Solve 48-16y+y^2

$$48 - 16 y + y ^ { 2 }$$

Factor

$$\left(y-12\right)\left(y-4\right)$$

Solution Steps

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$y^{2}-16y+48$$

Factor the expression by grouping. First, the expression needs to be rewritten as y^{2}+ay+by+48. To find a and b, set up a system to be solved.

$$a+b=-16 ab=1\times 48=48$$

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 48.

$$-1,-48 -2,-24 -3,-16 -4,-12 -6,-8$$

Calculate the sum for each pair.

$$-1-48=-49 -2-24=-26 -3-16=-19 -4-12=-16 -6-8=-14$$

The solution is the pair that gives sum -16.

$$a=-12 b=-4$$

Rewrite y^{2}-16y+48 as \left(y^{2}-12y\right)+\left(-4y+48\right).

$$\left(y^{2}-12y\right)+\left(-4y+48\right)$$

Factor out y in the first and -4 in the second group.

$$y\left(y-12\right)-4\left(y-12\right)$$

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

$$\left(y-12\right)\left(y-4\right)$$

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.

$$y^{2}-16y+48=0$$

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.

$$y=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 48}}{2}$$

Square -16.

$$y=\frac{-\left(-16\right)±\sqrt{256-4\times 48}}{2}$$

Multiply -4 times 48.

$$y=\frac{-\left(-16\right)±\sqrt{256-192}}{2}$$

Add 256 to -192.

$$y=\frac{-\left(-16\right)±\sqrt{64}}{2}$$

Take the square root of 64.

$$y=\frac{-\left(-16\right)±8}{2}$$

The opposite of -16 is 16.

$$y=\frac{16±8}{2}$$

Now solve the equation y=\frac{16±8}{2} when ± is plus. Add 16 to 8.

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

Divide 24 by 2.

$$y=12$$

Now solve the equation y=\frac{16±8}{2} when ± is minus. Subtract 8 from 16.

$$y=\frac{8}{2}$$

Divide 8 by 2.

$$y=4$$

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 4 for x_{2}.

$$y^{2}-16y+48=\left(y-12\right)\left(y-4\right)$$

Evaluate

$$\left(y-12\right)\left(y-4\right)$$