Solve 4x^3-3x^2-1=0

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

Solve for x (complex solution)

$$x=\frac{-\sqrt{15}i-1}{8}\approx -0.125-0.484122918i$$

$$x=1$$

$$x=\frac{-1+\sqrt{15}i}{8}\approx -0.125+0.484122918i$$

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 -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}-3x^{2}-1 by x-1 to get 4x^{2}+x+1. Solve the equation where the result equals to 0.

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

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

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

Do the calculations.

$$x=\frac{-1±\sqrt{-15}}{8}$$

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

$$x=\frac{-\sqrt{15}i-1}{8} x=\frac{-1+\sqrt{15}i}{8}$$

List all found solutions.

$$x=1 x=\frac{-\sqrt{15}i-1}{8} x=\frac{-1+\sqrt{15}i}{8}$$

Solve for x

$$x=1$$

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 -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}-3x^{2}-1 by x-1 to get 4x^{2}+x+1. Solve the equation where the result equals to 0.

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

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

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

Do the calculations.

$$x=\frac{-1±\sqrt{-15}}{8}$$

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.

$$x=1$$