How do you factor #x^4-8x^2+16#?

3 Answers
Kha L.
Mar 13, 2018

#(x^2-4)^2#

Explanation:

Make a dumby variable, in this case I will use #u#.
Let #u=x^2#
You have #x^4-8x^2+16#.
This is equivalent to #(x^2)^2-8x^2+16#.
You can now plug in #u# wherever #x^2# is.
You get #u^2-8u+16#.
Now, you can factor this like a normal polynomial.
#(u-4)(u-4)#
or
#(u-4)^2#.
Finally, you plug in #u# back into the expression.
The factored form will be #(x^2-4)^2#.

Richard
Mar 13, 2018

#(x+2)^2(x-2)^2#

Explanation:

#=(x^4-8x^2+16)#
#=(x^2-4)(x^2-4)#
#=(x+2)(x-2)(x+2)(x-2)#
#=(x+2)^2(x-2)^2#

R. Nguyen
Mar 13, 2018

Notice how this is almost a simple quadratic trinomial. In fact, it can be thought of as a quadratic of #x^2#.

Let #color(blue)(u = x^2)#

#x^4 - 8x^2 + 16#

#= (x^2)^2 - 8(x^2) + 16#

#= color(blue)u^2 - 8color(blue)u + 16#

We can easily factor this quadratic.

#= (color(blue)u - 4)^2#

#= (color(blue)(x^2) - 4)^2#

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