Solve x ^ 3 - 3x ^ 2 - 8x - 4 = 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

$$x^{3}-3x^{2}-8x-4=0$$

Solve for x

$x=2\sqrt{2}+2\approx 4.828427125$
$x=2-2\sqrt{2}\approx -0.828427125$
$x=-1$

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 $-4$ and $q$ divides the leading coefficient $1$. List all candidates $\frac{p}{q}$.

$$±4,±2,±1$$

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.

$$x=-1$$

By Factor theorem, $x-k$ is a factor of the polynomial for each root $k$. Divide $x^{3}-3x^{2}-8x-4$ by $x+1$ to get $x^{2}-4x-4$. Solve the equation where the result equals to $0$.

$$x^{2}-4x-4=0$$

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 $a$, $-4$ for $b$, and $-4$ for $c$ in the quadratic formula.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\left(-4\right)}}{2}$$

Do the calculations.

$$x=\frac{4±4\sqrt{2}}{2}$$

Solve the equation $x^{2}-4x-4=0$ when $±$ is plus and when $±$ is minus.

$$x=2-2\sqrt{2}$$ $$x=2\sqrt{2}+2$$

List all found solutions.

$$x=-1$$ $$x=2-2\sqrt{2}$$ $$x=2\sqrt{2}+2$$