# How do you find all the zeros of f(x) = x^4 – x^3 + 7x^2 – 9x – 18?

May 30, 2016

${p}_{4} \left(x\right) = \left(x + 1\right) \left(x - 2\right) \left(x + 3 i\right) \left(x - 3 i\right)$

#### Explanation:

The constant coefficient divided by the maximum power coefficient in a polynomial, gives the product of all its roots.

So $\frac{18}{1} = {x}_{1} {x}_{2} {x}_{3} {x}_{4}$ for a polynomial represented as

${p}_{4} \left(x\right) = \left(x - {x}_{1}\right) \left(x - {x}_{2}\right) \left(x - {x}_{3}\right) \left(x - {x}_{4}\right)$

Supposing integer roots and testing for $\pm 1 , \pm 2 , \pm 3 , \pm 6$
we obtain:

${p}_{4} \left(- 1\right) = {p}_{4} \left(2\right) = 0$ so we have at least a root for each $1$ and $- 1$
then we can do

${p}_{4} \left(x\right) = \left(x + 1\right) \left(x - 2\right) \left(x - {x}_{3}\right) \left(x - {x}_{4}\right) = {x}^{4} - {x}^{3} + 7 {x}^{2} - 9 x - 18$

Now dividing $\frac{{x}^{4} - {x}^{3} + 7 {x}^{2} - 9 x - 18}{\left(x + 1\right) \left(x - 2\right)} = {x}^{2} + 9$

so all the roots are

${p}_{4} \left(x\right) = \left(x + 1\right) \left(x - 2\right) \left(x + 3 i\right) \left(x - 3 i\right)$