# Thread: factor a cubic polynomial

1. ## factor a cubic polynomial

How do I factor
$y^3- 18y^2+107y-216$

to equal $(y-8)(y^2-10y+27)=0$

2. ## Re: factor a cubic polynomial

you have $P(y) = y^3- 18y^2+107y-216$

Then find $a$ such that $P(a) = 0$ meaning $y-a$ will be a factor, then divide $P(y)$ by $y-a$

Spoiler:
Try $y=8$

3. ## Re: factor a cubic polynomial

Unfortunately, I'm not sure what you mean. I am familiar with factoring by grouping and MCD. Do these two methods not apply?

4. ## Re: factor a cubic polynomial

Are you familiar with the Rational Roots Theorem? Because of the Rational Roots Theorem, the only possible roots of the given polynomial are factors of 216. One of those factors is 8, which turns out to be a root and the reason you can factor it as you describe.

5. ## Re: factor a cubic polynomial

We need to use the factor theorem when faced with polynomials of order 3 or higher.

It states that for a polynomial $\displaystyle P(y)$ if $\displaystyle P(a)=0$ , then $\displaystyle y-a$ is a linear factor of $\displaystyle P(y)$.

In your case $\displaystyle P(y) = y^3- 18y^2+107y-216$ substitute $\displaystyle y=8$ gives $\displaystyle P(8) = 512-1152+856-216 = 0$ therefore $\displaystyle y-8$ is a factor.

So now all you need to do is find $\displaystyle \frac{y^3- 18y^2+107y-216}{y-8} = \cdots$

and you will have your solution.

6. ## Re: factor a cubic polynomial

We are given to factor:

$y^3-18y^2+107y-216$

Looking at the coefficient of the squared term, we find that if the coefficient of the linear term is 80, then we can factor the first 3 terms. We should also observe that 107 - 80 = 27 and 216/27 = 8, which is one of the factors of 80 we are using.

$y^3-18y^2+80y+27y-216$

$y(y-8)(y-10)+27(y-8)$

$(y-8)\left(y(y-10)+27\right)$

$(y-8)\left(y^2-10y+27\right)$

7. ## Re: factor a cubic polynomial

Originally Posted by icemanfan
Are you familiar with the Rational Roots Theorem? Because of the Rational Roots Theorem, the only possible roots of the given polynomial are factors of 216. One of those factors is 8, which turns out to be a root and the reason you can factor it as you describe.
Originally Posted by pickslides
We need to use the factor theorem when faced with polynomials of order 3 or higher.

It states that for a polynomial $\displaystyle P(y)$ if $\displaystyle P(a)=0$ , then $\displaystyle y-a$ is a linear factor of $\displaystyle P(y)$.

In your case $\displaystyle P(y) = y^3- 18y^2+107y-216$ substitute $\displaystyle y=8$ gives $\displaystyle P(8) = 512-1152+856-216 = 0$ therefore $\displaystyle y-8$ is a factor.

So now all you need to do is find $\displaystyle \frac{y^3- 18y^2+107y-216}{y-8} = \cdots$

and you will have your solution.

Ah, I see now that I needed to use the Integer Value Theorem. Now what you saying is making sense