Solve x^3-2x^2-5=4

$${ x }^{ 3 } -2 { x }^{ 2 } -5 = 4$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{11}i-1}{2}\approx -0.5-1.658312395i$$

$$x=3$$

$$x=\frac{-1+\sqrt{11}i}{2}\approx -0.5+1.658312395i$$

Solution Steps

Subtract 4 from both sides.

$$x^{3}-2x^{2}-5-4=0$$

Subtract 4 from -5 to get -9.

$$x^{3}-2x^{2}-9=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -9 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±9,±3,±1$$

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.

$$x=3$$

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

$$x^{2}+x+3=0$$

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 1 for a, 1 for b, and 3 for c in the quadratic formula.

$$x=\frac{-1±\sqrt{1^{2}-4\times 1\times 3}}{2}$$

Do the calculations.

$$x=\frac{-1±\sqrt{-11}}{2}$$

Solve the equation x^{2}+x+3=0 when ± is plus and when ± is minus.

$$x=\frac{-\sqrt{11}i-1}{2} x=\frac{-1+\sqrt{11}i}{2}$$

List all found solutions.

$$x=3 x=\frac{-\sqrt{11}i-1}{2} x=\frac{-1+\sqrt{11}i}{2}$$

Solve for x

$$x=3$$

Solution Steps

Subtract 4 from both sides.

$$x^{3}-2x^{2}-5-4=0$$

Subtract 4 from -5 to get -9.

$$x^{3}-2x^{2}-9=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -9 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±9,±3,±1$$

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.

$$x=3$$

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

$$x^{2}+x+3=0$$

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 1 for a, 1 for b, and 3 for c in the quadratic formula.

$$x=\frac{-1±\sqrt{1^{2}-4\times 1\times 3}}{2}$$

Do the calculations.

$$x=\frac{-1±\sqrt{-11}}{2}$$

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.

$$x=3$$