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Factor the expression x^4-20x^2+64
^ means power
please help




much appriciated

2007-03-22 16:39:16 · 5 answers · asked by lonewolf 1 in Science & Mathematics Mathematics

5 answers

x^4 - 20x^2 + 64
= (x^2)^2 - 20(x^2) + 64

Let x^2 = a

Rewrite expression as:
a^2 - 20a + 64
= a^2 - 16a - 4a + 64
= a(a - 16) - 4(a - 16)
= (a - 16)(a - 4)

Substitute x^2 back into the factors
(a - 16)(a - 4) = (x^2 - 16)(x^2 - 4)
Apply the identity m^2 - n^2 = (m + n)(m - n) to both factors

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

2007-03-22 17:22:33 · answer #1 · answered by Akilesh - Internet Undertaker 7 · 0 0

First, find the factors for the last number.

1, 64
2, 32
4, 16 <

Which of these, when combined, will make the middle constant number (20)?

(x^2 - 4)(x^2 - 16)

you can further factor these:

(x + 2)(x - 2) and (x + 4)(x - 4)

There you have it!
Just remember to mind your signs.

Hope this helps!

2007-03-22 23:46:02 · answer #2 · answered by p37ry 5 · 1 0

The above may be correct, but here is the process.

If you are ever stumped with one of these higher power problems, use this solution: let another variable (e.g. 'y') = x^2. You then have

(x^2)^2 - 20(x^2) + 64
y^2 - 20y + 64
Factor this simple quadratic
(y - 16)(y - 4)
Substitute x^2 back in
(x^2 - 16)(x^2 - 4)
Factor the difference of two squares
(x + 4)(x - 4)(x + 2)(x - 2)

You're done.

2007-03-22 23:48:38 · answer #3 · answered by Anonymous · 1 0

(x-4)(x+4)(x-2)(x+2)

2007-03-22 23:43:18 · answer #4 · answered by Sahil C 2 · 0 0

are you missing a number?

x^2(x^2-20x+64)

2007-03-22 23:46:00 · answer #5 · answered by Justina 3 · 0 2

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