Question

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

Answer 1

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

`2j(8j^2+22j-63)`

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

`(-22+-sqrt(484-4(8)(-63)))/16`
`(-22+-50)/16`
`j_1=-9/2`, which is the same as the factor `2j+9=0`
`j_2=7/4`, which is the same as the factor `4j-7=0`

Thus,

`16j^3+44j^2-126j=2j(2j+9)(4j-7)`

Answer 2

`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)`