Rational Zeros Theorem

If $LaTeX: \frac{p}{q}\:$ $\frac{p}{q}\:$ is a rational number written in lowest terms, and if $LaTeX: \frac{p}{q}\:$ $\frac{p}{q}\:$ is a zero of f, a polynomial function with integer coefficients, then p is a factor of the constant term and q is a factor of the leading coefficient.

Example

Consider the polynomial function $LaTeX: f\left(x\right)=6x^4+7x^3-12x^2-3x+2$ $f\left(x\right)=6x^4+7x^3-12x^2-3x+2$

List all possible rational zeros.
Find all rational zeros and factor f(x) into linear factors.

Solutions

For a rational number $LaTeX: \frac{p}{q}\:$ $\frac{p}{q}\:$ to be a zero, p must be a factor of $LaTeX: a_0=2\:$ $a_0=2\:$ and q can be a factor of $LaTeX: a_4=6.\:$ $a_4=6.\:$ Thus, p can be $LaTeX: \pm1\:$ $\pm1\:$ or $LaTeX: \pm2,\:$ $\pm2,\:$ and q can be $LaTeX: \pm1,\:\pm2,\:\pm3,\:$ $\pm1,\:\pm2,\:\pm3,\:$ or $LaTeX: \pm6.\:$ $\pm6.\:$ The possible rational zeros, $LaTeX: \frac{p}{q}\:$ $\frac{p}{q}\:$ are $LaTeX: \pm1,\:\pm2,\:\pm\frac{1}{2},\:\pm\frac{1}{3},\:\pm\frac{1}{6},\:$ $\pm1,\:\pm2,\:\pm\frac{1}{2},\:\pm\frac{1}{3},\:\pm\frac{1}{6},\:$ and $LaTeX: \pm\frac{2}{3}.$ $\pm\frac{2}{3}.$
Use the remainder theorem to show that 1 is a zero.

The 0 remainder shows that 1 is a zero. The quotient is $LaTeX: 6x^3+13x^2+x-2,\:$ $6x^3+13x^2+x-2,\:$ so $LaTeX: f\left(x\right)=\left(x-1\right)\left(6x^3+13x^2+x-2\right)$ $f\left(x\right)=\left(x-1\right)\left(6x^3+13x^2+x-2\right)$ .

Now, use the quotient polynomial and synthetic division to find that −2 is a zero.

The new quotient polynomial is $LaTeX: 6x^2+x-1,\:$ $6x^2+x-1,\:$ which factors into $LaTeX: \left(3x-1\right)\left(2x+1\right).$ $\left(3x-1\right)\left(2x+1\right).$
Therefore, f(x) can now be completely factored. Here is f(x) factored into linear factors:
$LaTeX: f\left(x\right)=\left(x-1\right)\left(x+2\right)\left(3x-1\right)\left(2x+1\right)$ $f\left(x\right)=\left(x-1\right)\left(x+2\right)\left(3x-1\right)\left(2x+1\right)$
Setting each of the linear factors for f(x) to zero and solving gives all of the rational zeros, which in this case are $LaTeX: 1,\:-2,\:\frac{1}{3},\:$ $1,\:-2,\:\frac{1}{3},\:$ and $LaTeX: -\frac{1}{2}.$ $-\frac{1}{2}.$

Note:

In this example when we found the quadratic factor $LaTeX: 6x^2+x-1,\:$ $6x^2+x-1,\:$ we were able to complete the factoring to linear factors directly. Had this polynomial not been easily factorable, we could have used the quadratic formula to find the other zeros (and factors).

Caution The rational zeros theorem gives only possible rational zeros. It does not tell us whether these rational numbers are actual zeros. We must rely on other methods to determine whether or not they are indeed zeros. Furthermore, the function must have integer coefficients. To apply the rational zeros theorem to a polynomial with fractional coefficients, multiply through by the least common denominator of all the fractions. For example, any rational zeros of p(x) defined below will also be rational zeros of q(x).

$LaTeX: p\left(x\right)=x^4-\frac{1}{6}x^3+\frac{2}{3}x^2-\frac{1}{6}x-\frac{1}{3}$ $p\left(x\right)=x^4-\frac{1}{6}x^3+\frac{2}{3}x^2-\frac{1}{6}x-\frac{1}{3}$

Multiply the terms of p(x) by 6:

$LaTeX: q\left(x\right)=6x^4-x^3+4x^2-x-2$ $q\left(x\right)=6x^4-x^3+4x^2-x-2$