Factor Completely

mathxyz · 2024-05-20 06:33:17

Factor Completely

Problem 100; page 58.

x^6 + 2x^3 + 1

Wow! I never saw a factoring problem like this before.

Let x^6 = (x^3)^2

Let u = x^3

u^2 + 2u + 1

(u + 1)(u + 1)

Back-substitute for u.

(x^3 + 1)(x^3 + 1)

I think I can now apply the sum of cubes to one of the factors.

Yes?

Bob · 2024-05-20 07:34:05

Good so far. Sum of cubes? Try it.

Bob

mathxyz · 2024-05-20 09:34:22

Bob wrote:

Good so far. Sum of cubes? Try it.
Bob

The Sum of Cubes formula is on page 44. Check it out. I will work on this later tonight or tomorrow morning.

mathxyz · 2024-05-20 13:54:26

Bob wrote:

Good so far. Sum of cubes? Try it.
Bob

Given (x^3 + 1)(x^3 + 1), I can apply the sum of two cubes to each factor.

I did the work on paper and got this:

(x + 1)(x^2 - x + 1)(x + 1)(x^2 - x + 1)

You say?

Last edited by mathxyz (2024-05-20 15:10:03)

Bob · 2024-05-20 19:09:02

That looks good.

Bob

mathxyz · 2024-05-21 00:57:34

Bob wrote:

That looks good.
Bob

Thanks. By the way, only highlighted problems by Sullivan will be posted WHEN I GET STUCK. If no help is needed, the problem will not be posted.

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