## Factoring 7n^5+7n^4-3n^2-6n-3 completely

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
BrowardMom
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### Factoring 7n^5+7n^4-3n^2-6n-3 completely

How to factor 7n^5+7n^4-3n^2-6n-3 completely?

I got it down to:

7n^4(n+1)-3(n+1)(n+1) or, of course (n+1)^2 at the end.

Do I pull out the (n+1)? If yes, I can't bring the 7n^4 and the -3 together then, can I, what with the extra (n+1)?

Thanks so much for your help!

stapel_eliz
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BrowardMom wrote:How to factor 7n^5+7n^4-3n^2-6n-3 completely?

To factor, you use many of the same techniques that you would otherwise use to find the zeros, or solutions, of the equation "(polynomial) equal to (zero)".

The first step is usually to apply the Rational Roots Test. This will give you a listing of possible "nice" (fractional or whole-number) values to try.

Then you test these values by using synthetic division.

In your case, the possible zeroes or roots are:

. . . . .$\pm\, \frac{1,\, 3}{1,\, 7}\, =\, \pm\, \frac{1}{7},\, \frac{3}{7},\, 1,\, 3$

Whole numbers are usually nicer, so try -1 and +1 first:

-1 | 7  7 0 -3 -6 -3   |   -7 0  0  3  3   +----------------     7  0 0 -3 -3  0

Since this has a zero remainder, then n = -1 is a zero, so n + 1 is a factor.

Looking at the graph of the original polynomial, it appears that n = -1 is a zero twice, so:

-1 | 7  0 0  -3 -3   |   -7 7  -7 10   +--------------     7 -7 7 -10  7

Okay, so there must be something else going on at n = -1; clearly, it is not a duplicated root.

If you try the various other potential roots, you'll discover that none works. So n = -1 is the only rational root, and n + 1 is the only "nice" factor. This leaves you with:

. . . . .$(n\, +\, 1)(7n^4\, -\, 3n\, -\, 3)$

I don't see anything that can be done with this, though...

BrowardMom
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### Re: Factoring 7n^5+7n^4-3n^2-6n-3 completely

Thanks for your reply, but it includes things my kid hasn't learned yet! (This is for a 7th grader taking Algebra I Honors.) She hasn't learned the Rational Roots test or synthetic division yet.

She was given homework, and this was the one I was stuck on. The instruction is simply "Factor the expression completely." I took the problem:

7n^5+7n^4-3n^2-6n-3

and grouped it into two parts, beginning with

7n^5+7n^4

which I factored to 7n^4(n+1). The second part,

-3n^2-6n-3

factored to -3(n^2+2n+1) which could be further factored to -3(n+1)(n+1). Not sure if that is simplified completely or if it should go to -3(n+1)^2...?

So, when you bring both factored sections together, you get

7n^4(n+1)-3(n+1)(n+1)

If that extra (n+1) was NOT there, she's been taught to redistribute to (7n^4-3)(n+1). Can one still do something similar? I'm guessing not, but have no idea since there were no other questions like this on the homework. But then, there was no mention of negative numbers under the radical symbol at all, either, and thanks to PurpleMath, I learned those are "no real number solution." (Thanks!)

stapel_eliz
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If your "student" hasn't learned the Rational Roots Test, etc, then she can't possibly be expected to factor the quartic. So taking the (n + 1) factor out front, and then simplifying the rest, is as far as she can go.

And it was pretty impressive that, given her background, she found that one factor!

BrowardMom
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### Re: Factoring 7n^5+7n^4-3n^2-6n-3 completely

Her teacher told the kids to ignore that problem - guess she realized it really shouldn't be in the book at this point...just as you already knew!

Thanks again for all your help -- it's been 30 years since ol' Mom did this stuff, but bring on the logs and limits...now that I have Purplemath!