# How do you find all the zeros of f(x)=x^4+5x^3+5x^2-5x-6?

Mar 15, 2016

Look at coefficient sums and divide by the factors found to simplify the problem and find zeros:

$x = 1$, $x = - 1$, $x = - 2$ and $x = - 3$

#### Explanation:

First note that the sum of the coefficients is zero.

That is: $1 + 5 + 5 - 5 - 6 = 0$

So $f \left(1\right) = 0$ and $\left(x - 1\right)$ is a factor:

${x}^{4} + 5 {x}^{3} + 5 {x}^{2} - 5 x - 6 = \left(x - 1\right) \left({x}^{3} + 6 {x}^{2} + 11 x + 6\right)$

Next note the if you reverse the signs of the terms of the remaining cubic factor with odd degree then the sum of the coefficients is zero.

That is $- 1 + 6 - 11 + 6 = 0$

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

${x}^{3} + 6 {x}^{2} + 11 x + 6 = \left(x + 1\right) \left({x}^{2} + 5 x + 6\right)$

Then note that $2 + 3 = 5$ and $2 \times 3 = 6$, so the remaining quadratic factor factorises as follows:

${x}^{2} + 5 x + 6 = \left(x + 2\right) \left(x + 3\right)$

Putting this all together, we find:

$f \left(x\right) = \left(x - 1\right) \left(x + 1\right) \left(x + 2\right) \left(x + 3\right)$

with zeros $x = 1$, $x = - 1$, $x = - 2$ and $x = - 3$