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

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

Factor Completely

Factor the polynomial completely.

x^6 + 2x^3 + 1


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

Bob
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Registered: 2010-06-20
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Re: Factor Completely

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


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
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#3 2024-09-20 10:55:54

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

Re: Factor Completely

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?


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

Bob
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Registered: 2010-06-20
Posts: 10,610

Re: Factor Completely

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


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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

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

Re: Factor Completely

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.


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