Solve 4x^3-9x^2+6x-8=0

$$4 { x }^{ 3 } -9 { x }^{ 2 } +6x-8=0$$

Solve for x (complex solution)

$$x=\frac{-3\sqrt{7}i+1}{8}\approx 0.125-0.992156742i$$

$$x=2$$

$$x=\frac{1+3\sqrt{7}i}{8}\approx 0.125+0.992156742i$$

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

$$±2,±4,±8,±1,±\frac{1}{2},±\frac{1}{4}$$

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.

$$x=2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 4x^{3}-9x^{2}+6x-8 by x-2 to get 4x^{2}-x+4. Solve the equation where the result equals to 0.

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

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

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

Do the calculations.

$$x=\frac{1±\sqrt{-63}}{8}$$

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

$$x=\frac{-3i\sqrt{7}+1}{8} x=\frac{1+3i\sqrt{7}}{8}$$

List all found solutions.

$$x=2 x=\frac{-3i\sqrt{7}+1}{8} x=\frac{1+3i\sqrt{7}}{8}$$

Solve for x

$$x=2$$

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 -8 and q divides the leading coefficient 4. List all candidates \frac{p}{q}.

$$±2,±4,±8,±1,±\frac{1}{2},±\frac{1}{4}$$

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.

$$x=2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 4x^{3}-9x^{2}+6x-8 by x-2 to get 4x^{2}-x+4. Solve the equation where the result equals to 0.

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

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

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

Do the calculations.

$$x=\frac{1±\sqrt{-63}}{8}$$

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.

$$x=2$$