Solve (3x^2-2x)^2+5=6(3x^2-2x)

$$( 3 x ^ { 2 } - 2 x ) ^ { 2 } + 5 = 6 ( 3 x ^ { 2 } - 2 x )$$

Solve for x

$$x=-1$$

$$x = \frac{5}{3} = 1\frac{2}{3} \approx 1.666666667$$

$$x=1$$

$$x=-\frac{1}{3}\approx -0.333333333$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x^{2}-2x\right)^{2}.

$$9\left(x^{2}\right)^{2}-12x^{2}x+4x^{2}+5=6\left(3x^{2}-2x\right)$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$9x^{4}-12x^{2}x+4x^{2}+5=6\left(3x^{2}-2x\right)$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$9x^{4}-12x^{3}+4x^{2}+5=6\left(3x^{2}-2x\right)$$

Use the distributive property to multiply 6 by 3x^{2}-2x.

$$9x^{4}-12x^{3}+4x^{2}+5=18x^{2}-12x$$

Subtract 18x^{2} from both sides.

$$9x^{4}-12x^{3}+4x^{2}+5-18x^{2}=-12x$$

Combine 4x^{2} and -18x^{2} to get -14x^{2}.

$$9x^{4}-12x^{3}-14x^{2}+5=-12x$$

Add 12x to both sides.

$$9x^{4}-12x^{3}-14x^{2}+5+12x=0$$

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

$$9x^{4}-12x^{3}-14x^{2}+12x+5=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 5 and q divides the leading coefficient 9. List all candidates \frac{p}{q}.

$$±\frac{5}{9},±\frac{5}{3},±5,±\frac{1}{9},±\frac{1}{3},±1$$

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 9x^{4}-12x^{3}-14x^{2}+12x+5 by x-1 to get 9x^{3}-3x^{2}-17x-5. Solve the equation where the result equals to 0.

$$9x^{3}-3x^{2}-17x-5=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -5 and q divides the leading coefficient 9. List all candidates \frac{p}{q}.

$$±\frac{5}{9},±\frac{5}{3},±5,±\frac{1}{9},±\frac{1}{3},±1$$

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 9x^{3}-3x^{2}-17x-5 by x+1 to get 9x^{2}-12x-5. Solve the equation where the result equals to 0.

$$9x^{2}-12x-5=0$$

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 9 for a, -12 for b, and -5 for c in the quadratic formula.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 9\left(-5\right)}}{2\times 9}$$

Do the calculations.

$$x=\frac{12±18}{18}$$

Solve the equation 9x^{2}-12x-5=0 when ± is plus and when ± is minus.

$$x=-\frac{1}{3} x=\frac{5}{3}$$

List all found solutions.

$$x=1 x=-1 x=-\frac{1}{3} x=\frac{5}{3}$$