Chapter 6 FaCtoring and the distributive ProPerty 163EXAMPLE
Factor the quadratic polynomial.
4 x^2 – 4x – 15
The possibilities that give us 41 x^2 _____− 5 are
(a) (4x + 15)(x − 1)
(b) (4x − 15)(x + 1)
(c) (4x − 1)(x + 15)
(d) (4x + 1)(x − 15)
(e) (4x + 5)(x − 3)
(f ) (4x − 5)(x + 3)
(g) (4x + 3)(x − 5)
(h) (4x − 3)(x + 5)
(i) (2x + 15)(2x − 1)
(j) (2x − 15)(2x + 1)
(k) (2x + 5)(2x − 3)
(l) (2x − 5)(2x + 3)
We have chosen these combinations to force the first and last terms of the
quadratic to be 4x^2 and –15, respectively. We only need to check the com-
bination that will give a middle term of –4x (if there is one).
(a) −4x + 15x = 11x
(b) 4x −15x = −11x
(c) 60x − x = 59x
(d) −60x + x = −59x
(e) −12x + 5x = −7x
(f ) 12x − 5x = 7x
(g) −20x + 3x = −17x
(h) 20x − 3x = 17x
(i) −2x + 30x = 28x
(j) 2x − 30x = −28x
(k) −6x + 10x = 4x
(l) 6x − 10x = −4x
Because the sum of the middle terms is −4x, Combination (l) is the correct
factorization: 4x^2 − 4x − 15 = (2x − 5)(2x + 3).
EXAMPLE
Factor the quadratic polynomial.