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

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

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Answer 1 · 2015-05-25T22:35:52

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

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

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

$\frac{- 22 \pm \sqrt{484 - 4 (8) (- 63)}}{16}$ $(-22+-sqrt(484-4(8)(-63)))/16$
$\frac{- 22 \pm 50}{16}$ $(-22+-50)/16$
$j_{1} = - \frac{9}{2}$ $j_1=-9/2$ , which is the same as the factor $2 j + 9 = 0$ $2j+9=0$
$j_{2} = \frac{7}{4}$ $j_2=7/4$ , which is the same as the factor $4 j - 7 = 0$ $4j-7=0$

Thus,

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

Answer 2 · 2015-05-26T03:55:38

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

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

Converted trinomial: $x^2 + 22x - 63 = (x - p')(x - q').$

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

Then $p = (p')/a = -14/8 = -7/4$ and $q = (q')/a = 36/8 = 9/2.$

Factored form: $f(x) = 2(x - 7/4)(x + 9/2) = 2(4x - 7)(2x + 9)$

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

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How do you factor 16j^3 + 44j^2 - 126j= 016j3+44j2−126j=016j^3 + 44j^2 - 126j= 0?

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