Solve 3x ^ 3 - 2x + 7; x = - 8 3 | 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

$$3 x ^ { 3 } - 2 x + 7 \quad x = - 8$$

Solve for x (complex solution)

$x=-\frac{\sqrt{87}i}{6}+\frac{1}{2}\approx 0.5-1.554563176i$
$x=-1$
$x=\frac{\sqrt{87}i}{6}+\frac{1}{2}\approx 0.5+1.554563176i$

Solution Steps

Combine $-2x$ and $7x$ to get $5x$.

$$3x^{3}+5x=-8$$

Add $8$ to both sides.

$$3x^{3}+5x+8=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form $\frac{p}{q}$, where $p$ divides the constant term $8$ and $q$ divides the leading coefficient $3$. List all candidates $\frac{p}{q}$.

$$±\frac{8}{3},±8,±\frac{4}{3},±4,±\frac{2}{3},±2,±\frac{1}{3},±1$$

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 $3x^{3}+5x+8$ by $x+1$ to get $3x^{2}-3x+8$. Solve the equation where the result equals to $0$.

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

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 $3$ for $a$, $-3$ for $b$, and $8$ for $c$ in the quadratic formula.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 3\times 8}}{2\times 3}$$

Do the calculations.

$$x=\frac{3±\sqrt{-87}}{6}$$

Solve the equation $3x^{2}-3x+8=0$ when $±$ is plus and when $±$ is minus.

$$x=-\frac{\sqrt{87}i}{6}+\frac{1}{2}$$ $$x=\frac{\sqrt{87}i}{6}+\frac{1}{2}$$

List all found solutions.

$$x=-1$$ $$x=-\frac{\sqrt{87}i}{6}+\frac{1}{2}$$ $$x=\frac{\sqrt{87}i}{6}+\frac{1}{2}$$

Solve for x

$x=-1$

Solution Steps

Combine $-2x$ and $7x$ to get $5x$.

$$3x^{3}+5x=-8$$

Add $8$ to both sides.

$$3x^{3}+5x+8=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form $\frac{p}{q}$, where $p$ divides the constant term $8$ and $q$ divides the leading coefficient $3$. List all candidates $\frac{p}{q}$.

$$±\frac{8}{3},±8,±\frac{4}{3},±4,±\frac{2}{3},±2,±\frac{1}{3},±1$$

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 $3x^{3}+5x+8$ by $x+1$ to get $3x^{2}-3x+8$. Solve the equation where the result equals to $0$.

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

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 $3$ for $a$, $-3$ for $b$, and $8$ for $c$ in the quadratic formula.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 3\times 8}}{2\times 3}$$

Do the calculations.

$$x=\frac{3±\sqrt{-87}}{6}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$x\in \emptyset$$

List all found solutions.

$$x=-1$$