Solve y^2-16y+60

$$y ^ { 2 } - 16 y + 60$$

Factor

$$\left(y-10\right)\left(y-6\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as y^{2}+ay+by+60. To find a and b, set up a system to be solved.

$$a+b=-16 ab=1\times 60=60$$

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 60.

$$-1,-60 -2,-30 -3,-20 -4,-15 -5,-12 -6,-10$$

Calculate the sum for each pair.

$$-1-60=-61 -2-30=-32 -3-20=-23 -4-15=-19 -5-12=-17 -6-10=-16$$

The solution is the pair that gives sum -16.

$$a=-10 b=-6$$

Rewrite y^{2}-16y+60 as \left(y^{2}-10y\right)+\left(-6y+60\right).

$$\left(y^{2}-10y\right)+\left(-6y+60\right)$$

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

$$y\left(y-10\right)-6\left(y-10\right)$$

Factor out common term y-10 by using distributive property.

$$\left(y-10\right)\left(y-6\right)$$

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.

$$y^{2}-16y+60=0$$

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.

$$y=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 60}}{2}$$

Square -16.

$$y=\frac{-\left(-16\right)±\sqrt{256-4\times 60}}{2}$$

Multiply -4 times 60.

$$y=\frac{-\left(-16\right)±\sqrt{256-240}}{2}$$

Add 256 to -240.

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

Take the square root of 16.

$$y=\frac{-\left(-16\right)±4}{2}$$

The opposite of -16 is 16.

$$y=\frac{16±4}{2}$$

Now solve the equation y=\frac{16±4}{2} when ± is plus. Add 16 to 4.

$$y=\frac{20}{2}$$

Divide 20 by 2.

$$y=10$$

Now solve the equation y=\frac{16±4}{2} when ± is minus. Subtract 4 from 16.

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

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 10 for x_{1} and 6 for x_{2}.

$$y^{2}-16y+60=\left(y-10\right)\left(y-6\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.

$$x ^ 2 -16x +60 = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = 16 rs = 60$$

Two numbers r and s sum up to 16 exactly when the average of the two numbers is \frac{1}{2}*16 = 8. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = 8 - u s = 8 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 60

$$(8 - u) (8 + u) = 60$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$64 - u^2 = 60$$

Simplify the expression by subtracting 64 on both sides

$$-u^2 = 60-64 = -4$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = 4 u = \pm\sqrt{4} = \pm 2$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =8 - 2 = 6 s = 8 + 2 = 10$$

Evaluate

$$\left(y-10\right)\left(y-6\right)$$