# Thread: factor each of the following completely

1. ## factor each of the following completely

can somone help me with these 2 problems

1st problem 12x^3+28x^2-27x-63

2nd problem 216x^3+1000

if you could just show me step by step how its solved so i can figure out the other few i gota do and not use confusing math terms

2. Hints:

12x^3+28x^2-27x-63=4x^2(3x+7)-9(3x+7)

216x^3+1000=(6x)^3 + 10^3

3. I'm assuming the question is to solve these equations equal to zero?
For the first equation you have $\displaystyle 12x^3+28x^2-27x-63=0$ yes?
You can factorise this into $\displaystyle (2x-3)(6x^2+23x+21)$ and then solve this either by factorising, or using the quadratic formula.

For the second, you have $\displaystyle 216x^3-1000=0$ which you can reduce to $\displaystyle 27x^3-125=0$ or $\displaystyle 27x^3=125$. Then take the cube root of each side.
Hope this helps.

4. Originally Posted by worc3247
I'm assuming the question is to solve these equations equal to zero?
Why would you assume that? The post said "factor", why not factor?

For the first equation you have $\displaystyle 12x^3+28x^2-27x-63=0$ yes?
You can factorise this into $\displaystyle (2x-3)(6x^2+23x+21)$ and then solve this either by factorising, or using the quadratic formula.

For the second, you have $\displaystyle 216x^3-1000=0$ which you can reduce to $\displaystyle 27x^3-125=0$ or $\displaystyle 27x^3=125$. Then take the cube root of each side.
Hope this helps.
Actually, it was $\displaystyle 216x^3+ 1000$ but it is still true that
$\displaystyle a^3+ b^3= (a+ b)(a^2- ab+ b^2)$

6. Sometimes you can use the regroup and factor method on these 3rd order equations:

grouping the left 2 members then the right 2 members: (12x^3 + 28x^2) - (27x + 63)
factoring 4x^2 out of the left side and 9 out of the right side: 4x^2(3x +7) - 9(3x+7)
factoring (3x+7) from both sides: (3x+7)(4x^2-9)
factoring the quadratic is like when you factor (a^2-b^2) so you have: (3x+7)(2x+3)(2x-3)

and it looks like HallsofIvy gave you the formula for the x^3+y^3 one.. hope this is a bit clearer on the mechanics... grin