Question

How do you factor `9x ^ { 2} + 21x - 18`?

Answer 1

To factor `9x2+21x−18`

Factor out the `3`: `3(3x^2+7x-6)`
Continue factoring: `3(3x-2)(x+3)`

Answer 2

`3(3x - 2)(x + 3)`

Explanation:

`f(x) = 9x^2 + 21x - 18 = 3(3x^2 + 7x - 6)`

Factor the trinomial in parentheses by the new AC Method (Socratic, Google Search)

`y = 3x^2 + 7x - 6 = 3(x + p)(x + q)`

Factor converted trinomial:
`y' = x^2 + 7x - 18 =` (x + p')(x + q').

Find `p' and q'`

knowing the sum `(b = 7)` and the product `(ac = - 18).`

They are: `p' = -2 and q' = 9 rarr ["sum " = 7 and " product " = - 18]`

Back to y:

`p = (p')/a = -2/3` , and `q = (q')/a = 9/3 = 3`

Factored form of `f(x):`

`f(x) = 3(3)(x - 2/3)(x + 3) = 3(3x - 2)(x + 3)`