# Factoring again

• Mar 24th 2008, 12:42 AM
mt_lapin
Factoring again
Factor:

$\displaystyle 27x^2+64$

I am completely lost...
• Mar 24th 2008, 12:53 AM
a.a
$\displaystyle 27x^2+64$

this is a binomial expression- an expression with two terms, the variable, x is only present in the first term and thus you cannot factor it. We are only left with the constants 27 and 64. These numbers do not have a common favtor hence we cannot factor again. Thus the expression canot be factored
• Mar 24th 2008, 01:36 AM
mr fantastic
Quote:

Originally Posted by mt_lapin
Factor:

$\displaystyle 27x^2+64$

I am completely lost...

I'll assume you're required to factorise over the complex number field. Otherwise, as has been previously correctly noted, it can't be done and the question is pointless.

Recall: $\displaystyle A^2 + B^2 = (A + iB)(A - iB)$.

In your case, $\displaystyle A = \sqrt{27} \, x = 3 \sqrt{3} \, x$ and B = 8 ....
• Mar 24th 2008, 01:54 AM
mt_lapin
Oh dear. I meant $\displaystyle 27^3+64$. Does that make it solvable now? Really sorry :/
• Mar 24th 2008, 02:05 AM
mr fantastic
Quote:

Originally Posted by mt_lapin
Oh dear. I meant $\displaystyle 27^3+64$. Does that make it solvable now? Really sorry :/

$\displaystyle = (27)^3 + (4)^3$.

Factorise using the sum of two cubes formula.
• Mar 24th 2008, 02:16 AM
mt_lapin
I keep missing things. It's supposed to be $\displaystyle 27x^3+64$

I promise this is correct.
• Mar 24th 2008, 02:20 AM
mt_lapin
Would I then need $\displaystyle (3x)^3+4^3$ and to factorise using the sum of two cubes formula?
• Mar 24th 2008, 02:26 AM
mt_lapin
The answer would be: $\displaystyle (3x+4)(9x^2-12x+16)$

Is this correct?
• Mar 24th 2008, 03:36 AM
mr fantastic
Quote:

Originally Posted by mt_lapin
The answer would be: $\displaystyle (3x+4)(9x^2-12x+16)$

Is this correct?

Yes.