By LearningExpress Editors
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.
 x^{2} + 12x + 35
 x^{2} – 6x + 8
 x^{2} – x – 12
 2x^{2} + 3x + 1
 3x^{2} + 8x – 11
Practice 2
Factor each expression.
 x^{2} – 16
 9x^{2} – 81
 49x^{4} – 100
Solutions
Practice 1
1.  To factor x^{2} + 12x + 35, begin by listing the positive and negative factors of the first and last terms: 
x^{2}: –x, x  
35: –35, –7, –5, –1, 1, 5, 7, 35  
x^{2} 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) = x^{2}  
Outside: (x)(7) = 7x  
Inside: (5)(x) = 5x  
Last: (5)(7) = 35  
x^{2} + 7x + 5x + 35 = x^{2} + 12x + 35  
2.  To factor x^{2} – 6x + 8, begin by listing the positive and negative factors of the first and last terms: 
x^{2}: –x, x  
8: –8, –4, –2, –1, 1, 2, 4, 8  
x^{2} 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) = x^{2}  
Outside: (x)(–4) = –4x  
Inside: (–2)(x) = –2x  
Last: (–2)(–4) = 8  
x^{2} – 4x – 2x + 8 = x^{2} – 6x + 8  
3.  To factor x^{2} – x – 12, begin by listing the positive and negative factors of the first and last terms: 
x^{2}: –x, x  
12: –12, –6, –4, –3, –2, –1, 1, 2, 3, 4, 6, 12  
x^{2} 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) = x^{2}  
Outside: (x)(3) = 3x  
Inside: (–4)(x) = –4x  
Last: (–4)(3) = –12  
x^{2} + 3x – 4x – 12 = x^{2} – x – 12  
4.  To factor 2x^{2} + 3x + 1, begin by listing the positive and negative factors of the first and last terms: 
2x^{2}: –2, –1, 1, 2, –x, x  
1: –1, 1  
2x^{2} 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) = 2x^{2}  
Outside: (2x)(1) = 2x  
Inside: (1)(x) = x  
Last: (1)(1) = 1  
2x^{2} + 2x + x + 1 = 2x^{2} + 3x + 1  
5.  To factor 3x^{2} + 8x – 11, begin by listing the positive and negative factors of the first and last terms: 
3x^{2}: –3, –1, 1, 3, –x, x  
11: –11, –1, 1, 11  
3x^{2} 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) = 3x^{2}  
Outside: (3x)(1) = 3x  
Inside: (–11)(x) = –11x  
Last: (–11)(1) = –11  
3x^{2} + 3x – 11x – 11 = 3x^{2} – 8x – 11, not 3x^{2} + 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) = 3x^{2}  
Outside: (3x)(–1) = –3x  
Inside: (11)(x) = 11x  
Last: (11)(–1) = –11  
3x^{2} – 3x + 11x – 11 = 3x^{2} + 8x – 11  
Therefore, 3x^{2} + 8x – 11 factors to (3x + 11)(x – 1). 

1
 2
View Full Article
From Algebra in 15 Minutues A Day. Copyright © 2009 by LearningExpress, LLC. All Rights Reserved.
Post a Comment
 No comments so far
Ask a Question
Have questions about this article or topic? Ask150 Characters allowed
Related Questions
See More QuestionsPopular Articles
Wondering what others found interesting? Check out our most popular articles.
 Kindergarten Sight Words List
 First Grade Sight Words List
 10 Fun Activities for Children with Autism
 Definitions of Social Studies
 Grammar Lesson: Complete and Simple Predicates
 Child Development Theories
 Signs Your Child Might Have Asperger's Syndrome
 How to Practice Preschool Letter and Name Writing
 Social Cognitive Theory
 Theories of Learning