Angelina factored (x - 4^8 ) and wrote that it was equal to (x^2 - 4^2)^2(x^2 + 4^2). Use complete sentences to explain how you could confirm whether Angela’s solution is correct. Prove your conclusion by multiplying out Angela’s solution.
Are you sure the first part isn't supposed to be (x^8 - 4^8) maybe? It just doesn't make much sense with the way it's written now. It's just so trivial.
that's how it's written in my question..
You can take the end result,\[\large\rm (x^2 - 4^2)^2(x^2 + 4^2)\]expand it out (which is the opposite of factoring), and you should end up with the starting result, \(\large\rm (x-4^8)\) If you don't, then the statement about the starting thing factoring into the other stuff is false.
So let's start here,\[\large\rm \color{orangered}{\left(\color{black}{x^2-4^2}\right)^2}\]We want to expand out this square, the orange part. Remember how to expand a square binomial?\[\large\rm (a-b)^2=(a-b)(a-b)\]Where you multiply all the stuff and the things?
yeah I remember how to do that
so it would be\[(x)^2-(2)^2\] , right?
Crapppp sorry OpenStudy didn't give me the notification >.<
When you expand out a square binomial, \[\large\rm (a-b)^2=(a-b)(a-b)\]you get four terms,\[\large\rm =a^2-ab-ba+b^2\]But the middle two terms will always combine,\[\large\rm =a^2-2ab+b^2\]
its fine :)
So if we expand out our thing,\[\large\rm \color{orangered}{\left(\color{black}{x^2-4^2}\right)^2}\quad=\quad (x^2)^2-x^24^2-4^2x^2+(4^2)^2\]lot of squares floating around! :O
The middle terms combine,\[\large\rm =(x^2)^2-2(4^2x^2)+(4^2)^2\]
\[\large\rm \color{orangered}{(x^2 - 4^2)^2}(x^2 + 4^2)\]So here is what we've done so far, we've turned this orange thing into three terms by expanding it,\[\large\rm \color{orangered}{\left((x^2)^2-2(4^2x^2)+(4^2)^2\right)}(x^2 + 4^2)\]
And then we would need to multiply these sets of brackets out. It's going to be a big mess :d
Oh let's simplify some of the orange stuff a sec,\[\large\rm \color{orangered}{\left(x^4-2(4^2x^2)+4^4\right)}(x^2 + 4^2)\]
|dw:1481850766401:dw|Then we would expand out these brackets, ending up with 6 terms.
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