# Need help on factor theorem.

• Feb 9th 2010, 04:56 PM
zxazsw
Need help on factor theorem.
I am supposed to use the Factor Theorem to either factor the polynomial completely or to prove that it has no linear factors with integer coefficients.
I think I haven't completely understood the whole concept yet, which is why I couldn't get my brain working on the following problems.

a. P(x)= x^4+4x^3-7x^2-34x-24
b. P(x)= x^6-6x^5+15x^4-20x^3+15x^2-6x+1

I think these are harder than the regular ones like P(x)=x^3+2x^2-9x+3, etc.

Thanks.(Clapping)
• Feb 9th 2010, 05:22 PM
Quacky
A very long method
The key here is to look at the constant at the end.
For part a, this is $-24$. Think about a quadratic. When factorizing $x^2-7x-8=0$, you'd look for factors of 8, and come up with:
$(x+1)(x-8)=0$
So, $x=8, x=-1$

To factorize, though the co-efficients of x are important, it's the constant that determines the possible factors, which in this case are (+ or -) 1,2,4,8. The co-efficients just allow you to select which specific factors will work.

Now in part a, just apply this to a quartic:

$P(x)= x^4+4x^3-7x^2-34x-24$
Assume p(x)=0
Then
$x^4+4x^3-7x^2-34x-24=0$
Now work out what the factors of 24 are, and you'll get $1,2,3,4,6,8,12,24$
(Yuck: Including negatives, there're a lot to choose from. Perhaps someone at a more advanced level will know of an easier way)

To start, choose $x=1$

Now think back to the example of the quadratic, which does not apply to your question but I am using as an example:
$(x+1)(x-8)=0$
So $x=8, x=-1$

So, if x=1 gives a result of P(x)=0, then (x-1) is a factor of P(x). Now let's go back to your example.

So, if when we substitute x=1 into p(x) and the result is equal to 0, then (x-1) is a factor of p(x)

$P(x)= x^4+4x^3-7x^2-34x-24$
$
P(1)= (1)^4+4(1)^3-7(1)^2-34(1)-24$

This gives a result which clearly does not equal zero, so $(x-1)$ is not a factor. You'd then try x=-1, then x=2, x=-2, etc...

As P(x) is a quartic, there will be a maximum of four factors. I'm afraid I don't know of a faster approach.

Once you find a factor, partially factorize the expression. This way, once you have found two factors, you'll be left with an easy quadratic.
• Feb 9th 2010, 06:07 PM
Krahl
So we agree that the roots of P(x)=0 are in $

\{\pm 1,\pm 2,\pm 3,\pm 4,\pm 6, \pm 8,\pm 12,\pm 24 \}
$

$P(1) \neq 0$
$P(-1) = 0$
$P(2) \neq 0$
$P(-2) = 0$

this rules out 24, -24 and 12

$P(3) = 0$

this rules out -3, 8,-8,6,-6

only ones left are 4,-4

P(-4)=0