# Without the use of the solve function of a calculator how do I solve the equation: x^4-5x^3-x^2+11x-30=0?

## Given that $\left(x - 5\right)$ and $\left(x + 2\right)$ are factors of $f \left(x\right)$

May 14, 2016

The zeros are $x = 5$, $x = - 2$, $x = 1 \pm \sqrt{2} i$

#### Explanation:

$f \left(x\right) = {x}^{4} - 5 {x}^{3} - {x}^{2} + 11 x - 30$

We are told that $\left(x - 5\right)$ is a factor, so separate it out:

${x}^{4} - 5 {x}^{3} - {x}^{2} + 11 x - 30 = \left(x - 5\right) \left({x}^{3} - x + 6\right)$

We are told that $\left(x + 2\right)$ is also a factor, so separate that out:

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

The discriminant of the remaining quadratic factor is negative, but we can still use the quadratic formula to find the Complex roots:

${x}^{2} - 2 x + 3$ is in the form $a {x}^{2} + b x + c$ with $a = 1$, $b = - 2$ and $c = 3$.

The roots are given by the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$= \frac{2 \pm \sqrt{{\left(- 2\right)}^{2} - \left(4 \cdot 1 \cdot 3\right)}}{2 \cdot 1}$

$= \frac{2 \pm \sqrt{4 - 12}}{2}$

$= \frac{2 \pm \sqrt{- 8}}{2}$

$= \frac{2 \pm \sqrt{8} i}{2}$

$= \frac{2 \pm 2 \sqrt{2} i}{2}$

$= 1 \pm \sqrt{2} i$

May 15, 2016

Let us try without knowing that $\left(x - 5\right)$ and $\left(x + 2\right)$ are factors.
The constant term equals the roots product,so
$30 = {r}_{1} \cdot {r}_{2} \cdot {r}_{3} \cdot {r}_{4}$.

This coefficient is an integer value whose factors are $\pm 1 , \pm 2 , \pm 5 , \pm 3$ Trying those values we can see that
$p \left(- 2\right) = p \left(5\right) = 0$ obtaining two roots.
We can represent the polynomial as
x^4 - 5 x^3 - x^2 + 11 x - 30=(x-5)(x+2)(x² + a x + b)

Calculating the right side and comparing both sides we obtain
$- 5 = a - 3$
$- 1 = b - 3 a - 10$
$11 = - 10 a - 3 b$
$- 30 = - 10 b$

Solving for $\left(a , b\right)$ we get $a = - 2 , b = 3$
Evaluating the roots of ${x}^{2} - 2 x + 3 = 0$ we get $1 - i \sqrt{2} , 1 + i \sqrt{2}$