How do you find all the zeros of f(x)=x^4+4x^3-6x^2-36x-27?

Aug 13, 2016

$f \left(x\right)$ has zeros:

$- 1$ (with multiplicity $1$)

$3$ (with multiplicity $1$)

$- 3$ (with multiplicity $2$)

Explanation:

$f \left(x\right) = {x}^{4} + 4 {x}^{3} - 6 {x}^{2} - 36 x - 27$

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 $- 27$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1 , \pm 3 , \pm 9 , \pm 27$

We find:

$f \left(- 1\right) = 1 - 4 - 6 + 36 - 27 = 0$

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

${x}^{4} + 4 {x}^{3} - 6 {x}^{2} - 36 x - 27 = \left(x + 1\right) \left({x}^{3} + 3 {x}^{2} - 9 x - 27\right)$

In the remaining cubic, the ratio of the first and second terms is the same as that between the third and fourth terms. So this cubic will factor by grouping:

${x}^{3} + 3 {x}^{2} - 9 x - 27$

$= \left({x}^{3} + 3 {x}^{2}\right) - \left(9 x + 27\right)$

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

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

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

$= \left(x - 3\right) \left(x + 3\right) \left(x + 3\right)$

So the remaining zeros are $x = 3$ (with multiplicity $1$) and $x = - 3$ with multiplicity $2$.