Solve 12x^3+8x^2=x+1

$$12 x ^ { 3 } + 8 x ^ { 2 } = x + 1$$

Solve for x

$$x=-\frac{1}{2}=-0.5$$

$$x=\frac{1}{3}\approx 0.333333333$$

Solution Steps

Subtract x from both sides.

$$12x^{3}+8x^{2}-x=1$$

Subtract 1 from both sides.

$$12x^{3}+8x^{2}-x-1=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -1 and q divides the leading coefficient 12. List all candidates \frac{p}{q}.

$$±\frac{1}{12},±\frac{1}{6},±\frac{1}{4},±\frac{1}{3},±\frac{1}{2},±1$$

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.

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

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 12x^{3}+8x^{2}-x-1 by 3\left(x-\frac{1}{3}\right)=3x-1 to get 4x^{2}+4x+1. Solve the equation where the result equals to 0.

$$4x^{2}+4x+1=0$$

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 4 for a, 4 for b, and 1 for c in the quadratic formula.

$$x=\frac{-4±\sqrt{4^{2}-4\times 4\times 1}}{2\times 4}$$

Do the calculations.

$$x=\frac{-4±0}{8}$$

Solutions are the same.

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

List all found solutions.

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