# How do you find the zeros of y=x^3-2x^2-11x+12?

Mar 31, 2018

The zeros are $1$, $4$ and $- 3$

#### Explanation:

Given:

${x}^{3} - 2 {x}^{2} - 11 x + 12$

Sum of coefficients shortcut

Note that the sum of the coefficients is $0$. That is:

$1 - 2 - 11 + 12 = 0$

Hence, $x = 1$ is a zero and $\left(x - 1\right)$ a factor:

${x}^{3} - 2 {x}^{2} - 11 x + 12 = \left(x - 1\right) \left({x}^{2} - x - 12\right)$

Fishing for factors

To factor the remaining quadratic we want a pair of factors of $12$ that differ by $1$. The pair $4 , 3$ works, so we find:

${x}^{2} - x - 12 = \left(x - 4\right) \left(x + 3\right)$

So the zeros of the given cubic are $1$, $4$ and $- 3$

So the given example cubic proved to be not too difficult, but what might we do if the sum of coefficients shortcut didn't apply?

Rational zeros theorem

Given any polynomial with integer coefficients in standard form, any rational zeros are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term and $q$ a divisor of the coefficient of the leading term.

In our example, any rational zeros would take the form $\frac{p}{q}$ with $p$ a divisor of $12$ and $q$ a divisor of $1$, so one of:

$\pm 1 , \pm 2 , \pm 3 , \pm 4 , \pm 6 , \pm 12$

We could try substituting each in turn until we find a zero, then divide the polynomial by the corresponding factor as before.