Solve 2x ^ 3 - x ^ 2 - 7x + 6 = 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

$$2x^{3}-x^{2}-7x+6=0$$

Solve for x

$x=1$
$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$
$x=-2$

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 $6$ and $q$ divides the leading coefficient $2$. List all candidates $\frac{p}{q}$.

$$±3,±6,±\frac{3}{2},±1,±2,±\frac{1}{2}$$

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 $2x^{3}-x^{2}-7x+6$ by $x-1$ to get $2x^{2}+x-6$. Solve the equation where the result equals to $0$.

$$2x^{2}+x-6=0$$

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 $2$ for $a$, $1$ for $b$, and $-6$ for $c$ in the quadratic formula.

$$x=\frac{-1±\sqrt{1^{2}-4\times 2\left(-6\right)}}{2\times 2}$$

Do the calculations.

$$x=\frac{-1±7}{4}$$

Solve the equation $2x^{2}+x-6=0$ when $±$ is plus and when $±$ is minus.

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

List all found solutions.

$$x=1$$ $$x=-2$$ $$x=\frac{3}{2}$$