Solve ((x-3)-5)left((x-3)+1right)^2=x^2-10x+16

$$((x-3)-5) { \left((x-3)+1 \right) }^{ 2 } = { x }^{ 2 } -10x+16$$

Solve for x

$$x=3$$

$$x=8$$

$$x=2$$

Solution Steps

Subtract 5 from -3 to get -8.

$$\left(x-8\right)\left(x-3+1\right)^{2}=x^{2}-10x+16$$

Add -3 and 1 to get -2.

$$\left(x-8\right)\left(x-2\right)^{2}=x^{2}-10x+16$$

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

$$\left(x-8\right)\left(x^{2}-4x+4\right)=x^{2}-10x+16$$

Use the distributive property to multiply x-8 by x^{2}-4x+4 and combine like terms.

$$x^{3}-12x^{2}+36x-32=x^{2}-10x+16$$

Subtract x^{2} from both sides.

$$x^{3}-12x^{2}+36x-32-x^{2}=-10x+16$$

Combine -12x^{2} and -x^{2} to get -13x^{2}.

$$x^{3}-13x^{2}+36x-32=-10x+16$$

Add 10x to both sides.

$$x^{3}-13x^{2}+36x-32+10x=16$$

Combine 36x and 10x to get 46x.

$$x^{3}-13x^{2}+46x-32=16$$

Subtract 16 from both sides.

$$x^{3}-13x^{2}+46x-32-16=0$$

Subtract 16 from -32 to get -48.

$$x^{3}-13x^{2}+46x-48=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -48 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±48,±24,±16,±12,±8,±6,±4,±3,±2,±1$$

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.

$$x=2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-13x^{2}+46x-48 by x-2 to get x^{2}-11x+24. Solve the equation where the result equals to 0.

$$x^{2}-11x+24=0$$

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, -11 for b, and 24 for c in the quadratic formula.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 1\times 24}}{2}$$

Do the calculations.

$$x=\frac{11±5}{2}$$

Solve the equation x^{2}-11x+24=0 when ± is plus and when ± is minus.

$$x=3 x=8$$

List all found solutions.

$$x=2 x=3 x=8$$