Factoring

**mt_lapin** · March 23rd 2008, 09:38 PM

Factor:

$x^3+4$

?

**o_O** · March 23rd 2008, 09:56 PM

Sum of cubes:

$a^{3} + b^{3} = (a + b)\left(a^{2} - ab + b^{2}\right)$

**mt_lapin** · March 23rd 2008, 10:00 PM

Ok. So what would be the answer? Would $(x+4)(x^2-4x+16)$ be correct ?

**o_O** · March 23rd 2008, 10:01 PM

Not quite. Notice how 4 represents $b^{3}$ . If that's the case, what is b?

**mt_lapin** · March 23rd 2008, 10:12 PM

I don't get it. :/

**mt_lapin** · March 23rd 2008, 10:15 PM

b seems to be 4 to me.

**o_O** · March 23rd 2008, 10:28 PM

This is what I'm saying:
$b^{3} = 4$

Solve for b.

If $b = 4$ , then $b^{3} = 64 \neq 4$

**mt_lapin** · March 23rd 2008, 10:41 PM

but $b^{3} = 64$ does equal 4.

Can you please fully explain the problem out to me. I foiled my answer 5 times and every time I got $x^3+64$

**mt_lapin** · March 23rd 2008, 10:58 PM

Oh! I see where I messed up. What I needed factored was $x^3+64$ not $x^3+4$ which I posted. I'm really sorry.

**o_O** · March 23rd 2008, 10:58 PM

In that case, your original answer was right.

**mt_lapin** · March 23rd 2008, 11:02 PM

Thank you. Sorry again for the confusion.

Thread: Factoring