Solve 63x ^ 3 + 6x ^ 2 - 12x + 8 = 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

$$63x^{3}+6x^{2}-12x+8=0$$

Solve for x (complex solution)

$x=-\frac{4\sqrt{3}i}{21}+\frac{2}{7}\approx 0.285714286-0.32991444i$
$x=-\frac{2}{3}\approx -0.666666667$
$x=\frac{4\sqrt{3}i}{21}+\frac{2}{7}\approx 0.285714286+0.32991444i$

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 $8$ and $q$ divides the leading coefficient $63$. List all candidates $\frac{p}{q}$.

$$±\frac{8}{63},±\frac{8}{21},±\frac{8}{9},±\frac{8}{7},±\frac{8}{3},±8,±\frac{4}{63},±\frac{4}{21},±\frac{4}{9},±\frac{4}{7},±\frac{4}{3},±4,±\frac{2}{63},±\frac{2}{21},±\frac{2}{9},±\frac{2}{7},±\frac{2}{3},±2,±\frac{1}{63},±\frac{1}{21},±\frac{1}{9},±\frac{1}{7},±\frac{1}{3},±1$$

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.

$$x=-\frac{2}{3}$$

By Factor theorem, $x-k$ is a factor of the polynomial for each root $k$. Divide $63x^{3}+6x^{2}-12x+8$ by $3\left(x+\frac{2}{3}\right)=3x+2$ to get $21x^{2}-12x+4$. Solve the equation where the result equals to $0$.

$$21x^{2}-12x+4=0$$

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 $21$ for $a$, $-12$ for $b$, and $4$ for $c$ in the quadratic formula.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 21\times 4}}{2\times 21}$$

Do the calculations.

$$x=\frac{12±\sqrt{-192}}{42}$$

Solve the equation $21x^{2}-12x+4=0$ when $±$ is plus and when $±$ is minus.

$$x=-\frac{4i\sqrt{3}}{21}+\frac{2}{7}$$ $$x=\frac{4i\sqrt{3}}{21}+\frac{2}{7}$$

List all found solutions.

$$x=-\frac{2}{3}$$ $$x=-\frac{4i\sqrt{3}}{21}+\frac{2}{7}$$ $$x=\frac{4i\sqrt{3}}{21}+\frac{2}{7}$$

Solve for x

$x=-\frac{2}{3}\approx -0.666666667$

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 $8$ and $q$ divides the leading coefficient $63$. List all candidates $\frac{p}{q}$.

$$±\frac{8}{63},±\frac{8}{21},±\frac{8}{9},±\frac{8}{7},±\frac{8}{3},±8,±\frac{4}{63},±\frac{4}{21},±\frac{4}{9},±\frac{4}{7},±\frac{4}{3},±4,±\frac{2}{63},±\frac{2}{21},±\frac{2}{9},±\frac{2}{7},±\frac{2}{3},±2,±\frac{1}{63},±\frac{1}{21},±\frac{1}{9},±\frac{1}{7},±\frac{1}{3},±1$$

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.

$$x=-\frac{2}{3}$$

By Factor theorem, $x-k$ is a factor of the polynomial for each root $k$. Divide $63x^{3}+6x^{2}-12x+8$ by $3\left(x+\frac{2}{3}\right)=3x+2$ to get $21x^{2}-12x+4$. Solve the equation where the result equals to $0$.

$$21x^{2}-12x+4=0$$

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 $21$ for $a$, $-12$ for $b$, and $4$ for $c$ in the quadratic formula.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 21\times 4}}{2\times 21}$$

Do the calculations.

$$x=\frac{12±\sqrt{-192}}{42}$$

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.

$$x=-\frac{2}{3}$$