Find the zeroes of the polynomial 11x^3 - x^2 - 77x + 7

What we are looking for

The zeroes of a polynomial are the $x$-values that make the expression equal to $0$. So we want to solve $$11x^3-x^2-77x+7=0.$$

Testing for a rational root

Possible rational roots come from factors of $7$ over factors of $11$: $$\pm 1,\ \pm 7,\ \pm \frac{1}{11},\ \pm \frac{7}{11}.$$ Try $x=\frac{1}{11}$: $$ 11\left(\frac{1}{11}\right)^3-\left(\frac{1}{11}\right)^2-77\left(\frac{1}{11}\right)+7 =\frac{1}{121}-\frac{1}{121}-7+7=0. $$ So $x=\frac{1}{11}$ is a root, meaning $(11x-1)$ is a factor.

Dividing out the factor

Divide $11x^3-x^2-77x+7$ by $(11x-1)$ (synthetic division with $r=\frac{1}{11}$ works quickly) to get: $$11x^2-77.$$ So $$11x^3-x^2-77x+7=(11x-1)(11x^2-77).$$

Solving the remaining quadratic factor

Set the remaining factor to zero: $$11x^2-77=0 \quad\Rightarrow\quad x^2=7 \quad\Rightarrow\quad x=\pm\sqrt{7}.$$

So the zeroes are: $$x=\frac{1}{11},\ \sqrt{7},\ -\sqrt{7}.$$