# Solve the inequality -3x-2+-2x^2+x^4%0

## {x^4-2x^2-3x-2}\geq {0}

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$-x^{2}-x-2+x^{4}-x^{5}+2x^2+2x^{3}\geq 0$

## Step by step solution

Problem

${x^4-2x^2-3x-2}\geq {0}$
1

We can factor the polynomial $-2-3x-2x^2+x^4$ using synthetic division (Ruffini's rule). We search for a root in the factors of the constant term $-2$ and we found that $-1$ is a root of the polynomial

$-2-1\left(-3\right)+{\left(-1\right)}^2\left(-2\right)+{\left(-1\right)}^4=0$
2

Let's divide the polynomial by $x+1$ using synthetic division. First, write the coefficients of the terms of the numerator in descending order. Then, take the first coefficient $1$ and multiply by the factor $-1$. Add the result to the second coefficient and then multiply this by $-1$ and so on

$\left|\begin{array}{c}1 & 0 & -2 & -3 & -2 \\ & -1 & 1 & 1 & 2 \\ 1 & -1 & -1 & -2 & 0\end{array}\right|-1$
3

In the last row of the division appear the new coefficients, with remainder equals zero. Now, rewrite the polynomial (a degree less) with the new coefficients, and multiplied by the factor $x+1$

$\left(1+x\right)\left(x^{3}-x^{2}-x-2\right)\geq 0$
4

Multiplying polynomials $-x\cdot x$ and $-2+-x$

$x^{3}-x^{2}-x-2+x\cdot xx^{2}-x\cdot xx^{3}+xx^2+2x^2\geq 0$
5

When multiplying exponents with same base you can add the exponents

$x^{3}-x^{2}-x-2+xx^{\left(1+2\right)}-xx^{\left(1+3\right)}+x^{3}+2x^2\geq 0$
6

Add the values $3$ and $1$

$x^{3}-x^{2}-x-2+xx^{3}-xx^{4}+x^{3}+2x^2\geq 0$
7

When multiplying exponents with same base you can add the exponents

$x^{3}-x^{2}-x-2+x^{\left(1+3\right)}-x^{5}+x^{3}+2x^2\geq 0$
8

Add the values $3$ and $1$

$x^{3}-x^{2}-x-2+x^{4}-x^{5}+x^{3}+2x^2\geq 0$
9

Adding $x^{3}$ and $x^{3}$

$-x^{2}-x-2+x^{4}-x^{5}+2x^2+2x^{3}\geq 0$

$-x^{2}-x-2+x^{4}-x^{5}+2x^2+2x^{3}\geq 0$

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