How do you factor a^8-17a^4+16?

Question

Please explain step by step because I'm quite confused.

Answer 1

a^8 = a^4^2, so

a^8-17a^4+16 = a^4^2 - 17a^4 + 16.
This might seem pointless, but notice that the function is actually quadratic with respect to a^4. We can solve quadratic functions. It shouldn't take too much effort to see that -16*-1 = 16, and -16 + -1 =-17, so this has integer roots.
a^8-17a^4+16 = (a^4 - 1)(a^4 - 16)

Now (a^4 - 1) is a difference of squares, and is equal to (a^2 - 1)(a^2 + 1). Similarly (a^4 - 16) = (a^2 - 4)(a^2 + 4)
(a^2 + 1) and (a^2 + 4) are irreducible, but (a^2 - 1) and (a^2 - 4) are also differences of squares, so
a^8-17a^4+16 = (a^4 - 1)(a^4 - 16) = (a^2 - 1)(a^2 + 1)(a^2 - 4)(a^2 + 4) = (a - 1)(a + 1)(a^2 + 1)(a - 2)(a+2)(a^2 + 4)

Answer 2

(a^4 - 1)(a^4 - 16)
= (a^2 - 1)(a^2 + 1)(a^2 - 4)(a^2 + 4)
= (a + 1)(a - 1)(a^2 + 1)(a + 2)(a - 2)(a^2 + 4)

Answer 3

pretend that a^4=x...plug that into your original equation...

a^8-17a^4+16
x^2-17x+16

now...this is what you are used to seeing. now solve this eq like you normally would...

x^2-17x+16
find out what 2 numbers multiply to get a positive 16 (meaning both positive or both negative) and what 2 numbers subtract to get 17

try -1 and -7

so...
x^2-17x+16
(x-1)*(x-16)

now...just plug x=a^4 back in

(a^4-1)*(a^4-16)
if you are solving for a..

a^4=1
so a=+/-1

a^4=16
2*2*2*2=16
-2*-2*-2*-2=16
a=+/-2

Answer 4

= a^8 - a^4 - 16a^4 + 16
= a^4(a^4 - 1) - 16(a^4 + 1)
= (a^4-16)(a^4-1)
= (a^2-4)(a^2+4)(a^2-1)(a^2+1)
= (a-2)(a+2)(a^2+4)(a-1)(a+1)(a^2+1)

Answer 5

Treat it like a quadratic, because it is in the same form:

(a^4-16)(a^4-1)
(a^2+4)(a^2-4)(a^2+1)(a^2-1)
(a^2+2i)(a^2-2i)(a-2)(a+2)(a+i)(a-i)(a+1)(a-1)

Answer 6

(a^4 - 16)(a^4 - 1)

Answer 7

(a^4-16)(a^4-1)

Answer 8

(a^4-16)(a^4-1)

(a^2-4)(a^2+4)(a^2-1)(a^2+1)
(a-2)(a+2)(a^2+4)(a-1)(a+1)(a^2+1)

Answer 9

(a^4-16)(a^4-1)
(a^2-4)(a^2+4)(a^2-1)(a^2+1)
(a-2)(a+2)(a^2+4)(a-1)(a+1)(a^2+)