Question

Which correctly factored form of the function


`f(x) = 36 {x}^(2) + 15x - 6`
can be used to identify the zeros?
A.

`f(x) = (4x - 1)(3x + 2)`
B.

`f(x) = (12x - 2)(3x + 3)`
C.

`f(x) = 3(4x - 1)(3x + 2)`
D.

`f(x) = 3(12x - 2)(3x + 3)`
​

Answer 1

ƒ(x) = 3(4x - 1)(3x + 2)

Explanation:

Your function is: ƒ(x) = 36x² + 15x - 6

1. Remove the common factor

36x² + 15x - 6 = 3(12x² + 5x - 2)

2. Factor the quadratic

(a) Multiply the leading coefficient and the constant

12 × (-2) = -24

(b) Find two numbers that multiply to give -24 and add to give 5.

Possible pairs are 1, 24; 2, 12; 3, 8; 4, 6

One of the numbers must be negative. Start with the numbers near the end of the list.

By trial and error, you will find that 8 and -3 work:

-3 × 8 = -24 and -3 + 8 = 5

(b) Rewrite 5x as -3x + 8x

12x² - 3x + 8x - 2

(c) Factor by grouping the first two and the last two terms

(3x)(4x - 1) + 2(4x - 1) = (4x + 1)(3x + 2)

ƒ(x) = 3(4x -1)(3x + 2)

This is the correctly factored form that you can use to find the zeros.