# How do you find the roots of x^3-x^2-34x-56=0?

Nov 8, 2016

This cubic equation has roots $7 , - 2 , - 4$.

#### Explanation:

$f \left(x\right) = {x}^{3} - {x}^{2} - 34 x - 56$

By the rational roots theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 56$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1 , \pm 2 , \pm 4 , \pm 7 , \pm 8 , \pm 14 , \pm 28 , \pm 56$

In addition, note that the pattern of signs of the coefficients of the terms is $+ - - -$. By Descartes' Rule of Signs, since this has only one change of sign, $f \left(x\right)$ has exactly one positive Real zero.

If $x \ge 1$ then ${x}^{2} + 34 x + 56 \ge 1 + 34 + 56 = 91$. So that positive Real zero must satisfy ${x}^{3} \ge 91$. So we can immediately rule out $1 , 2 , 4$ as possibilities. Try $x = 7$:

$f \left(7\right) = {7}^{3} - {7}^{2} - 34 \left(7\right) - 56 = 343 - 49 - 238 - 56 = 0$

So $x = 7$ is a zero and $\left(x - 7\right)$ a factor:

${x}^{3} - {x}^{2} - 34 x - 56 = \left(x - 7\right) \left({x}^{2} + 6 x + 8\right)$

We can factor the remaining quadratic by finding a pair of numbers with sum $6$ and product $8$. The pair $2 , 4$ works, so we have:

${x}^{2} + 6 x + 8 = \left(x + 2\right) \left(x + 4\right)$

Putting it all together:

${x}^{3} - {x}^{2} - 34 x - 56 = \left(x - 7\right) \left(x + 2\right) \left(x + 4\right)$

with zeros $7 , - 2 , - 4$