Solve (x ^ 3 - 8)/(x - 2) = 9 - 2x | 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

$$\frac{x^{3}-8}{x-2}=9-2x$$

Solve for x

$x=-5$
$x=1$

Solution Steps

Variable $x$ cannot be equal to $2$ since division by zero is not defined. Multiply both sides of the equation by $x-2$.

$$x^{3}-8=\left(x-2\right)\times 9-2x\left(x-2\right)$$

Use the distributive property to multiply $x-2$ by $9$.

$$x^{3}-8=9x-18-2x\left(x-2\right)$$

Use the distributive property to multiply $-2x$ by $x-2$.

$$x^{3}-8=9x-18-2x^{2}+4x$$

Combine $9x$ and $4x$ to get $13x$.

$$x^{3}-8=13x-18-2x^{2}$$

Subtract $13x$ from both sides.

$$x^{3}-8-13x=-18-2x^{2}$$

Subtract $-18$ from both sides.

$$x^{3}-8-13x-\left(-18\right)=-2x^{2}$$

The opposite of $-18$ is $18$.

$$x^{3}-8-13x+18=-2x^{2}$$

Add $2x^{2}$ to both sides.

$$x^{3}-8-13x+18+2x^{2}=0$$

Add $-8$ and $18$ to get $10$.

$$x^{3}+10-13x+2x^{2}=0$$

Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{3}+2x^{2}-13x+10=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form $\frac{p}{q}$, where $p$ divides the constant term $10$ and $q$ divides the leading coefficient $1$. List all candidates $\frac{p}{q}$.

$$±10,±5,±2,±1$$

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 $x^{3}+2x^{2}-13x+10$ by $x-1$ to get $x^{2}+3x-10$. Solve the equation where the result equals to $0$.

$$x^{2}+3x-10=0$$

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 $a$, $3$ for $b$, and $-10$ for $c$ in the quadratic formula.

$$x=\frac{-3±\sqrt{3^{2}-4\times 1\left(-10\right)}}{2}$$

Do the calculations.

$$x=\frac{-3±7}{2}$$

Solve the equation $x^{2}+3x-10=0$ when $±$ is plus and when $±$ is minus.

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

Remove the values that the variable cannot be equal to.

$$x=1\text{ or }x=-5$$

List all found solutions.

$$x=1$$ $$x=-5$$ $$x=2$$

Variable $x$ cannot be equal to $2$.

$$x=-5$$ $$x=1$$