Thread: 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.

Hello zxazsw

Originally Posted by zxazsw

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.

Two things you need to use here:

• The Factor Theorem itself: $\displaystyle (x-a)$ is a factor of $\displaystyle P(x)$ if and only if $\displaystyle P(a) = 0$.

• If the constant term in $\displaystyle P(x)$ is $\displaystyle k$, then in any factor of the form $\displaystyle (ax+b)$, $\displaystyle b$$\displaystyle k$. must be a factor of

So in part (a), you'll only need to try values of $\displaystyle x$ which are factors of $\displaystyle 24$ (not forgetting to try negative numbers as well), to see whether $\displaystyle P(x) = 0$. If my arithmetic is correct, none of them works, so there aren't any linear factors with integer coefficients.

By the same reasoning, in part (b), $\displaystyle (x \pm1)$ are the only possibilities. So try $\displaystyle x = 1$ first. And, even when you have found the first factor, keep on trying with the quotient until you're sure there are no factors left. (Hint: do you know anything about Pascal's triangle or the Binomial Theorem?)

Grandad