Solve x^3-2x^2/x^2+4x-12=-13x-42/x+6

$$\frac { x ^ { 3 } - 2 x ^ { 2 } } { x ^ { 2 } + 4 x - 12 } = \frac { - 13 x - 42 } { x + 6 }$$

Solve for x

$$x=-7$$

Solution Steps

Variable x cannot be equal to any of the values -6,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(x+6\right), the least common multiple of x^{2}+4x-12,x+6.

$$x^{3}-2x^{2}=\left(x-2\right)\left(-13x-42\right)$$

Use the distributive property to multiply x-2 by -13x-42 and combine like terms.

$$x^{3}-2x^{2}=-13x^{2}-16x+84$$

Add 13x^{2} to both sides.

$$x^{3}-2x^{2}+13x^{2}=-16x+84$$

Combine -2x^{2} and 13x^{2} to get 11x^{2}.

$$x^{3}+11x^{2}=-16x+84$$

Add 16x to both sides.

$$x^{3}+11x^{2}+16x=84$$

Subtract 84 from both sides.

$$x^{3}+11x^{2}+16x-84=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -84 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±84,±42,±28,±21,±14,±12,±7,±6,±4,±3,±2,±1$$

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.

$$x=2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+11x^{2}+16x-84 by x-2 to get x^{2}+13x+42. Solve the equation where the result equals to 0.

$$x^{2}+13x+42=0$$

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, 13 for b, and 42 for c in the quadratic formula.

$$x=\frac{-13±\sqrt{13^{2}-4\times 1\times 42}}{2}$$

Do the calculations.

$$x=\frac{-13±1}{2}$$

Solve the equation x^{2}+13x+42=0 when ± is plus and when ± is minus.

$$x=-7 x=-6$$

Remove the values that the variable cannot be equal to.

$$x=-7$$

List all found solutions.

$$x=2 x=-7 x=-6$$

Variable x cannot be equal to any of the values 2,-6.

$$x=-7$$