Question

x5 - 16x3 + 8x2 - 128
x3(x2 - 16) + 8(x2 - 16)
(x2 -16)(x3 + 8)

Answer 1

Has the polynomial below been factored correctly and completely? Explain your answer.

Answer 2

Yes, it had been factored correctly. One easy way to check this would be to multiply out the last set of terms, and see if such equals the very first expression.

Answer 3

Okay thank you very much

Answer 4

Oh, wait, but it has not been factored completely! Sorry, didn't catch that last part...

Answer 5

oh whats wrong with it?

Answer 6

Because, by the difference of squares we know that:
\[
a^2-b^2=(a+b)(a-b)
\]Try finding out which part is still missing

Answer 7

tbh i have no idea. thats why i'm on here i don't get it

Answer 8

Well, all right, so, we have the last term equals:
\[
(x^2-16)(x^3+8)
\]We know that:
\[
a^2-b^2=(a-b)(a+b)
\]And that \(x^2\) is a square, while \(16\) is also a square, thus:
\[
x^2-16=(x+4)(x-4)
\]We also notice that, the last term is:
\[
x^3+8=x^3+2^3
\]So, both are cubes. If we use:
\[
a^3+b^3=(a+b)(a^2-ab+b^2)
\]We then get:
\[
x^3+2^3=(x+2)(x^2-2x+4)
\]So, the expression factors down *completely* AND *correctly* to:
\[
(x+8)(x-8)(x+2)(x^2-2x+4)
\]

Answer 9

ohhh okay thanks a lot for helping me

Answer 10

And, I messed up.. again, ugh, it's \((x-4)(x+4)(x+2)(x^2-2x+4)\)

Answer 11

i figured that out tho

Answer 12

Oh, then, all right!