Solve x^2(x+6)-x(x+6)=30(x+6)

$$x ^ { 2 } ( x + 6 ) - x ( x + 6 ) = 30 ( x + 6 )$$

Solve for x

$$x=-5$$

$$x=6$$

$$x=-6$$

Solution Steps

Use the distributive property to multiply x^{2} by x+6.

$$x^{3}+6x^{2}-x\left(x+6\right)=30\left(x+6\right)$$

Use the distributive property to multiply x by x+6.

$$x^{3}+6x^{2}-\left(x^{2}+6x\right)=30\left(x+6\right)$$

To find the opposite of x^{2}+6x, find the opposite of each term.

$$x^{3}+6x^{2}-x^{2}-6x=30\left(x+6\right)$$

Combine 6x^{2} and -x^{2} to get 5x^{2}.

$$x^{3}+5x^{2}-6x=30\left(x+6\right)$$

Use the distributive property to multiply 30 by x+6.

$$x^{3}+5x^{2}-6x=30x+180$$

Subtract 30x from both sides.

$$x^{3}+5x^{2}-6x-30x=180$$

Combine -6x and -30x to get -36x.

$$x^{3}+5x^{2}-36x=180$$

Subtract 180 from both sides.

$$x^{3}+5x^{2}-36x-180=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -180 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±180,±90,±60,±45,±36,±30,±20,±18,±15,±12,±10,±9,±6,±5,±4,±3,±2,±1$$

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.

$$x=-5$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+5x^{2}-36x-180 by x+5 to get x^{2}-36. Solve the equation where the result equals to 0.

$$x^{2}-36=0$$

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, 0 for b, and -36 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\left(-36\right)}}{2}$$

Do the calculations.

$$x=\frac{0±12}{2}$$

Solve the equation x^{2}-36=0 when ± is plus and when ± is minus.

$$x=-6 x=6$$

List all found solutions.

$$x=-5 x=-6 x=6$$