# factor each of the following completely

• Mar 23rd 2011, 01:48 PM
factor each of the following completely
can somone help me with these 2 problems (Crying)

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
• Mar 23rd 2011, 01:57 PM
Also sprach Zarathustra
Hints:

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

216x^3+1000=(6x)^3 + 10^3
• Mar 23rd 2011, 01:58 PM
worc3247
I'm assuming the question is to solve these equations equal to zero?
For the first equation you have $12x^3+28x^2-27x-63=0$ yes?
You can factorise this into $(2x-3)(6x^2+23x+21)$ and then solve this either by factorising, or using the quadratic formula.

For the second, you have $216x^3-1000=0$ which you can reduce to $27x^3-125=0$ or $27x^3=125$. Then take the cube root of each side.
Hope this helps.
• Mar 23rd 2011, 02:14 PM
HallsofIvy
Quote:

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?

Quote:

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

For the second, you have $216x^3-1000=0$ which you can reduce to $27x^3-125=0$ or $27x^3=125$. Then take the cube root of each side.
Hope this helps.
Actually, it was $216x^3+ 1000$ but it is still true that
$a^3+ b^3= (a+ b)(a^2- ab+ b^2)$
• Mar 23rd 2011, 02:32 PM
worc3247
My bad, I didn't read the title.
• Mar 24th 2011, 12:56 PM
Zap
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