Question

Need help!

Answer 1

\[\left[\begin{matrix}x^2-& 4\\ x -& 8\end{matrix}\right] \over x-2\]

Answer 2

@Embryo

Answer 3

I'm confused as to what's going on in the top equation, is that a matrix in the top?

Answer 4

I think so @Embryo

Answer 5

well as far as i know, the top part of your matrix can be expressed as follows\[x^2-4=(x-2)(x+2)\]since it's a perfect square, meaning that it takes the form\[(x^2-b^2)=(x-b)(x+b)\]since 4 is really just\[2^2\]that means that we can re express our matrix on top as\[\left[\begin{matrix}(x-2)(x+2) & \\ (x-8) &\end{matrix}\right]\]
so when you divide this matrix by (x-2), your solution should come out as\[\left(\begin{matrix}(x+2) \\ \frac{ (x-8) }{ (x-2) }\end{matrix}\right)\]i feel like your have a typo, and that your question was supposed to be\[(x^2-8)\]and not\[(x-8)\]

Answer 6

it says it's just x-8

Answer 7

@Embryo

Answer 8

well, then the answer i showed you should be correct

Answer 9

ok thank you!!