Solve y^2-18y+72

$${ y }^{ 2 } -18y+72$$

Factor

$$\left(y-12\right)\left(y-6\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as y^{2}+ay+by+72. To find a and b, set up a system to be solved.

$$a+b=-18 ab=1\times 72=72$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 72.

$$-1,-72 -2,-36 -3,-24 -4,-18 -6,-12 -8,-9$$

Calculate the sum for each pair.

$$-1-72=-73 -2-36=-38 -3-24=-27 -4-18=-22 -6-12=-18 -8-9=-17$$

The solution is the pair that gives sum -18.

$$a=-12 b=-6$$

Rewrite y^{2}-18y+72 as \left(y^{2}-12y\right)+\left(-6y+72\right).

$$\left(y^{2}-12y\right)+\left(-6y+72\right)$$

Factor out y in the first and -6 in the second group.

$$y\left(y-12\right)-6\left(y-12\right)$$

Factor out common term y-12 by using distributive property.

$$\left(y-12\right)\left(y-6\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$y^{2}-18y+72=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$y=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 72}}{2}$$

Square -18.

$$y=\frac{-\left(-18\right)±\sqrt{324-4\times 72}}{2}$$

Multiply -4 times 72.

$$y=\frac{-\left(-18\right)±\sqrt{324-288}}{2}$$

Add 324 to -288.

$$y=\frac{-\left(-18\right)±\sqrt{36}}{2}$$

Take the square root of 36.

$$y=\frac{-\left(-18\right)±6}{2}$$

The opposite of -18 is 18.

$$y=\frac{18±6}{2}$$

Now solve the equation y=\frac{18±6}{2} when ± is plus. Add 18 to 6.

$$y=\frac{24}{2}$$

Divide 24 by 2.

$$y=12$$

Now solve the equation y=\frac{18±6}{2} when ± is minus. Subtract 6 from 18.

$$y=\frac{12}{2}$$

Divide 12 by 2.

$$y=6$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 12 for x_{1} and 6 for x_{2}.

$$y^{2}-18y+72=\left(y-12\right)\left(y-6\right)$$

Evaluate

$$\left(y-12\right)\left(y-6\right)$$