Solve (x^2-3x)(x-1)(x-2)-24=0

$$( { x }^{ 2 } -3x)(x-1)(x-2)-24=0$$

Solve for x (complex solution)

$$x=-1$$

$$x=4$$

$$x=\frac{-\sqrt{15}i+3}{2}\approx 1.5-1.936491673i$$

$$x=\frac{3+\sqrt{15}i}{2}\approx 1.5+1.936491673i$$

Solution Steps

Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.

$$\left(x-2\right)\left(x-1\right)\left(x^{2}-3x\right)-24=0$$

Multiply and combine like terms.

$$x^{4}-6x^{3}+11x^{2}-6x-24=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -24 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±24,±12,±8,±6,±4,±3,±2,±1$$

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 x^{4}-6x^{3}+11x^{2}-6x-24 by x+1 to get x^{3}-7x^{2}+18x-24. Solve the equation where the result equals to 0.

$$x^{3}-7x^{2}+18x-24=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -24 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±24,±12,±8,±6,±4,±3,±2,±1$$

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.

$$x=4$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-7x^{2}+18x-24 by x-4 to get x^{2}-3x+6. Solve the equation where the result equals to 0.

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

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, -3 for b, and 6 for c in the quadratic formula.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 1\times 6}}{2}$$

Do the calculations.

$$x=\frac{3±\sqrt{-15}}{2}$$

Solve the equation x^{2}-3x+6=0 when ± is plus and when ± is minus.

$$x=\frac{-\sqrt{15}i+3}{2} x=\frac{3+\sqrt{15}i}{2}$$

List all found solutions.

$$x=-1 x=4 x=\frac{-\sqrt{15}i+3}{2} x=\frac{3+\sqrt{15}i}{2}$$

Solve for x

$$x=-1$$

$$x=4$$

Solution Steps

Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.

$$\left(x-2\right)\left(x-1\right)\left(x^{2}-3x\right)-24=0$$

Multiply and combine like terms.

$$x^{4}-6x^{3}+11x^{2}-6x-24=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -24 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±24,±12,±8,±6,±4,±3,±2,±1$$

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 x^{4}-6x^{3}+11x^{2}-6x-24 by x+1 to get x^{3}-7x^{2}+18x-24. Solve the equation where the result equals to 0.

$$x^{3}-7x^{2}+18x-24=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -24 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±24,±12,±8,±6,±4,±3,±2,±1$$

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.

$$x=4$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-7x^{2}+18x-24 by x-4 to get x^{2}-3x+6. Solve the equation where the result equals to 0.

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

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, -3 for b, and 6 for c in the quadratic formula.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 1\times 6}}{2}$$

Do the calculations.

$$x=\frac{3±\sqrt{-15}}{2}$$

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.

$$x=-1 x=4$$