How to solve (x)^(3)-13x+12=0

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve (x)^(3)-13x+12=0 | Equatix

Basic Math

Pre-Algebra

Algebra

Trigonometry

Calculus

Precalculus

Statistics

Finite Math

Linear Algebra

Chemistry

Question

$${ x }^{ 3 } -13x+12=0$$

Solve for x

$x=1$
$x=3$
$x=-4$

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 $12$ and $q$ divides the leading coefficient $1$. List all candidates $\frac{p}{q}$.

$$±12,±6,±4,±3,±2,±1$$

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 $x^{3}-13x+12$ by $x-1$ to get $x^{2}+x-12$. Solve the equation where the result equals to $0$.

$$x^{2}+x-12=0$$

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 $a$, $1$ for $b$, and $-12$ for $c$ in the quadratic formula.

$$x=\frac{-1±\sqrt{1^{2}-4\times 1\left(-12\right)}}{2}$$

Do the calculations.

$$x=\frac{-1±7}{2}$$

Solve the equation $x^{2}+x-12=0$ when $±$ is plus and when $±$ is minus.

$$x=-4$$ $$x=3$$

List all found solutions.

$$x=1$$ $$x=-4$$ $$x=3$$

Show Solution