# (Pre-calc) Dealing with exponents [closed]

How do I solve equations like:

$-4x^2(x + 3) - 2x^2(x - 1)$

I mean, I can do it if the x didn't have the $^2$. How do I Deal with these? I know that something like $8x = (xxxxxxxx)$, but what would $8x^2$ look like?

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## closed as off-topic by user127.0.0.1, Claude Leibovici, Paul, Yiorgos S. Smyrlis, mauJan 25 '14 at 9:40

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Uh, correction: $8x = x+x+x+x+x+x+x+x$. On the other hand, $(xxxxxxxx) = x^8$. – Newb Jan 25 '14 at 6:55
Do you want to solve for $x$ such that your function be equal to zero ? If this is the case, add $= 0$ in your post. Otherwise, replace $solve$ by $simplify$ – Claude Leibovici Jan 25 '14 at 6:56

I'll walk you through it. Let me know if any part of this isn't clear.

\begin{align} -4x^2(x + 3) - 2x^2(x - 1)&= (-4x^2)(x+3) - (2x^2)(x-1)\\ &= ((-4x^2)x + (-4x^2)3) - ((2x^2)x-(2x^2)1)\\ &= ((-4)x^3+(-12)x^2)-((2x^3)-2x^2)\\ &= ((-4)x^3+(-12)x^2)-(2x^3)+2x^2\\ &= -4x^3-12x^2-2x^3+2x^2\\ &= -6x^3-10x^2 \end{align}

If you don't know why $a-(b-c)=a-b+c$ (which is necessary to get from the third to the fourth step), here's an explanation: \begin{align}a-(b-c) &=a+(-(b-c))\\ &= a+((-1)(b-c))\\ &= a+((-1)(b+((-1)c)))\\ &= a+((-1)b)+((-1)(-1)c)\\ &= a+(-b)+(1c)\\ &= a-b+c\end{align}

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This is a nice answer, but I really doubt that the asker knows what it means to be a function of $x$. – user61527 Jan 25 '14 at 7:04
You're right. I'll remove that. – Newb Jan 25 '14 at 7:05

HINT

Develop the terms. You will arrive to a simple cubic equation in $x$. Factor and ...solve.

I am sure you can take from here.

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"develop the terms" doesn't mean anything to someone learning elementary mathematics in English (it may have fifty years ago, but not anymore). – symplectomorphic Jan 25 '14 at 6:57
I think this answer is only clear to someone who already knows how to solve the problem. – Newb Jan 25 '14 at 7:14