Solve x ^ 3 + 5x ^ 2 - 4x - 20 = 0 . | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$x ^ { 3 } + 5 x ^ { 2 } - 4 x - 20 = 0$$

Solve for x

$x=-2$
$x=2$
$x=-5$

Solution Steps

By Rational Root Theorem, all rational roots of a polynomial are in the form $\frac{p}{q}$, where $p$ divides the constant term $-20$ and $q$ divides the leading coefficient $1$. List all candidates $\frac{p}{q}$.

$$±20,±10,±5,±4,±2,±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=2$$

By Factor theorem, $x-k$ is a factor of the polynomial for each root $k$. Divide $x^{3}+5x^{2}-4x-20$ by $x-2$ to get $x^{2}+7x+10$. Solve the equation where the result equals to $0$.

$$x^{2}+7x+10=0$$

All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. Substitute $1$ for $a$, $7$ for $b$, and $10$ for $c$ in the quadratic formula.

$$x=\frac{-7±\sqrt{7^{2}-4\times 1\times 10}}{2}$$

Do the calculations.

$$x=\frac{-7±3}{2}$$

Solve the equation $x^{2}+7x+10=0$ when $±$ is plus and when $±$ is minus.

$$x=-5$$ $$x=-2$$

List all found solutions.

$$x=2$$ $$x=-5$$ $$x=-2$$