Is there any faster way to factorize this polynomial $p(x) = x^3-8x^2+19x-12$?

Question

Let $p(x) = x^3-8x^2+19x-12$. By RRT it follows that $(x-1)$ is a factor. After that doing long division I get $$(x-1)(x^2-7x+12)$$ $$(x-1)(x-3)(x-4).$$

But is there any faster way to do it?

Answer 1

Since the sum of the coefficients equals zero, $x = 1$ is a solution.

Then we can take advantage of such fact to obtain the desired result: \begin{align*} p(x) & = x^{3} - 8x^{2} + 19x - 12\\\\ & = (x^{3} - x^{2}) - (7x^{2} - 7x) + (12x - 12)\\\\ & = x^{2}(x - 1) - 7x(x - 1) + 12(x - 1)\\\\ & = (x^{2} - 7x + 12)(x - 1)\\\\ & = [(x^{2} - 4x) - (3x - 12)](x - 1)\\\\ & = [x(x - 4) - 3(x - 4)](x - 1)\\\\ & = (x - 3)(x - 4)(x - 1) \end{align*}

This procedure has the benefit of not being based on the long division algorithm neither on the RRT.

This is the fastest way I can think of.

Answer 2

You can find all the factors using the rational root theorem (RRT). Recall that the theorem states that if $x = \frac{p}{q}$ is a rational solution, written in lowest terms so that $p$ and $q$ are relatively prime, then $p$ is a factor of the constant term and $q$ is a factor of the leading coefficient.

Note that the leading coefficient is $1$ and the only factor of $1$ is $1$. The possible factors of the constant term are $\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12$. At this point, you just have to try the roots. Luckily, if you find one that works such as $x = 1$, then you can factorise (maybe use synthetic division which is a little quicker than long division) to consider a polynomial of lower degree and reduce your problem.