Solve x^3-4x=-5x^2+20

$$x ^ { 3 } - 4 x = - 5 x ^ { 2 } + 20$$

Solve for x

$$x=-2$$

$$x=2$$

$$x=-5$$

Solution Steps

Add 5x^{2} to both sides.

$$x^{3}-4x+5x^{2}=20$$

Subtract 20 from both sides.

$$x^{3}-4x+5x^{2}-20=0$$

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

$$x^{3}+5x^{2}-4x-20=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -20 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±20,±10,±5,±4,±2,±1$$

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.

$$x=2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+5x^{2}-4x-20 by x-2 to get x^{2}+7x+10. Solve the equation where the result equals to 0.

$$x^{2}+7x+10=0$$

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, 7 for b, and 10 for c in the quadratic formula.

$$x=\frac{-7±\sqrt{7^{2}-4\times 1\times 10}}{2}$$

Do the calculations.

$$x=\frac{-7±3}{2}$$

Solve the equation x^{2}+7x+10=0 when ± is plus and when ± is minus.

$$x=-5 x=-2$$

List all found solutions.

$$x=2 x=-5 x=-2$$