Solve (x-2)(x^2+6x+9)=6(x+3)

$$( x - 2 ) ( x ^ { 2 } + 6 x + 9 ) = 6 ( x + 3 )$$

Solve for x

$$x=-3$$

$$x=3$$

$$x=-4$$

Solution Steps

Use the distributive property to multiply x-2 by x^{2}+6x+9 and combine like terms.

$$x^{3}+4x^{2}-3x-18=6\left(x+3\right)$$

Use the distributive property to multiply 6 by x+3.

$$x^{3}+4x^{2}-3x-18=6x+18$$

Subtract 6x from both sides.

$$x^{3}+4x^{2}-3x-18-6x=18$$

Combine -3x and -6x to get -9x.

$$x^{3}+4x^{2}-9x-18=18$$

Subtract 18 from both sides.

$$x^{3}+4x^{2}-9x-18-18=0$$

Subtract 18 from -18 to get -36.

$$x^{3}+4x^{2}-9x-36=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -36 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±36,±18,±12,±9,±6,±4,±3,±2,±1$$

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.

$$x=3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+4x^{2}-9x-36 by x-3 to get x^{2}+7x+12. Solve the equation where the result equals to 0.

$$x^{2}+7x+12=0$$

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 12 for c in the quadratic formula.

$$x=\frac{-7±\sqrt{7^{2}-4\times 1\times 12}}{2}$$

Do the calculations.

$$x=\frac{-7±1}{2}$$

Solve the equation x^{2}+7x+12=0 when ± is plus and when ± is minus.

$$x=-4 x=-3$$

List all found solutions.

$$x=3 x=-4 x=-3$$