Solve x*(x-2)^2*(x-4)=45

$$x \cdot ( x - 2 ) ^ { 2 } \cdot ( x - 4 ) = 45$$

Solve for x (complex solution)

$$x=-1$$

$$x=5$$

$$x=-\sqrt{5}i+2\approx 2-2.236067977i$$

$$x=2+\sqrt{5}i\approx 2+2.236067977i$$

Solution Steps

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

$$x\left(x^{2}-4x+4\right)\left(x-4\right)=45$$

Use the distributive property to multiply x by x^{2}-4x+4.

$$\left(x^{3}-4x^{2}+4x\right)\left(x-4\right)=45$$

Use the distributive property to multiply x^{3}-4x^{2}+4x by x-4 and combine like terms.

$$x^{4}-8x^{3}+20x^{2}-16x=45$$

Subtract 45 from both sides.

$$x^{4}-8x^{3}+20x^{2}-16x-45=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -45 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±45,±15,±9,±5,±3,±1$$

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^{4}-8x^{3}+20x^{2}-16x-45 by x+1 to get x^{3}-9x^{2}+29x-45. Solve the equation where the result equals to 0.

$$x^{3}-9x^{2}+29x-45=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -45 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±45,±15,±9,±5,±3,±1$$

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.

$$x=5$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-9x^{2}+29x-45 by x-5 to get x^{2}-4x+9. Solve the equation where the result equals to 0.

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

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, -4 for b, and 9 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{4±\sqrt{-20}}{2}$$

Solve the equation x^{2}-4x+9=0 when ± is plus and when ± is minus.

$$x=-\sqrt{5}i+2 x=2+\sqrt{5}i$$

List all found solutions.

$$x=-1 x=5 x=-\sqrt{5}i+2 x=2+\sqrt{5}i$$

Solve for x

$$x=-1$$

$$x=5$$

Solution Steps

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

$$x\left(x^{2}-4x+4\right)\left(x-4\right)=45$$

Use the distributive property to multiply x by x^{2}-4x+4.

$$\left(x^{3}-4x^{2}+4x\right)\left(x-4\right)=45$$

Use the distributive property to multiply x^{3}-4x^{2}+4x by x-4 and combine like terms.

$$x^{4}-8x^{3}+20x^{2}-16x=45$$

Subtract 45 from both sides.

$$x^{4}-8x^{3}+20x^{2}-16x-45=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -45 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±45,±15,±9,±5,±3,±1$$

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^{4}-8x^{3}+20x^{2}-16x-45 by x+1 to get x^{3}-9x^{2}+29x-45. Solve the equation where the result equals to 0.

$$x^{3}-9x^{2}+29x-45=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -45 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±45,±15,±9,±5,±3,±1$$

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.

$$x=5$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-9x^{2}+29x-45 by x-5 to get x^{2}-4x+9. Solve the equation where the result equals to 0.

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

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, -4 for b, and 9 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{4±\sqrt{-20}}{2}$$

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.

$$x=-1 x=5$$