How to factor a four term polynomial without grouping?

Question

$$2x^3 + 9x^2 +7x -6$$

This equation doesn't factor by grouping, and other than that I have no idea how to solve this problem. Will someone please help?

Answer 1

The way to factor a four-term polynomial like this is to apply Rational Root Theorem along with synthetic division or substitution to determine whether a rational root works for the polynomial or not.

Here is how Rational Root Theorem works. The number of rational roots is determined by the quotient of the factors of the last term and the factors of the first term. That is:

$$\{\pm \text{factors of last term} / \text{factors of first term}\} = \{\pm 1, 2, 3, 6 / 1, 2\}$$

So the possible rational roots are:

$$\pm 1, \pm2, \pm3, \pm6, \pm1/2, \pm3/2$$

Select one of the rational roots and apply synthetic division. The number of times you need to test a root for the polynomial depends on whether you get $0$ for evaluating the polynomial or factored polynomial at some rational root or not. This is done computationally.

For example, suppose that $x = -3$ is one of the rational roots. Evaluate the polynomial at $x = -3$, so we have $0$. This means that the given polynomial has the factor $(x + 3)$, so we can apply synthetic division to obtain the factored polynomial, which is $2x^2 + 3x - 2$.

You can either work out via Rational Root Theorem or by simply brain work of factorizing that polynomial. Try this one with Rational Root Theorem and check that factored polynomial has two rational roots.

Answer 2

Use the Rational Root Theorem to find possible candidates for roots. For example, $-2$ is a candidate, and it is indeed a zero of your polynomial (the polynomial is equal to zero when $x = -2$), hence $(x - (-2)) = (x + 2)$ is a factor.

Now use synthetic division on the polynomial, call it $p(x)$, to obtain the quotient when divided by $(x + 2)$.

You may simply be able to simply "eye up" the polynomial-quotient, this time with the resulting quadratic equation to $p(x)/(x+2).\;$ If the factorization of the quadratic is readily apparent, then you'll have three factors in all: $(x+ 2)$ and the factors of the quadratic.

Otherwise, use the coefficients of the quadratic and the Rational Root Theorem to find a second zero/root, $x = b$, and use polynomial division to divide the quadratic by the second factor $(x - b)$, to find the resulting quotient, which will be the "last" factor. If all had gone well, (and it should, in this case), you'll now havethree factors in all!

Answer 3

A quick way to determine if a polynomial like this one has any rational roots is by the rather easily proven theorem that for a polynomial with integer coefficients, a rational root $\frac{p}{q}$, in lowest terms, must satisfy that $p$ divides the free coefficient, and $q$ must divide the leading coefficient. This gives you a very short list of candidates for rational roots, and you can just try them one-by-one. If you find a root, then proceed to divide the polynomial and try to find more roots. If none of the rational candidates you found is a root (this is not the case for the polynomial you are considering now) then some other methods can be used.

There are general criteria for irreducibility as well as the (rather nasty, but workable) formulas for the solution of third and fourth degree polynomials with complex (hence real (hence integer)) coefficients. Again, you don't need any of that for your polynomial.