Question

Factor Fully.
x^7-8x^4-16x^3+128

So far, I have this but I'm unsure of what to do next since the factor theorem doesn't allow me to go further.

P(x)= x^7-8x^4-16x^3+128
= (x+2)(x-2)^2(x^4+2x^3+8x^2+8x+16)

Answer 1

Try to factor by grouping; in particular, you'll want to group the terms as follows:
\(\large x^7-8x^4-16x^3+128 = (x^7-8x^4)+(-16x^3+128)=\ldots\)

Can you take things from here?

Answer 2

Is there any reason that you would group them this way? The like terms in this scenario aren't that clear to me.

Answer 3

Oh, shouldn't it be (-8x^4 - 16x^3) + (x^7+128) instead? Haha since 128 is 2^3 and you can factor out -18 in the first set.

Answer 4

-8 I mean.

Answer 5

You get better with these kind of things with practice. But the key thing to note with this particular problem is that if you factor out \(\large x^4\) from the first two terms, you're left with a cubic (i.e. \(\large x^3-8\)). But then we note that when we group the last two terms together, we also get a cubic that looks something like this. In particular, you want to note that \(\large -16x^3+128 = -16(x^3-8)\). Hence, if you group them as I did originally, we have that \[\large \begin{aligned} x^7-8x^4-16x^3+128 &= (x^7-8x^4)+(-16x^3+128)\\ &= x^4(x^3-8) -16(x^3-8)\\ &= (x^4-16)(x^3-8)\end{aligned}\]I hope this makes sense! :-)

Answer 6

Obviously, this is not the completely factored form, but once you overcome the first hurdle (which is what I showed above), then everything else shouldn't be too bad from here.

Answer 7

Ohh, I see it now. I can take it from there, thank you very much! (:

Answer 8

I went straight for the factor theorem method so I got stuck midway. lol

Answer 9

I'll just let you know that the completely factored form will have 3 linear terms (2 of which are the same) and 2 irreducible (prime) quadratics. :-)