# Thread: powers problems

1. ## powers problems

Hi all, working on this problem and have hit a roadblock, I think there is a rule I can use to make it simpler and I am sure I am just not seeing it, any help?

$\frac {q^2}{56}(56-15q+q^2)-[\frac{q}{56}(56-15q+q^2)]^2$

can I pull the preceding $\frac{q^2}{56}$ and put that together with the second terms $\frac{q}{56}$ and then put that in front of the whole or is that not allowed? Thanks for the help!

2. Hello, Raezputin!

I hope I read the problem correctly . . .

Smplify: . $\frac{q^2}{56}(56-15q+q^2) - \bigg[\frac{q}{56}(56-15q+q^2)\bigg]^2$

We have: . $\frac{q^2}{56}\left(56 - 15q + q^2\right) - \frac{q^2}{56^2}\left(56 - 15q + q^2\right)^2$

We can factor out: . $\frac{q^2}{56^2}(56-15q + q^2)$

. . $\frac{q^2}{56^2}(56-15q+q^2)\cdot \bigg[56 - (56 - 15q + q^2)\bigg] \;\;=\;\;\frac{q^2}{56^2}(56 -15q-q^2)\cdot\left(15q - q^2\right)$

. . $= \;\;\frac{q^2}{56^2}(56-15q+q^2)\cdot q(15-q) \;\;=\;\;\frac{q^3}{56^2}(56-15q+q^2)(15-q)$

. . $= \;\;\frac{q^3}{3136}(7-q)(8-q)(15-q)$