Solve 2x^3-12x^2+12/left(x-4right)^2=3

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

Solve for x (complex solution)

$$x=\frac{-\sqrt{39}i+3}{4}\approx 0.75-1.5612495i$$

$$x=6$$

$$x=\frac{3+\sqrt{39}i}{4}\approx 0.75+1.5612495i$$

Solution Steps

Variable x cannot be equal to 4 since division by zero is not defined. Multiply both sides of the equation by \left(x-4\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

$$2x^{3}-12x^{2}+12=3\left(x^{2}-8x+16\right)$$

Use the distributive property to multiply 3 by x^{2}-8x+16.

$$2x^{3}-12x^{2}+12=3x^{2}-24x+48$$

Subtract 3x^{2} from both sides.

$$2x^{3}-12x^{2}+12-3x^{2}=-24x+48$$

Combine -12x^{2} and -3x^{2} to get -15x^{2}.

$$2x^{3}-15x^{2}+12=-24x+48$$

Add 24x to both sides.

$$2x^{3}-15x^{2}+12+24x=48$$

Subtract 48 from both sides.

$$2x^{3}-15x^{2}+12+24x-48=0$$

Subtract 48 from 12 to get -36.

$$2x^{3}-15x^{2}-36+24x=0$$

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

$$2x^{3}-15x^{2}+24x-36=0$$

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 2. List all candidates \frac{p}{q}.

$$±18,±36,±9,±6,±12,±\frac{9}{2},±3,±2,±4,±\frac{3}{2},±1,±\frac{1}{2}$$

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.

$$x=6$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{3}-15x^{2}+24x-36 by x-6 to get 2x^{2}-3x+6. Solve the equation where the result equals to 0.

$$2x^{2}-3x+6=0$$

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 2 for a, -3 for b, and 6 for c in the quadratic formula.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 2\times 6}}{2\times 2}$$

Do the calculations.

$$x=\frac{3±\sqrt{-39}}{4}$$

Solve the equation 2x^{2}-3x+6=0 when ± is plus and when ± is minus.

$$x=\frac{-\sqrt{39}i+3}{4} x=\frac{3+\sqrt{39}i}{4}$$

List all found solutions.

$$x=6 x=\frac{-\sqrt{39}i+3}{4} x=\frac{3+\sqrt{39}i}{4}$$

Solve for x

$$x=6$$

Solution Steps

Variable x cannot be equal to 4 since division by zero is not defined. Multiply both sides of the equation by \left(x-4\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

$$2x^{3}-12x^{2}+12=3\left(x^{2}-8x+16\right)$$

Use the distributive property to multiply 3 by x^{2}-8x+16.

$$2x^{3}-12x^{2}+12=3x^{2}-24x+48$$

Subtract 3x^{2} from both sides.

$$2x^{3}-12x^{2}+12-3x^{2}=-24x+48$$

Combine -12x^{2} and -3x^{2} to get -15x^{2}.

$$2x^{3}-15x^{2}+12=-24x+48$$

Add 24x to both sides.

$$2x^{3}-15x^{2}+12+24x=48$$

Subtract 48 from both sides.

$$2x^{3}-15x^{2}+12+24x-48=0$$

Subtract 48 from 12 to get -36.

$$2x^{3}-15x^{2}-36+24x=0$$

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

$$2x^{3}-15x^{2}+24x-36=0$$

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 2. List all candidates \frac{p}{q}.

$$±18,±36,±9,±6,±12,±\frac{9}{2},±3,±2,±4,±\frac{3}{2},±1,±\frac{1}{2}$$

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.

$$x=6$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{3}-15x^{2}+24x-36 by x-6 to get 2x^{2}-3x+6. Solve the equation where the result equals to 0.

$$2x^{2}-3x+6=0$$

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 2 for a, -3 for b, and 6 for c in the quadratic formula.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 2\times 6}}{2\times 2}$$

Do the calculations.

$$x=\frac{3±\sqrt{-39}}{4}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$x\in \emptyset$$

List all found solutions.

$$x=6$$