How should you go about simplying cubic polynomial: $y(x) = x^3+12x^2+21x+10$

Question

Claim: $$y(x) = x^3+12x^2+21x+10$$

Can be factored into $$(x+1)^2(x+10)$$

But what is the quickest way to see this?

Answer 1

You can use the rational roots theorem, in this case it yields that the roots must be integers that divide 10. The problem is then that there are quite a few possible roots. However, you are free to define a new variable $t$ by putting $t = x - u$ for arbitrary $u$ and apply the rational roots theorem to the polynomial as a function of $t$. Adding $u$ to these possible roots yields the possible roots of the original polynomial as a function of $x$, however the list you find is not the same as the original one. This then means that only candidates that appear on both list can be the roots of the polynomial.

An efficient way to go about this is to test the root candidates, in case you find a root $p$, you perform a synthetic division to factor out $x-p$ from the polynomial, if the root candidate doesn't work and the value of the polynomial is $y\neq 0$, you factor $y$ and add the value of the root candidate to the factors and take the intersection of the old list of root candidates to get to a new list of root candidates.

So, in this case, we have the root candidates:

$$x = 1, -1, 2, -2, 5,-5,10,-10$$

If we try $x = 1$ which we know cannot be a root since it will yields a positive $y$, we find that it yields $y = 44$

Adding $1$ to the list of factors of $44$ yields:

$x = 3, -1, 5, -3, 12, -10$

The intersection with the old list is:

$x = -1, 5, -10$

Then obviously $x = 5$ is not going to be a root, so you are left with $x = -1$ and $x = -10$. If it is given that all roots are rational roots, then since the polynomial is of third degree, one of them has to be a double or triple root. Since in this case the constant term equals 10, you must have the factorization $(x+1)^2 (x+10)$.

You can also use that in case of a double root, the derivative will also be zero at that point. Because y'(0) = 21, we see that the double root must be a divisor of 21, and that can thus only be $x = -1$.

Answer 2

In general, if we're trying to look for rational roots, we can use the Rational Root Theorem to get a set of possible roots of the polynomial and then test which one are actually roots.

For example, if you use the rational root theorem for the given polynomial, you get that if a rational root exists, it must be one of the following:

$$x=\pm\frac{1,2,5,10}{1}$$

Now, use the Remainder-Factor Theorem to test each of these to get which one are actually the roots. Now, if we get that $x=\dfrac{p}{q}$ is one rational root of the polynomial, we have a factor of the polynomial as $(qx-p)$ by the Remainder-Factor Theorem.

Note: It works only for rational roots.