How do I factor the polynomial x^5-4x^3-8x^2+32 completely?
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We have to factor x^5 - 4x^3 - 8x^2 + 32
x^5 - 4x^3 - 8x^2 + 32
=> x^3( x^2 - 4) - 8(x^2 - 4)
factor out x^2 - 4
=> (x^2 - 4)(x^3 - 8)
use the relation a^2 - b^2 = (a - b)(a + b)
=> (x - 2)(x + 2)(x^3 - 8)
use a^3 - b^3 = (a - b)(a^2 + ab + b^2)
=> (x - 2)(x + 2)(x - 2)(x^2 + 2x + 4)
=> (x - 2)^2*(x + 2)(x^2 + 2x + 4)
The required factorization of the polynomial is (x - 2)^2*(x + 2)(x^2 + 2x + 4)
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I'll suggest to group the 1st term with the last one.
We'll use the formula:
a^n + b^n = (a+b)(a^n-1 - a^n-2*b + ... + b^n)
x^5 + 32 = x^5 + 2^5 = (x+2)(x^4 - 2x^3 + 4x^2 - 8x + 16)
We'll group the middle terms:
-4x^3-8x^2 = -4x^2(x + 2)
We'll re-write the polynomial:
(x+2)(x^4 + 2x^3 + 4x^2 + 8x + 16) - 4x^2(x + 2)
We'll factorize by x+2:
(x+2)(x^4 - 2x^3 + 4x^2 - 8x + 16 - 4x^2)
(x+2)(x^4 - 2x^3 - 8x + 16)
We'll regroup x^4 - 2x^3 = x^3(x-2)
-8x + 16 = -8(x-2)
(x+2)(x^3(x-2) - 8(x-2)) = (x+2)*(x-2)*(x^3 - 8)
But x^3 - 8 = (x-2)(x^2 + 2x + 4)
The polynomial will become:
(x+2)*(x-2)*(x^3 - 8) = (x+2)*(x-2)^2*[(x^2 + 2x + 4)]
(x+2)*(x-2)*(x^3 - 8) = (x-2)*(x^2 - 4)*(x^2 + 2x + 4)
The factorized polynomial is (x-2)*(x^2 - 4)*(x^2 + 2x + 4).
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