Factor Completely

Oculus8596 · 2024-09-19 02:13:37

Factor the polynomial completely.

x^6 + 2x^3 + 1

Bob · 2024-09-19 20:03:29

You might spot an immediate factorisation but if not here's how to force it.

Let y = x^3 and the expression becomes y^2 + 2y +1 which can be factored into two identical factors,.

Putting x back into one of those and you're down to a cubic. Substitute x = 1 and the expression is zero so x-1 nust be a factor. Extracting that leaves a quadratic which cannot be factored further.

Bob

Oculus8596 · 2024-09-20 10:55:54

Bob wrote:

You might spot an immediate factorisation but if not here's how to force it.
Let y = x^3 and the expression becomes y^2 + 2y +1 which can be factored into two identical factors,.
Putting x back into one of those and you're down to a cubic. Substitute x = 1 and the expression is zero so x-1 nust be a factor. Extracting that leaves a quadratic which cannot be factored further.
Bob

Can x^6 be expressed as (x^3)^2?

I can then say y = x^3.

The given polynomial becomes y^2 + 2y + 1.

y^2 + 2y + 1 factors out to be (y + 1)(y + 1).

I can back-substitute to get (x^3 + 1)(x^3 + 1),which equals (x^3 + 1)^2.

Can the sum of cubes be applied to further factor x^3 + 1?

Bob · 2024-09-20 19:28:17

Consider x^3 + 1. The factor theorem can be used here. Set x = -1 and x^3 + 1 evaluates to zero. This means (x + 1) is a factor.

x^3 + 1 = (x+1)(x^2 -x +1)

Bob

Oculus8596 · 2024-09-20 21:09:52

Bob wrote:

Consider x^3 + 1. The factor theorem can be used here. Set x = -1 and x^3 + 1 evaluates to zero. This means (x + 1) is a factor.

x^3 + 1 = (x+1)(x^2 -x +1)
Bob

The sum of cubes works just the same.

Math Is Fun Forum

#1 2024-09-19 02:13:37

Factor Completely

#2 2024-09-19 20:03:29

Re: Factor Completely

#3 2024-09-20 10:55:54

Re: Factor Completely

#4 2024-09-20 19:28:17

Re: Factor Completely

#5 2024-09-20 21:09:52

Re: Factor Completely

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