How do i factor 15AsquaredBcubed+10AsquaredBsquared-5ABcubed?

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- Jul 20th 2008, 10:51 AMeuclid2factoring
How do i factor 15AsquaredBcubed+10AsquaredBsquared-5ABcubed?

- Jul 20th 2008, 02:05 PMMatt Westwood
$\displaystyle 15A^2B^3+10A^2B^2+5AB^3$

In each bit you can divide by 5 without leaving any remainder.

You can also see there's an $\displaystyle A$ in each bit.

Also, there's a $\displaystyle B$ in each bit. In fact, there's at least 2 $\displaystyle B$'s in each bit.

So you can extract a 5, an $\displaystyle A$ and two $\displaystyle B$'s.

This gives, for each bit:

$\displaystyle 5AB^2 \times 3AB$

$\displaystyle 5AB^2 \times 2A$

$\displaystyle 5AB^2 \times B$

So putting it all together we get:

$\displaystyle 5AB^2 (3AB + 2A + B)$. - Jul 21st 2008, 05:16 AMtutor
Hi,

=15A^2B^3+10A^2B^2-5AB^3

t taking common of 5AB^2

=5AB^2(3AB+2A-B) is the answer(Rofl)