# Math Help - Factorization...Again.

1. ## Factorization...Again.

Good-day to everyone. Again, i have to factorize the given below (not solve for x), just factorize. Thanks

$
36x^2 - 24xy + 48y^2
$

$\\ 6( 6x^2 -4xy +8y^2)$
$
6(6x^2 + 8y^2 -4xy)
$

$
6[( \sqrt6x + \sqrt8y )^2 - 4xy)]
$

$
6[( \sqrt6x + \sqrt8y )( \sqrt6x + \sqrt8y) - 17.86xy]
$
*Completing the square

2. Originally Posted by BobBali
Good-day to everyone. Again, i have to factorize the given below (not solve for x), just factorize. Thanks

$
36x^2 - 24xy + 48y^2
$

$\\ 6( 6x^2 -4xy +8y^2)$
$
6(6x^2 + 8y^2 -4xy)
$

$
6[( \sqrt6x + \sqrt8y )^2 - 4xy)]
$

$
6[( \sqrt6x + \sqrt8y )( \sqrt6x + \sqrt8y) - 17.86xy]
$
*Completing the square
The HCF of 36, 24 and 48 is 12 rather than 6 (which you see in the next step because you can factor out a 2 from your second line)

$12(3x^2-2xy+4y^2)$

That quadratic term is prime so I'd just leave it at that

3. Originally Posted by BobBali
Good-day to everyone. Again, i have to factorize the given below (not solve for x), just factorize. Thanks

$
36x^2 - 24xy + 48y^2
$

$\\ 6( 6x^2 -4xy +8y^2)$
$
6(6x^2 + 8y^2 -4xy)
$

$
6[( \sqrt6x + \sqrt8y )^2 - 4xy)]$
I believe you have made this same mistake before:
$(a+ b)^2\ne a^2+ b^2$
so that $(\sqrt{6}x+ \sqrt{8}y)^2\ne 6x^2+ 8y^2$

$
6[( \sqrt6x + \sqrt8y )( \sqrt6x + \sqrt8y) - 17.86xy]
$
*Completing the square

4. Ok Thanks, so it cannot be factorized any further than this?? > [tex] 12(3x^2 -2xy + 4y^2) [math\]

5. No, this cannot be further factorised.

6. Thank you ;-)