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**Oculus8596****Banned**- From: Great Lakes,Illinois
- Registered: 2024-09-18
- Posts: 126

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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