Solve 1(x-3)(x-4)(x-5)=(x-2)(x-4)(x-5)

$$1 ( x - 3 ) ( x - 4 ) ( x - 5 ) = ( x - 2 ) ( x - 4 ) ( x - 5 )$$

Solve for x

$$x=4$$

$$x=5$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 1 by x-3.

$$\left(x-3\right)\left(x-4\right)\left(x-5\right)=\left(x-2\right)\left(x-4\right)\left(x-5\right)$$

Use the distributive property to multiply x-3 by x-4 and combine like terms.

$$\left(x^{2}-7x+12\right)\left(x-5\right)=\left(x-2\right)\left(x-4\right)\left(x-5\right)$$

Use the distributive property to multiply x^{2}-7x+12 by x-5 and combine like terms.

$$x^{3}-12x^{2}+47x-60=\left(x-2\right)\left(x-4\right)\left(x-5\right)$$

Use the distributive property to multiply x-2 by x-4 and combine like terms.

$$x^{3}-12x^{2}+47x-60=\left(x^{2}-6x+8\right)\left(x-5\right)$$

Use the distributive property to multiply x^{2}-6x+8 by x-5 and combine like terms.

$$x^{3}-12x^{2}+47x-60=x^{3}-11x^{2}+38x-40$$

Subtract x^{3} from both sides.

$$x^{3}-12x^{2}+47x-60-x^{3}=-11x^{2}+38x-40$$

Combine x^{3} and -x^{3} to get 0.

$$-12x^{2}+47x-60=-11x^{2}+38x-40$$

Add 11x^{2} to both sides.

$$-12x^{2}+47x-60+11x^{2}=38x-40$$

Combine -12x^{2} and 11x^{2} to get -x^{2}.

$$-x^{2}+47x-60=38x-40$$

Subtract 38x from both sides.

$$-x^{2}+47x-60-38x=-40$$

Combine 47x and -38x to get 9x.

$$-x^{2}+9x-60=-40$$

Add 40 to both sides.

$$-x^{2}+9x-60+40=0$$

Add -60 and 40 to get -20.

$$-x^{2}+9x-20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 9 for b, and -20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-9±\sqrt{9^{2}-4\left(-1\right)\left(-20\right)}}{2\left(-1\right)}$$

Square 9.

$$x=\frac{-9±\sqrt{81-4\left(-1\right)\left(-20\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-9±\sqrt{81+4\left(-20\right)}}{2\left(-1\right)}$$

Multiply 4 times -20.

$$x=\frac{-9±\sqrt{81-80}}{2\left(-1\right)}$$

Add 81 to -80.

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

Take the square root of 1.

$$x=\frac{-9±1}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-9±1}{-2}$$

Now solve the equation x=\frac{-9±1}{-2} when ± is plus. Add -9 to 1.

$$x=-\frac{8}{-2}$$

Divide -8 by -2.

$$x=4$$

Now solve the equation x=\frac{-9±1}{-2} when ± is minus. Subtract 1 from -9.

$$x=-\frac{10}{-2}$$

Divide -10 by -2.

$$x=5$$

The equation is now solved.

$$x=4 x=5$$

Steps for Completing the Square

Use the distributive property to multiply 1 by x-3.

$$\left(x-3\right)\left(x-4\right)\left(x-5\right)=\left(x-2\right)\left(x-4\right)\left(x-5\right)$$

Use the distributive property to multiply x-3 by x-4 and combine like terms.

$$\left(x^{2}-7x+12\right)\left(x-5\right)=\left(x-2\right)\left(x-4\right)\left(x-5\right)$$

Use the distributive property to multiply x^{2}-7x+12 by x-5 and combine like terms.

$$x^{3}-12x^{2}+47x-60=\left(x-2\right)\left(x-4\right)\left(x-5\right)$$

Use the distributive property to multiply x-2 by x-4 and combine like terms.

$$x^{3}-12x^{2}+47x-60=\left(x^{2}-6x+8\right)\left(x-5\right)$$

Use the distributive property to multiply x^{2}-6x+8 by x-5 and combine like terms.

$$x^{3}-12x^{2}+47x-60=x^{3}-11x^{2}+38x-40$$

Subtract x^{3} from both sides.

$$x^{3}-12x^{2}+47x-60-x^{3}=-11x^{2}+38x-40$$

Combine x^{3} and -x^{3} to get 0.

$$-12x^{2}+47x-60=-11x^{2}+38x-40$$

Add 11x^{2} to both sides.

$$-12x^{2}+47x-60+11x^{2}=38x-40$$

Combine -12x^{2} and 11x^{2} to get -x^{2}.

$$-x^{2}+47x-60=38x-40$$

Subtract 38x from both sides.

$$-x^{2}+47x-60-38x=-40$$

Combine 47x and -38x to get 9x.

$$-x^{2}+9x-60=-40$$

Add 60 to both sides.

$$-x^{2}+9x=-40+60$$

Add -40 and 60 to get 20.

$$-x^{2}+9x=20$$

Divide both sides by -1.

$$\frac{-x^{2}+9x}{-1}=\frac{20}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{9}{-1}x=\frac{20}{-1}$$

Divide 9 by -1.

$$x^{2}-9x=\frac{20}{-1}$$

Divide 20 by -1.

$$x^{2}-9x=-20$$

Divide -9, the coefficient of the x term, by 2 to get -\frac{9}{2}. Then add the square of -\frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-9x+\left(-\frac{9}{2}\right)^{2}=-20+\left(-\frac{9}{2}\right)^{2}$$

Square -\frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-9x+\frac{81}{4}=-20+\frac{81}{4}$$

Add -20 to \frac{81}{4}.

$$x^{2}-9x+\frac{81}{4}=\frac{1}{4}$$

Factor x^{2}-9x+\frac{81}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-\frac{9}{2}=\frac{1}{2} x-\frac{9}{2}=-\frac{1}{2}$$

Add \frac{9}{2} to both sides of the equation.

$$x=5 x=4$$