# rational root theorem to factor and find the zeroes of a polynomial

• Oct 24th 2013, 09:41 PM
Jonroberts74
rational root theorem to factor and find the zeroes of a polynomial
So the polynomial I have is

$\displaystyle x^3 - 3x + 2$

so then I find the factors of the constant and the leading coefficient

$\displaystyle \pm 1, 2$

So from here I can divide by $\displaystyle x-1, x+1, x-2, x+2$

If I start with $\displaystyle x-1$ that means I divide through the polynomial with 1 not - 1. this is where I start to get mixed up.

I divide 1 through and get a 0 and whats left is $\displaystyle x^2 + x -2$. so because this is now in standard form I can factor from here to get $\displaystyle (x-1)(x+2)$ and combined with my original divisor of $\displaystyle x -1$, I get $\displaystyle (x-1)^2(x+2)$

The problem is I was told that because I get $\displaystyle (x-1)(x-1)(x+2) \Rightarrow (x-1)^2(x+2)$ I should be able to divide by x-1 once and then through another again to find another zero because I have two factors of $\displaystyle (x-1)$.

I got a pro wolfram|alpha account but the step by step solution didn't explain if very well, it explained it in a way that assumes you already understand $\displaystyle x =1$ and $\displaystyle x = -2$.
• Oct 24th 2013, 11:15 PM
Prove It
Re: rational root theorem to factor and find the zeroes of a polynomial
No, you CAN'T divide by all of those things. You can only divide by the ones of those which are factors. You know which ones are factors using the remainder and factor theorems, which state that for a polynomial function P(x), if P(a) = 0 then (x - a) is a factor.
• Oct 25th 2013, 12:10 AM
Jonroberts74
Re: rational root theorem to factor and find the zeroes of a polynomial
let me reword,

the possible rational factors are $\displaystyle \pm 1, 2$ which means that I can try to divide $\displaystyle x^3-3x+2$ by the following $\displaystyle x-1, x+1, x+2,x-2$, but the the two that result in zero are $\displaystyle x-1, x+2$ which means $\displaystyle x=1$ or $\displaystyle x = -2$

That's not what I was trying to understand though,

I was reviewing an example and then asked someone else who confirmed it. In the polynomial the person divided by the factor 3, then they were able to divide the resulting polynomial by 3 again and then they ended up with a standard form polynomial which they further solved for two more solutions of x = -4 and x = 3. So they ended up with $\displaystyle (x-3)(x-3)(x-3)(+4) \Rightarrow (x-3)^3(x+4)$ as the factored version of the polynomial. So I was under the impression that I should have been able to divide $\displaystyle x^3-3x+2$ by (x-1) then the resulting polynomial by (x-1).

As I am replying I see that would be the case if the factor resulted in $\displaystyle (x-1)^3(x+2)$ but not for $\displaystyle (x-1)^2(x+2)$ because it leaves a polynomial in standard form--which already has two factors (when factored) and my answer results in three (x-1)(x-1)(x+2).
• Oct 25th 2013, 04:36 AM
votan
Re: rational root theorem to factor and find the zeroes of a polynomial
do you realize that [(x - 1)^3](x + 2) will give you a polynomial in x^4? Which is not what you wanted to get.
• Oct 25th 2013, 04:46 AM
topsquark
Re: rational root theorem to factor and find the zeroes of a polynomial
Quote:

Originally Posted by Jonroberts74
let me reword,

the possible rational factors are $\displaystyle \pm 1, 2$ which means that I can try to divide $\displaystyle x^3-3x+2$ by the following $\displaystyle x-1, x+1, x+2,x-2$, but the the two that result in zero are $\displaystyle x-1, x+2$ which means $\displaystyle x=1$ or $\displaystyle x = -2$

That's not what I was trying to understand though,

I was reviewing an example and then asked someone else who confirmed it. In the polynomial the person divided by the factor 3, then they were able to divide the resulting polynomial by 3 again and then they ended up with a standard form polynomial which they further solved for two more solutions of x = -4 and x = 3. So they ended up with $\displaystyle (x-3)(x-3)(x-3)(+4) \Rightarrow (x-3)^3(x+4)$ as the factored version of the polynomial. So I was under the impression that I should have been able to divide $\displaystyle x^3-3x+2$ by (x-1) then the resulting polynomial by (x-1).

As I am replying I see that would be the case if the factor resulted in $\displaystyle (x-1)^3(x+2)$ but not for $\displaystyle (x-1)^2(x+2)$ because it leaves a polynomial in standard form--which already has two factors (when factored) and my answer results in three (x-1)(x-1)(x+2).

The example you have given is correct. But the way you have explained your problem doesn't link up. If you are saying that you used the rational root test and found that x - 1 is a factor of the original polynomial, then divided it by x -1 and then found that x - 1 is a factor of the dividend then yes, it works. (You never said you did that.)

-Dan