FACTORING IS THE REVERSE of multiplying. Skill in factoring, then, depends upon skill in multiplying: Lesson 16. As for a quadratic trinomial --
2x² + 9x − 5
-- it will be factored as a product of binomials:
(? ?)(? ?)
The first term of each binomial will be the factors of 2x², and the second term will be the factors of 5. Now, how can we produce 2x²? There is only one way: 2x· x :
(2x ?)(x ?)
And how can we produce 5? Again, there is only one way: 1· 5. But does the 5 go with 2x --
(2x 5)(x 1)
or with x --
(2x 1)(x 5) ?
Notice: We have not yet placed any signs
How shall we decide between these two possibilities? It is the combination that will correctly give the middle term, 9x :
2x² + 9x − 5.
Consider the first possibility:
(2x 5)(x 1)
Is it possible to produce 9x by combining the outers and the inners: 2x· 1 with 5· x ?
No, it is not. Therefore, we must eliminate that possibility and consider the other:
(2x 1)(x 5)
Can we produce 9x by combining 10x with x ?
Yes -- if we choose +5 and −1:
(2x − 1)(x + 5)
(2x − 1)(x + 5) = 2x² + 9x − 5.
Skill in factoring depends on skill in multiplying -- particularly in picking out the middle term
Problem 1. Place the correct signs to give the middle term.
a) 2x² + 7x − 15 = (2x − 3)(x + 5)
b) 2x² − 7x − 15 = (2x + 3)(x − 5)
c) 2x² − x − 15 = (2x + 5)(x − 3)
d) 2x² − 13x + 15 = (2x − 3)(x − 5)
Note: When the constant term is negative, as in parts a), b), c), then the signs in each factor will be different. But when the constant term is positive, as in part d), the signs will be the same. Usually, however, that happens by itself.
Nevertheless, can you correctly factor the following?
2x² − 5x + 3 = (2x − 3)(x − 1)
Problem 2. Factor these trinomials.
a) 3x² + 8x + 5 = (3x + 5)(x + 1)
b) 3x² + 16x + 5 = (3x + 1)(x + 5)
c) 2x² + 9x + 7 = (2x + 7)(x + 1)
d) 2x² + 15x + 7 = (2x + 1)(x + 7)
e) 5x² + 8x + 3 = (5x + 3)(x + 1)
f) 5x² + 16x + 3 = (5x + 1)(x + 3)
Problem 3. Factor these trinomials.
a) 2x² − 7x + 5 = (2x − 5)(x − 1)
b) 2x² − 11x + 5 = (2x − 1)(x − 5)
c) 3x² + x − 10 = (3x − 5)(x + 2 )
d) 2x² − x − 3 = (2x − 3)(x + 1)
e) 5x² − 13x + 6 = (5x − 3)(x − 2)
f) 5x² − 17x + 6 = (5x − 2)(x − 3)
g) 2x² + 5x − 3 = (2x − 1)(x + 3)
h) 2x² − 5x − 3 = (2x + 1)(x − 3)
i) 2x² + x − 3 = (2x + 3)(x − 1)
j) 2x² − 13x + 21 = (2x − 7 )(x −3)
k) 5x² − 7x − 6 = (5x + 3)(x − 2)
i) 5x² − 22x + 21 = (5x − 7)(x − 3)
Example 1. 1 the coefficient of x². Factor x² + 3x − 10.
Solution. The binomial factors will have this form:
(x a)(x b)
What are the factors of 10? Let us hope that they are 2 and 5:
x² + 3x − 10 = (x 2)(x 5).
We must now choose the signs so that the coefficient of the middle term -- the sum of the outers plus the inners -- will be +3. Choose +5 and −2.
x² + 3x − 10 = (x − 2)(x + 5).
Note: When 1 is the coefficient of x², the order of the factors does not matter.
(x − 2)(x + 5) = (x + 5) (x − 2).
Example 2. Factor x² − x − 12.
Solution. We must find factors of 12 whose algebraic sum will be the coefficient of x : −1. Choose −4 and + 3:
x² − x − 12 = (x − 4 )(x + 3).
2x² + 9x − 5
-- it will be factored as a product of binomials:
(? ?)(? ?)
The first term of each binomial will be the factors of 2x², and the second term will be the factors of 5. Now, how can we produce 2x²? There is only one way: 2x· x :
(2x ?)(x ?)
And how can we produce 5? Again, there is only one way: 1· 5. But does the 5 go with 2x --
(2x 5)(x 1)
or with x --
(2x 1)(x 5) ?
Notice: We have not yet placed any signs
How shall we decide between these two possibilities? It is the combination that will correctly give the middle term, 9x :
2x² + 9x − 5.
Consider the first possibility:
(2x 5)(x 1)
Is it possible to produce 9x by combining the outers and the inners: 2x· 1 with 5· x ?
No, it is not. Therefore, we must eliminate that possibility and consider the other:
(2x 1)(x 5)
Can we produce 9x by combining 10x with x ?
Yes -- if we choose +5 and −1:
(2x − 1)(x + 5)
(2x − 1)(x + 5) = 2x² + 9x − 5.
Skill in factoring depends on skill in multiplying -- particularly in picking out the middle term
Problem 1. Place the correct signs to give the middle term.
a) 2x² + 7x − 15 = (2x − 3)(x + 5)
b) 2x² − 7x − 15 = (2x + 3)(x − 5)
c) 2x² − x − 15 = (2x + 5)(x − 3)
d) 2x² − 13x + 15 = (2x − 3)(x − 5)
Note: When the constant term is negative, as in parts a), b), c), then the signs in each factor will be different. But when the constant term is positive, as in part d), the signs will be the same. Usually, however, that happens by itself.
Nevertheless, can you correctly factor the following?
2x² − 5x + 3 = (2x − 3)(x − 1)
Problem 2. Factor these trinomials.
a) 3x² + 8x + 5 = (3x + 5)(x + 1)
b) 3x² + 16x + 5 = (3x + 1)(x + 5)
c) 2x² + 9x + 7 = (2x + 7)(x + 1)
d) 2x² + 15x + 7 = (2x + 1)(x + 7)
e) 5x² + 8x + 3 = (5x + 3)(x + 1)
f) 5x² + 16x + 3 = (5x + 1)(x + 3)
Problem 3. Factor these trinomials.
a) 2x² − 7x + 5 = (2x − 5)(x − 1)
b) 2x² − 11x + 5 = (2x − 1)(x − 5)
c) 3x² + x − 10 = (3x − 5)(x + 2 )
d) 2x² − x − 3 = (2x − 3)(x + 1)
e) 5x² − 13x + 6 = (5x − 3)(x − 2)
f) 5x² − 17x + 6 = (5x − 2)(x − 3)
g) 2x² + 5x − 3 = (2x − 1)(x + 3)
h) 2x² − 5x − 3 = (2x + 1)(x − 3)
i) 2x² + x − 3 = (2x + 3)(x − 1)
j) 2x² − 13x + 21 = (2x − 7 )(x −3)
k) 5x² − 7x − 6 = (5x + 3)(x − 2)
i) 5x² − 22x + 21 = (5x − 7)(x − 3)
Example 1. 1 the coefficient of x². Factor x² + 3x − 10.
Solution. The binomial factors will have this form:
(x a)(x b)
What are the factors of 10? Let us hope that they are 2 and 5:
x² + 3x − 10 = (x 2)(x 5).
We must now choose the signs so that the coefficient of the middle term -- the sum of the outers plus the inners -- will be +3. Choose +5 and −2.
x² + 3x − 10 = (x − 2)(x + 5).
Note: When 1 is the coefficient of x², the order of the factors does not matter.
(x − 2)(x + 5) = (x + 5) (x − 2).
Example 2. Factor x² − x − 12.
Solution. We must find factors of 12 whose algebraic sum will be the coefficient of x : −1. Choose −4 and + 3:
x² − x − 12 = (x − 4 )(x + 3).