Solve 0=3(x-5)(x-1)/8

$$0 = \frac { 3 ( x - 5 ) ( x - 1 ) } { 8 }$$

Solve for x

$$x=1$$

$$x=5$$

Steps Using Factoring By Grouping

Multiply both sides by 8. Anything times zero gives zero.

$$0=3\left(x-5\right)\left(x-1\right)$$

Use the distributive property to multiply 3 by x-5.

$$0=\left(3x-15\right)\left(x-1\right)$$

Use the distributive property to multiply 3x-15 by x-1 and combine like terms.

$$0=3x^{2}-18x+15$$

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

$$3x^{2}-18x+15=0$$

Divide both sides by 3.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=-6 ab=1\times 5=5$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-5 b=-1$$

Rewrite x^{2}-6x+5 as \left(x^{2}-5x\right)+\left(-x+5\right).

$$\left(x^{2}-5x\right)+\left(-x+5\right)$$

Factor out x in the first and -1 in the second group.

$$x\left(x-5\right)-\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(x-1\right)$$

To find equation solutions, solve x-5=0 and x-1=0.

$$x=5 x=1$$

Steps Using the Quadratic Formula

Multiply both sides by 8. Anything times zero gives zero.

$$0=3\left(x-5\right)\left(x-1\right)$$

Use the distributive property to multiply 3 by x-5.

$$0=\left(3x-15\right)\left(x-1\right)$$

Use the distributive property to multiply 3x-15 by x-1 and combine like terms.

$$0=3x^{2}-18x+15$$

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

$$3x^{2}-18x+15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -18 for b, and 15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 3\times 15}}{2\times 3}$$

Square -18.

$$x=\frac{-\left(-18\right)±\sqrt{324-4\times 3\times 15}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-18\right)±\sqrt{324-12\times 15}}{2\times 3}$$

Multiply -12 times 15.

$$x=\frac{-\left(-18\right)±\sqrt{324-180}}{2\times 3}$$

Add 324 to -180.

$$x=\frac{-\left(-18\right)±\sqrt{144}}{2\times 3}$$

Take the square root of 144.

$$x=\frac{-\left(-18\right)±12}{2\times 3}$$

The opposite of -18 is 18.

$$x=\frac{18±12}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{18±12}{6}$$

Now solve the equation x=\frac{18±12}{6} when ± is plus. Add 18 to 12.

$$x=\frac{30}{6}$$

Divide 30 by 6.

$$x=5$$

Now solve the equation x=\frac{18±12}{6} when ± is minus. Subtract 12 from 18.

$$x=\frac{6}{6}$$

Divide 6 by 6.

$$x=1$$

The equation is now solved.

$$x=5 x=1$$

Steps for Completing the Square

Multiply both sides by 8. Anything times zero gives zero.

$$0=3\left(x-5\right)\left(x-1\right)$$

Use the distributive property to multiply 3 by x-5.

$$0=\left(3x-15\right)\left(x-1\right)$$

Use the distributive property to multiply 3x-15 by x-1 and combine like terms.

$$0=3x^{2}-18x+15$$

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

$$3x^{2}-18x+15=0$$

Subtract 15 from both sides. Anything subtracted from zero gives its negation.

$$3x^{2}-18x=-15$$

Divide both sides by 3.

$$\frac{3x^{2}-18x}{3}=-\frac{15}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\left(-\frac{18}{3}\right)x=-\frac{15}{3}$$

Divide -18 by 3.

$$x^{2}-6x=-\frac{15}{3}$$

Divide -15 by 3.

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

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -3.

$$x^{2}-6x+9=-5+9$$

Add -5 to 9.

$$x^{2}-6x+9=4$$

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add 3 to both sides of the equation.

$$x=5 x=1$$