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

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

Solve for x

$$x=3$$

$$x=2$$

Solution Steps

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

$$\left(x^{2}-4x+4\right)\left(x-3\right)=0$$

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

$$x^{3}-7x^{2}+16x-12=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -12 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±12,±6,±4,±3,±2,±1$$

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}-7x^{2}+16x-12 by x-2 to get x^{2}-5x+6. Solve the equation where the result equals to 0.

$$x^{2}-5x+6=0$$

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 1 for a, -5 for b, and 6 for c in the quadratic formula.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 1\times 6}}{2}$$

Do the calculations.

$$x=\frac{5±1}{2}$$

Solve the equation x^{2}-5x+6=0 when ± is plus and when ± is minus.

$$x=2 x=3$$

List all found solutions.

$$x=2 x=3$$