Solve x ^ 3 - 4x ^ 2 - 9x + 36 = 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

$$x^{3}-4x^{2}-9x+36=0$$

Solve for x

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

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 $36$ and $q$ divides the leading coefficient $1$. List all candidates $\frac{p}{q}$.

$$±36,±18,±12,±9,±6,±4,±3,±2,±1$$

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.

$$x=3$$

By Factor theorem, $x-k$ is a factor of the polynomial for each root $k$. Divide $x^{3}-4x^{2}-9x+36$ by $x-3$ 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.

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

Do the calculations.

$$x=\frac{1±7}{2}$$

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

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

List all found solutions.

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