# How do you solve x^4-10x^3+27x^2-2x-40>=0?

Jul 21, 2016

$p \left(x\right) \ge 0$ for $\left(x \le - 1\right) \mathmr{and} \left(2 \le x \le 4\right) \mathmr{and} \left(5 < x\right)$

#### Explanation:

Analyzing the constant term which is the product of all roots we get

$40 = 1 \cdot {2}^{3} \cdot 5$. Trying for $\pm 1 , \pm 2 , \pm 4 , \pm 5$ we get

$p \left(x\right) = {x}^{4} - 10 {x}^{3} + 27 {x}^{2} - 2 x - 40$
$p \left(x\right) = \left(x + 1\right) \left(x - 2\right) \left(x - 4\right) \left(x - 5\right) \ge 0$

Ordering the roots

$- 1 , 2 , 4 , 5$

and knowing that for $x < - 1 \to p \left(x\right) > 0$ we have by continuity

$- 1 < x < 2 \to p \left(x\right) < 0$
$2 < x < 4 \to p \left(x\right) > 0$
$4 < x < 5 \to p \left(x\right) < 0$
$5 < x \to p \left(x\right) > 0$

so

$p \left(x\right) \ge 0$ for $\left(x \le - 1\right) \mathmr{and} \left(2 \le x \le 4\right) \mathmr{and} \left(5 < x\right)$

Jul 21, 2016

Solution set for ${x}^{4} - 10 {x}^{3} + 27 {x}^{2} - 2 x - 40 \ge$ is $x \le - 1$ and $2 \le x \le 4$ and $5 < x$.

#### Explanation:

Let us first factorize $f \left(x\right) = {x}^{4} - 10 {x}^{3} + 27 {x}^{2} - 2 x - 40$. As $- 1$ is a zero of the function as

$f \left(0\right) = {\left(- 1\right)}^{4} - 10 {\left(- 1\right)}^{3} + 27 {\left(- 1\right)}^{2} - 2 \left(- 1\right) - 40 = 1 + 10 + 27 + 2 - 40 = 0$, $\left(x + 1\right)$ is a factor of $f \left(x\right)$.

Also $f \left(2\right) = {2}^{4} - 10 \cdot {2}^{3} + 27 \cdot {2}^{2} - 2 \cdot 2 - 40 = 16 - 80 + 108 - 4 - 40 = 0$, hence $\left(x - 2\right)$ to is a factor of $f \left(x\right)$.

Dividing f(x))=x^4-10x^3+27x^2-2x-40 by $\left(x + 1\right)$ and then by $\left(x - 2\right)$, we get ${x}^{2} - 9 x + 20$ which can be further factorized to $\left(x - 4\right) \left(x - 5\right)$ ad hence, the inequality we have is

$\left(x + 1\right) \left(x - 2\right) \left(x - 4\right) \left(x - 5\right) \ge 0$

Now, these zeros divide real number line in five parts

(1) $x < - 1$ - Here all the terms are negative, hence $f \left(x\right)$ is positive. This forms part of solution.

(2) $- 1 < x < 2$ - Here while first term $\left(x + 1\right)$ is positive, all other terms are negative and hence $f \left(x\right)$ is negative. Hence, this does not form part of solution.

(3) $2 < x < 4$ - Here while first two terms $\left(x + 1\right)$ and $\left(x - 2\right)$ are positive, other two terms are negative and hence $f \left(x\right)$ is positive. Hence, this forms part of solution.

(4) $4 < x < 5$ - Here while first three term are positive, the last term $\left(x - 5\right)$ is negative and hence $f \left(x\right)$ is negative. Hence, this does not form part of solution.

(5) $5 < x$ - Here all the terms are positive and hence $f \left(x\right)$ is positive. Hence, this forms part of solution.

Hence solution set for ${x}^{4} - 10 {x}^{3} + 27 {x}^{2} - 2 x - 40 \ge$ is $x \le - 1$ and $2 \le x \le 4$ and $5 < x$.

graph{x^4-10x^3+27x^2-2x-40 [-10, 10, -40, 40]}