Solve 0=(x-3)^2(x-4)

$$0 = ( x - 3 ) ^ { 2 } ( x - 4 )$$

Solve for x

$$x=4$$

$$x=3$$

Solution Steps

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

$$0=\left(x^{2}-6x+9\right)\left(x-4\right)$$

Use the distributive property to multiply x^{2}-6x+9 by x-4 and combine like terms.

$$0=x^{3}-10x^{2}+33x-36$$

Swap sides so that all variable terms are on the left hand side.

$$x^{3}-10x^{2}+33x-36=0$$

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

$$±36,±18,±12,±9,±6,±4,±3,±2,±1$$

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.

$$x=3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-10x^{2}+33x-36 by x-3 to get x^{2}-7x+12. Solve the equation where the result equals to 0.

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

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, -7 for b, and 12 for c in the quadratic formula.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 1\times 12}}{2}$$

Do the calculations.

$$x=\frac{7±1}{2}$$

Solve the equation x^{2}-7x+12=0 when ± is plus and when ± is minus.

$$x=3 x=4$$

List all found solutions.

$$x=3 x=4$$