Part 1 Create a difference of squares binomial or a perfect square trinomial that can be factored multiple times. Part 2 Provide the factors of this polynomial. Part 3 Explain, in complete sentences, the process you used to create the binomial/trinomial. @texaschic101
@Abhisar @iPwnBunnies @sweetburger
Factored multiple times? How bout something like this?\[\Large\rm x^4-16\]=o
oh okay I didn't seem to understand "factored multiple times part"
Or do they want difference of only squares, written as a binomial power?\[\Large\rm (x^2-1)^2\]
written as a binomial to some power*
That would be another option we could look at.
oh okay. is it possible to explain that also?
So the way I'm starting this is by thinking of my difference of squares formula:\[\Large\rm a^2-b^2=(a-b)(a+b)\]So I know that if I can write both numbers as perfect squares (with subtraction between them) then they'll factor nicely using this identity.\[\Large\rm x^2-1=x^2-1^2\]Factoring to,\[\Large\rm x^2-1^2=(x-1)(x+1)\]Yes? Maybe if we stuck with the 16 it would be easier to work with.
Maybe I think the 16 will be easier
Don't you think?
Fine fine fine :)\[\Large\rm x^2-16 \quad=\quad x^2-4^2\]Do you understand how we could use the formula from here?
yes so far. :)
The reason I put the square on the outside of the brackets is because that's an easy way to get some factors!! c: Remember a square tells you to multiply the `thing` by itself.
\[\Large\rm (x^2-16)^2=(x^2-16)(x^2-16)\]
oh okay would I factor that further?
since it's difference of two squares binomial
Writing 16 as 4^2 shows us that it's a perfect square. \[\Large\rm (x^2-16)^2=(x^2-4^2)(x^2-4^2)\]
Then we can use our difference of squares formula on each set.\[\Large\rm =(x-4)(x+4)(x-4)(x+4)\]
I hope I didn't misread part 1.. The directions are a little strange :d
lol
thank you!
Yah I hope that made some sense to you! :o
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