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Factor each polynomial completely. If the polynomial cannot be factored, say it is prime.
A. x^6 + 2x^3 + 1
B. 3 - 27x^2
Question A
Note: x^6 = (x^3)^2
Let u = x^3
I now get u^2 + 2u + 1.
(u + 1)(u + 1) = (u + 1)^2
Back-substitute for u.
(x^3 + 1)^2 = (x^3 + 1)(x^3 + 1).
The expression x^3 + 1 can be factored using the sum of cubes.
Doing so, I get (x + 1)(x + 1)(x^2 - x + 1)(x^2 - x + 1).
Answer: (x + 1)^2 (x^2 - x + 1)^2
Do you agree?
Question B
This expression can be factored using the difference of cubes.
3(1 - 9x^3)
The difference of cubes:
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
Let a = 1
Let b = 3x
3(1 - 3x)(1 + 3x + (3x)^2)
3(1 - 3x)(1 + 3x + 9x^2)
Do you agree?
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