1. Re: Factoring Polynomials

Assuming the third term ought to be

if you had a typo on the 4th term also, (+16 instead of -16)
you would have a polynomial which has 3 factors (some repeated)
and 3 different integer roots for f(x) = 0.
so that would be:

x^5 - 6x^4 + 5x^3 + 16x^2 - 12x - 16

The factors of the first three terms are:
(x^3)(x - 5)(x - 1)

I can't factorise the last three terms

2. Re: Factoring Polynomials

Originally Posted by nickgc
so that would be:

x^5 - 6x^4 + 5x^3 + 16x^2 - 12x - 16

The factors of the first three terms are:
(x^3)(x - 5)(x - 1)

I can't factorise the last three terms
factorise x&#94;5 - 6x&#94;4 &#43; 5x&#94;3 &#43; 16x&#94;2 - 12x - 16 - Wolfram|Alpha

To get the factors, guess a root, use the null factor law, use polynomial long division etc. etc.

eg. Guess x = 1 is a root. Doesn't work. Guess x = -1. It works therefore x + 1 is a factor. Divide x + 1 into x^5 - 6x^4 + 5x^3 + 16x^2 - 12x - 16. Get a quartic factor. Repeat the process.

Not difficult, just drudge work for you to do (not us).

3. Re: Factoring Polynomials

Originally Posted by nickgc
so that would be:

x^5 - 6x^4 + 5x^3 + 16x^2 - 12x - 16

The factors of the first three terms are:
(x^3)(x - 5)(x - 1)

I can't factorise the last three terms
It seldom helps to factor just part of a polynomial in the way you did the first 3 terms.

I believe Archie Meade suggested that the polynomial: x^5 - 6x^4 + 5x^3 + 16x^2 - 12x - 16 can be factored (with 'nice' results).

There are a couple of easy numbers you can plug into Archie's polynomial in order to check for some factors.

Plug 1 in for x. This gives -12 as a result, so the remainder theorem tells you that (x-1) is not a factor.

Next, plug -1 in for x. This gives -12 as a result, so the remainder theorem tells you that (x-(-1)) is a factor. Of course, (x-(-1)) = (x+1).

So, x+1 is a factor of Archie's polynomial.

Use synthetic division (or else long division) to find the other factor. --It's a degree 4 polynomial. See if -1 is also a root of that polynomial.

A way to find some other possible factors is to graph the polynomial with a graphing calculator or some graphing software. The roots of the polynomial will give information about factors.

4. Re: Factoring Polynomials

x^5 - 6x^4 + 5x^3 + 16x^2 - 12x - 16

= ( x + 1 ) ( x^4 - 7x^3 + 12x^2 + 4x - 16 )
= ( x + 1 ) ( x + 1 ) ( x^3 - 8x^2 + 20x - 16 )
= ( x + 1 ) ( x + 1 ) ( x - 2) ( x^2 - 6x + 8 )
= ( x + 1 ) ( x + 1 ) ( x - 2) ( x - 2) ( x - 4 )

is this correct?

5. Re: Factoring Polynomials

Originally Posted by nickgc
x^5 - 6x^4 + 5x^3 + 16x^2 - 12x - 16

= ( x + 1 ) ( x^4 - 7x^3 + 12x^2 + 4x - 16 )
= ( x + 1 ) ( x + 1 ) ( x^3 - 8x^2 + 20x - 16 )
= ( x + 1 ) ( x + 1 ) ( x - 2) ( x^2 - 6x + 8 )
= ( x + 1 ) ( x + 1 ) ( x - 2) ( x - 2) ( x - 4 )

is this correct?
factorise x&#94;5 - 6x&#94;4 &#43; 5x&#94;3 &#43; 16x&#94;2 - 12x - 16 - Wolfram|Alpha

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