# Horrible Factorization

• January 30th 2011, 02:00 AM
BobBali
Horrible Factorization
Again, to factorize; not solve..

$
x^(3a+2) + 3x^(2a+2) - 5x^2
$

$
-5x^2 + x^(3a+2) + x^(2a+2)
$

$
x^-0.5( -5x^(2.5) + 3^(2a+2.5) + 1^(3a+2.5))
$
*Not too sure about this step, went off-track here, i think..(Speechless)
• January 30th 2011, 02:40 AM
Plato
Quote:

Originally Posted by BobBali
Again, to factorize; not solve..
$x^{(3a+2)} + 3x^{(2a+2)} - 5x^2$
$-5x^2 + x^(3a+2) + x^(2a+2)
$

$x^{-0.5}( -5x^{(2.5)} + 3^{(2a+2.5)} + 1^{(3a+2.5)})
$
*Not too sure about this step, went off-track here, i think.

If you have more than one character in an exponent, set off the whole exponent in braces.
$$x^{(3a+2)} + 3x^{(2a+2)} - 5x^2$$ gives $
x^{(3a+2)} + 3x^{(2a+2)} - 5x^2
$

Please edit your post to correct exponents.
• January 30th 2011, 06:21 AM
Soroban
Hello, BobBali!

Didn't you PREVIEW your post?
You didn't see that it didn't come out clearly?

And how did you manage to factor out $x^{-\frac{1}{2}}$ ? . . . And why?

Like Plato, I'll guess at what you meant.

Quote:

$\text{Factor: }\;x^{3a+2} + 3x^{2a+2} - 5x^2$

Take another look at what we have: . $x^{3a}\!\cdot\!x^2 + 3\!\cdot\!x^{2a}\!\cdot\!x^2 - 5\!\cdot\!x^2$

Factor out $x^2\!:\;\;x^2\left(x^{3a} + 3x^{2a} - 5)$

. . and that's the best we can do.