Question

How do you factor the trinomial `3x^3 + 4x = -8x^2`?

Answer 1

`x(3x-2)(x-2)=0`

Explanation:

Move everything onto one side.

`3x^3+8x^2+4x=0`

Factor out the common term `x`.

`xoverbrace((3x^2-8x+4))^("factor this individually")=0`

In order to factor `3x^2-8x+6`, we need to look for terms that multiply to be the product of the first and last coefficients `(12)` and adds to be the coefficient of the middle term `(-8)`. These numbers are `-2` and `-6`.

`3x^2-8x+4=3x^2-6x-2x+4`
`=3x(x-2)-2(x-2)`
`=color(blue)((3x-2)(x-2)`

We can plug this back in to find the fully factored trinomial:

`x(3x-2)(x-2)=0`

You didn't ask to solve for `x`, but if we wanted to, we would set each multiplied term equal to `0` to find that `x=0,2/3,2`.

Answer 2

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

Explanation:

Given `3x^3 + 4x =-8x^2`
Add `8X^2` to both side to equate the equation to 0

`=> 3x^2 + 8x^2 + 4x=0`
`=> x(3x^2 + 8x+4)=0` Factor the greatest common factor

We need factor multiply to equal 12

`12=+-1*+-12 ; +-2*+-6; +-3*+-4`

But add up to equal to 8
`2*6 = 12` ; `2+6=8`

`=> x(3x^2 + 6x+2x+4)=0`

`=> x[(3x^2 + 6x)+(2x+4)]=0`

Factor greatest common factor by grouping

`=> x[3x(color(red)(x+2))+2color(red)((x+2)]]=0`

Factor greatest common factor by grouping

Answer : `=> x(x+2)(3x+2)=0`