Introduction
This set of practice questions will present algebraic expressions for you to factor. You can use three different techniques to factor polynomials. In the first technique, you look for common factors in the terms of the polynomial. In the second, you will factor polynomials that are the difference of two perfect squares. The third technique, called the trinomial factor method, will allow you to factor algebraic expressions that have three terms. The trinomial expressions in this chapter will be in the form of x^{2} ± ax ± b, where a and b are whole numbers. The problems will be presented in random order to give you practice at recognizing which method or combination of methods will be required to factor the polynomial. Complete explanations of the solutions will follow.
Tips for Factoring Polynomials
Factoring using the greatest common factor: Look for a factor common to every term in the polynomial. Put that factor outside a set of parentheses and the polynomial inside with the factor removed from each term, e.g. 2x^{2} + 8 = 2(x^{2} + 4)
Factoring using the difference of two perfect squares: Polynomials in the form x^{2} – y^{2} can be factored into two terms: (x + y)(x – y).
Factoring using the trinomial method: This method requires you to factor the first and third terms and put the factors into the following factored form: ([ ] ± [ ])([ ] ± [ ]). The factors of the first term go in the first position in the parentheses and the factors of the third term go in the second position in each factor, e.g. x^{2} + 2x + 1 = (x + 1)(x + 1).
Practice Questions
Factor the following polynomials.
- 9a + 15
- 3a^{2}x + 9ax
- x^{2} – 16
- 4a^{2} – 25
- 7n^{2} – 28n
- 7x4y^{2} – 35x^{2}y^{2} + 14x^{2}y^{4}
- x^{2} + 3x + 2
- 9r^{2} – 49
- x^{2} – 2x – 8
- x^{2} + 5x + 6
- x^{2} + x – 6
- b^{2} – 100
- x^{2} + 7x + 12
- x^{2} – 3x – 18
- b^{2} – 6b + 8
- b^{2} – 4b – 21
- a^{2} + 11a – 12
- x^{2} + 10x + 25
- 36y^{4} – 4z^{2}
- x^{2} + 20x + 99
- c^{2} – 12c + 32
- h^{2} – 12h + 11
- m^{2} – 11m + 18
- v^{4} – 13v^{2} – 48
- x^{2} – 20x + 36
Answers
Numerical expressions in parentheses like this [ ] are operations performed on only part of the original expression. The operations performed within these symbols are intended to show how to evaluate the various terms that make up the entire expression.
Expressions with parentheses that look like this ( ) contain either numerical substitutions or expressions that are part of a numerical expression. Once a single number appears within these parentheses, the parentheses are no longer needed and need not be used the next time the entire expression is written.
When two pair of parentheses appear side by side like this ( )( ), it means that the expressions within are to be multiplied.
Sometimes parentheses appear within other parentheses in numerical or algebraic expressions. Regardless of what symbol is used, ( ), { }, or [ ], perform operations in the innermost parentheses first and work outward.
Underlined expressions show the original algebraic expression as an equation with the expression equal to its simplified result.
1. The terms have a common factor of 3. Factor 3 out of each term and write the expression in factored form. | |
9a + 15 = 3(3a + 5) | |
2. The terms have a common factor of 3ax. Factor 3ax out of each term and write the expression in factored form. | |
3. Both terms in the polynomial are perfect squares. Use the form for factoring the difference of two perfect squares and put the roots of each factor in the proper place. | |
x^{2} – 16 = (x + 4)(x – 4) | |
Check using FOIL. | |
4. Both terms in the polynomial are perfect squares. 4a^{2} = (2a)^{2}, and 25 = 5^{2}. Use the form for factoring the difference of two perfect squares and put the roots of each factor in the proper place. | |
4a^{2} – 25 = (2a + 5)(2a – 5) | |
Check using FOIL. | |
5. The terms have a common factor of 7n. Factor 7n out of each term and write the expression in factored form. | |
6. The terms have a common factor of 7x^{2}y^{2}. Factor 7x^{2}y^{2} out of each term in the expression and write it in factored form. | |
7. This expression can be factored using the trinomial method. The factors of x2 are x and x, and the factors of 2 are 1 and 2. Place the factors into the trinomial factor form and check using FOIL. | |
The factors are correct. | |
8. Both terms in the polynomial are perfect squares. 9r^{2} = (3r)^{2} and 49 = 7^{2}. Use the form for factoring the difference of two perfect squares and put the roots of each factor in the proper place. | |
9. This expression can be factored using the trinomial method. The factors of x^{2} are x and x, and the factors of 8 are (1)(8) and (2)(4). You want the result of the O and I of the FOIL method for multiplying factors to add up to ^{–}2x. Only terms with opposite signs will result in a negative numerical term, which is what you need, since the third term is ^{–}8. Place the factors (2)(4) into the trinomial factor form and check using FOIL. | |
(x + 4)(x – 2) = x^{2} – 2x + 4x – 8 = x^{2} + 2x – 8 | |
Almost correct! Change the position of the factors of the numerical term and check using FOIL. | |
The factors of the trinomial are now correct. | |
10. This expression can be factored using the trinomial method. The factors of x^{2} are x and x, and the factors of 6 are (1)(6) and (2)(3). Since the numerical term of the polynomial is positive, the signs in the factor form for trinomials will be the same because only two like signs multiplied together will result in a positive. Now consider the second term in the trinomial. In order to add up to 5x, the result of multiplying the Inner and Outer terms of the trinomial factors will have to be positive. Try using two positive signs and the factors 2 and 3, which add up to 5. Check using FOIL. | |
11. This expression can be factored using the trinomial method. The factors of x^{2} are x and x, and the factors of 6 are (1)(6) and (2)(3). You want the result of the O and I of the FOIL method for multiplying factors to add up to ^{+}1x. Only terms with opposite signs will result in a negative numerical term that you need with the third term being a ^{–}6. Place the factors (2)(3) into the trinomial factor form and check using FOIL. | |
The factors of the trinomial are now correct. | |
12. Both terms in the polynomial are perfect squares. b^{2} = (b)^{2} and 100 = 10^{2}. Use the form for factoring the difference of two perfect squares and put the roots of each factor in the proper place. | |
b^{2} – 100 = (b + 10)(b – 10) | |
Check using FOIL. | |
13. This expression can be factored using the trinomial method. The factors of x^{2} are x and x, and the factors of 12 are (1)(12) or (2)(6) or (3)(4). You want the result of the O and I of the FOIL method for multiplying factors to add up to ^{+}7x. The factors (3)(4) would give terms that add up to 7. Since all signs are positive, use positive signs in the factored form for the trinomial. | |
The result is correct. This is not just luck. You can use logical guesses to find the correct combination of factors and signs. | |
14. This expression can be factored using the trinomial method. The factors of x^{2} are x and x, and the factors of 18 are (1)(18) or (2)(9) or (3)(6). Only the product of a positive and a negative numerical term will result in ^{–}18. The sum of the results of multiplying the Outer and Inner terms of the trinomial factors needs to add up to a ^{–}3x. So use (3)(6) in the trinomial factors form and check using FOIL. | |
15. This expression can be factored using the trinomial method. The factors of b^{2} are b and b, and the factors of 8 are (1)(8) or (2)(4). You want the result of the O and I of the FOIL method for multiplying factors to add up to ^{–}6b. The signs within the parentheses of the factorization of the trinomial must be the same to result in a positive numerical term in the trinomial. The middle term has a negative sign, so let's try two negative signs. How can you get 6 from adding the two of the factors of 8? Right! Use the (2)(4). | |
Check the answer using FOIL. | |
16. This expression can be factored using the trinomial method. The factors of b^{2} are b and b, and the factors of 21 are (1)(21) or (3)(7). You want the result of the O and I products of the FOIL method for multiplying factors to add up to –4b. Only the product of a positive and a negative numerical term will result in ^{–}21. So let's use (3) and (7) in the trinomial factors form because the difference between 3 and 7 is 4. | |
Check using FOIL. | |
17. This expression can be factored using the trinomial method. The factors of a^{2} are a and a, and the factors of 12 are (1)(12) or (2)(6) or (3)(4). You want the result of the O and I of the FOIL method for multiplying factors to add up to ^{+}11a. Only the product of a positive and a negative numerical term will result in ^{–}12. Since the signs in the factors must be one positive and one negative, use the factors 12 and 1 in the trinomial factors form. | |
Use FOIL to check the answer. | |
18. This expression can be factored using the trinomial method. The factors of x^{2} are x and x, and the factors of 25 are (1)(25) or (5)(5). To get a positive 25 after multiplying the factors of the trinomial expression, the signs in the two factors must both be positive or both be negative. The sum of the results of multiplying the Outer and Inner terms of the trinomial factors needs to add up to a ^{+}10x. So let's use (5)(5) in the trinomial factors form and check using FOIL. | |
19. Both terms in the polynomial are perfect squares. | |
36y^{2} = (6y^{2})^{2} and 4z^{2} = (2z)^{2} | |
Use the form for factoring the difference of two perfect squares and put the roots of each factor in the proper place. | |
(6y^{2} + 2z)(6y^{2} – 2z) | |
Check using FOIL. | |
20. This expression can be factored using the trinomial method. The factors of x^{2} are x and x, and the factors of 99 are (1)(99) or (3)(33) or (9)(11). To get a positive 99 after multiplying the factors of the trinomial expression, the signs in the two factors must both be positive or both are negative. The sum of the results of multiplying the Outer and Inner terms of the trinomial factors needs to add up to a ^{+}20x. So let's use (9)(11) in the trinomial factors form because 9 + 11 = 20. Check using FOIL. | |
21. This expression can be factored using the trinomial method. The factors of x^{2} are x and x, and the factors of 32 are (1)(32) or (2)(16) or (4)(8). The sign of the numerical term is positive, so the signs in the factors of our trinomial factorization must be the same. The sign of the first-degree term (the variable to the power of 1) is negative. This leads one to believe that the signs in the trinomial factors will both be negative. The only factors of 32 that add up to 12 are 4 and 8. Check using FOIL. | |
22. The factors of h^{2} are h and h, and the factors of 11 are (1)(11). The sign of the numerical term is positive, so the signs in the factors of our trinomial factorization must be the same. The sign of the first-degree term (the variable to the power of 1) is negative. So use negative signs in the trinomial factors. Check your answer. | |
23. This expression can be factored using the trinomial method. The factors of m^{2} are m and m, and the factors of 18 are (1)(18) or (2)(9) or (3)(6). The sign of the numerical term is positive, so the signs in the factors of our trinomial factorization must be the same. The sum of the results of multiplying the Outer and Inner terms of the trinomial factors needs to add up to –11m. The only factors of 18 that can be added or subtracted in any way to equal 11 are 2 and 9. Use them and two subtraction signs in the trinomial factor terms. Check your answer using FOIL. | |
24. This expression can be factored using the trinomial method. The factors of v^{4} are (v^{2})(v^{2}), and the factors of 48 are (1)(48) or (2)(24) or (3)(16) or (4)(12) or (6)(8). Only the product of a positive and a negative numerical term will result in ^{–}48. The only factors of 48 that can be added or subtracted in any way to equal 13 are 3 and 16. Use 3 and 16 and a positive and negative sign in the terms of the trinomial factors. Check your answer using FOIL. | |
(v^{2} + 3)(v^{2} – 16) = v^{4} – 16v^{2} + 3v^{2} – 48 = v^{4} – 13v – 48 | |
You may notice that one of the two factors of the trinomial expression can itself be factored. The second term is the difference of two perfect squares. Factor (v^{2} – 16) using the form for factoring the difference of two perfect squares. | |
(v + 4)(v – 4) = v^{2} – 4v + 4v – 16 = v^{2} – 16 | |
This now makes the complete factorization of | |
25. The factors of x^{2} are x and x, and the factors of 36 are (1)(36) or (2)(18) or (4)(9) or (6)(6). The sign of the numerical term is positive, so the signs in the factors of our trinomial factorization must be the same. The sign of the first-degree term (the variable to the power of 1) is negative. This leads one to believe that the signs in the trinomial factors will both be negative. The only factors of 36 that add up to 20 are 2 and 18. Use them and two negative signs in the trinomial factor form. Check your answer using FOIL. | |
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