Question

fully factor

x^4-5x^2+4

Answer 1

\[x^4-5x^2+4\]

Answer 2

What are two numbers that multiply to 4 and add to -5?

Answer 3

but dont we only use that method when it's a quadratic ? :S

Answer 4

You can think of this problem (quartic) in the same way as you do the quadratic as you begin factoring. So, what are the two numbers that multiply to 4 and add to -5?

Answer 5

-4,-1

Answer 6

Then, x^4-5x^2+4 = (x^2 - 4) (x^2 -1)
Now, factor (x^2 - 4) and next factor (x^2 -1).
You should end up with four linear factors for the factorization of x^4-5x^2+4.
Post what you get here.

Answer 7

(x+2)(x-2)(x+1)(x-1)

Answer 8

difference of squares

Answer 9

Correct.
Factor: x^4 - 1.

Answer 10

\[(x^2-1)(x^2+1)\]

Answer 11

\[x^2+1\] ^ that can't be factored further, right?

Answer 12

x^2 + 1 cannot be further factored over the set of Reals but x^2 -1 can.
x^4 - 1 = (x^2 +1)(x^2-1) and then ....

Answer 13

and then (x-1)(x+1)

Answer 14

Yes. On your paper, you would write the complete factorization:

x^4 - 1 = (x^2 +1)(x^2-1) = (x^2 +1)(x +1) (x - 1)

Answer 15

thanks for your help, i have another factoring question that i need help on, can you please help ?

Answer 16

@burhan101 Yes. Post it here.

Answer 17

http://openstudy.com/study#/updates/505400ade4b02986d370771e