Finding the zeroes in a function I have come across a problem on my trigonometry homework where we need to find the zeros of a function without the use of a calculator. 
The Equation:
Given one of the zeros is $x=5$
$f(x) = x^3+2x^2-23x-60$
Is there a more efficient way to go about this problem than simply factor it out?
 A: The sum of the roots is $-2$, so the sum of the missing roots is $-7$. The product of the roots is $60$, so the product of the missing roots is $12$. So the missing roots satisfy the equation $x^2+7x+12=0$.
Is this simpler? Probably not. But different.
A: Yes if one of the roots is $5$ it means $x-5$ is factor of the given polynomial.Dividing the polynomial by $x-5$ we get the quadratic equation $x^2+7x+12$. Now the roots of this quadratic equation are $\frac{-7+(49-4*12)^{\frac{1}{2}}}{2}$ and $\frac{-7-(49-4*12)^{\frac{1}{2}}}{2}$. Thus the remaining two roots are $-4$ and $-3$. In general if we have a quadratic equation $ax^2+bx+c$ it has roots $\frac{-b \pm \sqrt{b^2-4ac}}{2a}$.
For dividing $x^3+2x^2-23x-60$ by $x-5$ take a look at Polynomial Long Division.
A: Since you're given a root, factoring is probably the most efficient way to solve this problem: polynomial long division isn't going to be too tricky in this case, and since you're dividing a cubic $x^3 + 2x^2 -23x - 60$ by the linear polynomial $x-5$, you'll end up with a quadratic, which you could then use the quadratic formula on or factor to get the other two roots.
But, let's say you weren't given the root $x=5$, or just wanted to find another way to solve this.  What then?
Well, you could use the cubic equation, but that's probably overkill.  If it were me, I'd like to try integer (or, at worst, rational) roots before I got my hands messy with a bunch of radicals.  Fortunately, there's a way I can do just that: the rational root theorem tells me what any possible rational root will look like, so I just have to try those.  For your polynomial, the possible roots are $$x =\pm 60, \pm 30, \pm 20, \pm 15, \pm 10, \pm 6, \pm 5, \pm 4, \pm 3, \pm 2, \pm 1$$
Of course, this is a lot of roots to check, but you could still conceivably do it.  And you only need to find one root before you can factor down to a quadratic and use the quadratic formula (or just factor).
