Factor:

$\displaystyle 27x^2+64$

I am completely lost...

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- Mar 24th 2008, 12:42 AMmt_lapinFactoring again
Factor:

$\displaystyle 27x^2+64$

I am completely lost... - Mar 24th 2008, 12:53 AMa.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 AMmr fantastic
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 AMmt_lapin
Oh dear. I meant $\displaystyle 27^3+64$. Does that make it solvable now? Really sorry :/

- Mar 24th 2008, 02:05 AMmr fantastic
- Mar 24th 2008, 02:16 AMmt_lapin
I keep missing things. It's supposed to be $\displaystyle 27x^3+64$

I promise this is correct. - Mar 24th 2008, 02:20 AMmt_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 AMmt_lapin
The answer would be: $\displaystyle (3x+4)(9x^2-12x+16)$

Is this correct? - Mar 24th 2008, 03:36 AMmr fantastic