# 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 \$\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.
• 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 \$\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)\$
• Mar 23rd 2011, 02:32 PM
worc3247