1. The last terms of each binomial must multiply to –12 and add to –1, because –x is the sum of the products of the Outside and Inside terms. –x is negative, as is –12, so we are looking for one positive number and one negative number that multiply to –12 and add to –1.
(–12)(1) = –12, but –12 + 1 = –11
(12)(–1) = –12, but 12 + (–1) = 11
(–6)(2) = –12, but –6 + 2 = –4
(6)(–2) = –12, but 6 + (–2) = 4
(4)(–3) = –12, but 4 + (–3) = 1
(–4)(3) = –12, and –4 + 3 = –1
The constant of one binomial is –4 and the constant of the other binomial is 3:
(x – 4)(x + 3)
Check the answer using FOIL:
First: (x)(x) = x2
Outside: (x)(3) = 3x
Inside: (–4)(x) = –4x
Last: (–4)(3) = –12
x2 + 3x – 4x – 12 = x2x – 12
4. To factor 2x2 + 3x + 1, begin by listing the positive and negative factors of the first and last terms:
2x2: –2, –1, 1, 2, –x, x
1: –1, 1
2x2 could be (2x)(x) or (–2x)(–x). Begin by trying 2x as the first term in one binomial and x as the first term in the other binomial:
(2x + __)(x + __)
The coefficient of one x term is 2 and the coefficient of the other x term is 1. The last terms of each binomial must multiply to 1. One of those terms will be multiplied by 2x, and the other will be multiplied by 1x. The sum of those products will add to 3x. 3x and 1 are positive, so we are looking for two positive numbers that multiply to 1. The only possibility is 1 times 1, so the constant of each binomial is 1:
(2x + 1)(x + 1)
Check the answer using FOIL:
First: (2x)(x) = 2x2
Outside: (2x)(1) = 2x
Inside: (1)(x) = x
Last: (1)(1) = 1
2x2 + 2x + x + 1 = 2x2 + 3x + 1
5. To factor 3x2 + 8x – 11, begin by listing the positive and negative factors of the first and last terms:
3x2: –3, –1, 1, 3, –x, x
11: –11, –1, 1, 11
3x2 could be (3x)(x) or (–3x)(–x). Begin by trying 3x as the first term in one binomial and x as the first term in the other binomial:
(3x + __)(x + __)
The coefficient of one x term is 3, and the coefficient of the other x term is 1. The last terms of each binomial must multiply to –11. One of those terms will be multiplied by 3x and the other will be multiplied by 1x. The sum of those products will add to 8x. 8x is positive, but –11 is negative, so we are looking for a positive number and a negative number that multiply to –11. The possibilities are (–1)(11) and (1)(–11):
(3x – 11)(x + 1)
Check the answer using FOIL:
First: (3x)(x) = 3x2
Outside: (3x)(1) = 3x
Inside: (–11)(x) = –11x
Last: (–11)(1) = –11
3x2 + 3x – 11x – 11 = 3x2 – 8x – 11, not 3x2 + 8x – 11
Try changing the signs of the last terms of each binomial:
(3x + 11)(x – 1)
Check the answer using FOIL:
First: (3x)(x) = 3x2
Outside: (3x)(–1) = –3x
Inside: (11)(x) = 11x
Last: (11)(–1) = –11
3x2 – 3x + 11x – 11 = 3x2 + 8x – 11
Therefore, 3x2 + 8x – 11 factors to (3x + 11)(x – 1).
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