Solve y^2-8y-20

$${ y }^{ 2 } -8y-20$$

Factor

$$\left(y-10\right)\left(y+2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as y^{2}+ay+by-20. To find a and b, set up a system to be solved.

$$a+b=-8 ab=1\left(-20\right)=-20$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -20.

$$1,-20 2,-10 4,-5$$

Calculate the sum for each pair.

$$1-20=-19 2-10=-8 4-5=-1$$

The solution is the pair that gives sum -8.

$$a=-10 b=2$$

Rewrite y^{2}-8y-20 as \left(y^{2}-10y\right)+\left(2y-20\right).

$$\left(y^{2}-10y\right)+\left(2y-20\right)$$

Factor out y in the first and 2 in the second group.

$$y\left(y-10\right)+2\left(y-10\right)$$

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

$$\left(y-10\right)\left(y+2\right)$$

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.

$$y^{2}-8y-20=0$$

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.

$$y=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\left(-20\right)}}{2}$$

Square -8.

$$y=\frac{-\left(-8\right)±\sqrt{64-4\left(-20\right)}}{2}$$

Multiply -4 times -20.

$$y=\frac{-\left(-8\right)±\sqrt{64+80}}{2}$$

Add 64 to 80.

$$y=\frac{-\left(-8\right)±\sqrt{144}}{2}$$

Take the square root of 144.

$$y=\frac{-\left(-8\right)±12}{2}$$

The opposite of -8 is 8.

$$y=\frac{8±12}{2}$$

Now solve the equation y=\frac{8±12}{2} when ± is plus. Add 8 to 12.

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

Divide 20 by 2.

$$y=10$$

Now solve the equation y=\frac{8±12}{2} when ± is minus. Subtract 12 from 8.

$$y=-\frac{4}{2}$$

Divide -4 by 2.

$$y=-2$$

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 -2 for x_{2}.

$$y^{2}-8y-20=\left(y-10\right)\left(y-\left(-2\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$y^{2}-8y-20=\left(y-10\right)\left(y+2\right)$$

Evaluate

$$\left(y-10\right)\left(y+2\right)$$