Solve (3x-1)(x^2-9)-2x(x^2-9)=0

$$( 3 x - 1 ) ( x ^ { 2 } - 9 ) - 2 x ( x ^ { 2 } - 9 ) = 0$$

Solve for x

$$x=3$$

$$x=-3$$

$$x=1$$

Solution Steps

Use the distributive property to multiply 3x-1 by x^{2}-9.

$$3x^{3}-27x-x^{2}+9-2x\left(x^{2}-9\right)=0$$

Use the distributive property to multiply -2x by x^{2}-9.

$$3x^{3}-27x-x^{2}+9-2x^{3}+18x=0$$

Combine 3x^{3} and -2x^{3} to get x^{3}.

$$x^{3}-27x-x^{2}+9+18x=0$$

Combine -27x and 18x to get -9x.

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

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

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

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 9 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±9,±3,±1$$

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^{3}-x^{2}-9x+9 by x-1 to get x^{2}-9. Solve the equation where the result equals to 0.

$$x^{2}-9=0$$

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 -9 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\left(-9\right)}}{2}$$

Do the calculations.

$$x=\frac{0±6}{2}$$

Solve the equation x^{2}-9=0 when ± is plus and when ± is minus.

$$x=-3 x=3$$

List all found solutions.

$$x=1 x=-3 x=3$$