# How do you factor 16j^3 + 44j^2 - 126j= 0?

May 25, 2015

First, you factor out what's multiplying all your terms, $2 j$.

$2 j \left(8 {j}^{2} + 22 j - 63\right)$

Now, we factor the parenthesis by finding its roots and turning them into factors by equaling them to zero.

$\frac{- 22 \pm \sqrt{484 - 4 \left(8\right) \left(- 63\right)}}{16}$
$\frac{- 22 \pm 50}{16}$
${j}_{1} = - \frac{9}{2}$, which is the same as the factor $2 j + 9 = 0$
${j}_{2} = \frac{7}{4}$, which is the same as the factor $4 j - 7 = 0$

Thus,

$16 {j}^{3} + 44 {j}^{2} - 126 j = 2 j \left(2 j + 9\right) \left(4 j - 7\right)$

May 26, 2015

$f \left(x\right) = 2 \left(8 {x}^{2} + 22 x - 63\right) = 2 \left(x - p\right) \left(x - q\right) .$

I use the new AC method to factor trinomials (see Google ,Yahoo Search)

Converted trinomial: ${x}^{2} + 22 x - 63 = \left(x - p '\right) \left(x - q '\right) .$

Compose factor pairs of a.c = -504-->(-12, 42)(-14, 36). This sum is 22 = b. Then p' = -14, and q' = 36.

Then $p = \frac{p '}{a} = - \frac{14}{8} = - \frac{7}{4}$ and $q = \frac{q '}{a} = \frac{36}{8} = \frac{9}{2.}$

Factored form: $f \left(x\right) = 2 \left(x - \frac{7}{4}\right) \left(x + \frac{9}{2}\right) = 2 \left(4 x - 7\right) \left(2 x + 9\right)$