Question

Factor Completely:

7h4-4h3+28h2-16h


the 4,2, and 2 are superscripts.

Answer 1

here you go.

Answer 2

Just giving away the answer isn't that helpful imo.

The first step is clear: there is an h in every term, so:
\(7h^4-4h^3+28h^2-16h=h(7h^3-4h^2+28h-16)\).

Now we seem to be stuck. How to factor this one? There are four terms!
Now wait a minute...why not change the order of the terms:

\(h(7h^3-28h-4h^2-16)\).

Why? Just because the 7 and 28 seem to be related to each other in the same way the 4 and 16 are: 7*4=28 and 4*4=16.

Answer 3

(obviously, the second term in the brackets is is +28h).


Next step is to factor the four terms in two groups:

\(7h^3+28h=7h(h^2+4)\)
\(-4h^2-16=-4(h^2+4)\)

Can you guess where this is going to?

Answer 4

@kyliexanne_ : do you understand so far?

Answer 5

The first factor was \(h\).
So now we've found \(h^2+4\) is also a factor.
If we put everything together, we get:

\(7h^4-4h^3+28h^2-16h=h(h^2+4)(7h-4)\).

Done!

Answer 6

The first factor was \(h\).
So now we've found \(h^2+4\) is also a factor.
If we put everything together, we get:

\(7h^4-4h^3+28h^2-16h=h(h^2+4)(7h-4)\).

Done!