Solve (x-2)^3-25(x-2)=0

$$( x - 2 ) ^ { 3 } - 25 ( x - 2 ) = 0$$

Solve for x

$$x=2$$

$$x=7$$

$$x=-3$$

Solution Steps

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

$$x^{3}-6x^{2}+12x-8-25\left(x-2\right)=0$$

Use the distributive property to multiply -25 by x-2.

$$x^{3}-6x^{2}+12x-8-25x+50=0$$

Combine 12x and -25x to get -13x.

$$x^{3}-6x^{2}-13x-8+50=0$$

Add -8 and 50 to get 42.

$$x^{3}-6x^{2}-13x+42=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 42 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±42,±21,±14,±7,±6,±3,±2,±1$$

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}-6x^{2}-13x+42 by x-2 to get x^{2}-4x-21. Solve the equation where the result equals to 0.

$$x^{2}-4x-21=0$$

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 -21 for c in the quadratic formula.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\left(-21\right)}}{2}$$

Do the calculations.

$$x=\frac{4±10}{2}$$

Solve the equation x^{2}-4x-21=0 when ± is plus and when ± is minus.

$$x=-3 x=7$$

List all found solutions.

$$x=2 x=-3 x=7$$