How do I factor
$\displaystyle y^3- 18y^2+107y-216$
to equal $\displaystyle (y-8)(y^2-10y+27)=0$
you have $\displaystyle P(y) = y^3- 18y^2+107y-216$
Then find $\displaystyle a$ such that $\displaystyle P(a) = 0$ meaning $\displaystyle y-a$ will be a factor, then divide $\displaystyle P(y)$ by $\displaystyle y-a$
Spoiler:
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.
We need to use the factor theorem when faced with polynomials of order 3 or higher.
It states that for a polynomial $\displaystyle \displaystyle P(y)$ if $\displaystyle \displaystyle P(a)=0$ , then $\displaystyle \displaystyle y-a$ is a linear factor of $\displaystyle \displaystyle P(y)$.
In your case $\displaystyle \displaystyle P(y) = y^3- 18y^2+107y-216$ substitute $\displaystyle \displaystyle y=8$ gives $\displaystyle \displaystyle P(8) = 512-1152+856-216 = 0$ therefore $\displaystyle \displaystyle y-8$ is a factor.
So now all you need to do is find $\displaystyle \displaystyle \frac{y^3- 18y^2+107y-216}{y-8} = \cdots$
and you will have your solution.
We are given to factor:
$\displaystyle y^3-18y^2+107y-216$
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.
$\displaystyle y^3-18y^2+80y+27y-216$
$\displaystyle y(y-8)(y-10)+27(y-8)$
$\displaystyle (y-8)\left(y(y-10)+27\right)$
$\displaystyle (y-8)\left(y^2-10y+27\right)$