Solve 4x^3+4x^2-x=1

$$4 x ^ { 3 } + 4 x ^ { 2 } - x = 1$$

Solve for x

$$x=-1$$

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

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

Solution Steps

Subtract 1 from both sides.

$$4x^{3}+4x^{2}-x-1=0$$

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 4. List all candidates \frac{p}{q}.

$$±\frac{1}{4},±\frac{1}{2},±1$$

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.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 4x^{3}+4x^{2}-x-1 by x+1 to get 4x^{2}-1. Solve the equation where the result equals to 0.

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

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, 0 for b, and -1 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 4\left(-1\right)}}{2\times 4}$$

Do the calculations.

$$x=\frac{0±4}{8}$$

Solve the equation 4x^{2}-1=0 when ± is plus and when ± is minus.

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

List all found solutions.

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