# How do you find all the zeros of f(x)=x^4+7x^3-x^2-67x-60?

Jul 16, 2016

Zeros: $- 1 , 3 , - 4 , - 5$

#### Explanation:

$f \left(x\right) = {x}^{4} + 7 {x}^{3} - {x}^{2} - 67 x - 60$

By the rational root 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 $- 60$ and $q$ a divisor of the coefficient $1$ of the leading term.

So the only possible rational zeros are:

$\pm 1 , \pm 2 , \pm 3 , \pm 4 , \pm 5 , \pm 6 , \pm 10 , \pm 12 , \pm 15 , \pm 20 , \pm 30 , \pm 60$

We find:

$f \left(- 1\right) = 1 - 7 - 1 + 67 - 60 = 0$

So $- 1$ is a zero and $\left(x + 1\right)$ is a factor:

${x}^{4} + 7 {x}^{3} - {x}^{2} - 67 x - 60 = \left(x + 1\right) \left({x}^{3} + 6 {x}^{2} - 7 x - 60\right)$

Let $g \left(x\right) = {x}^{3} + 6 {x}^{2} - 7 x - 60$

We find:

$g \left(3\right) = 27 + 54 - 21 - 60 = 0$

So $3$ is a zero and $\left(x - 3\right)$ is a factor:

${x}^{3} + 6 {x}^{2} - 7 x - 60 = \left(x - 3\right) \left({x}^{2} + 9 x + 20\right)$

To factor the remaining quadratic, note that $4 + 5 = 9$ and $4 \times 5 = 20$, so we find:

${x}^{2} + 9 x + 20 = \left(x + 4\right) \left(x + 5\right)$

Hence remaining two zeros: $- 4$ and $- 5$.