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

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Answer 1 · 2017-07-21T02:29:31

To factor $9 x 2 + 21 x - 18$ $9x2+21x−18$

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

Answer 2 · 2017-09-10T05:15:05

$3 (3 x - 2) (x + 3)$ $3(3x - 2)(x + 3)$

$f (x) = 9 x^{2} + 21 x - 18 = 3 (3 x^{2} + 7 x - 6)$ $f(x) = 9x^2 + 21x - 18 = 3(3x^2 + 7x - 6)$

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

$y = 3 x^{2} + 7 x - 6 = 3 (x + p) (x + q)$ $y = 3x^2 + 7x - 6 = 3(x + p)(x + q)$

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 $f(x):$

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

How do you factor 9x ^ { 2} + 21x - 189x2+21x−189x ^ { 2} + 21x - 18?