Solve 2y^2+8y-640

$$2 { y }^{ 2 } +8y-640$$

Factor

$$2\left(y-16\right)\left(y+20\right)$$

Solution Steps

Factor out 2.

$$2\left(y^{2}+4y-320\right)$$

Consider y^{2}+4y-320. Factor the expression by grouping. First, the expression needs to be rewritten as y^{2}+ay+by-320. To find a and b, set up a system to be solved.

$$a+b=4 ab=1\left(-320\right)=-320$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -320.

$$-1,320 -2,160 -4,80 -5,64 -8,40 -10,32 -16,20$$

Calculate the sum for each pair.

$$-1+320=319 -2+160=158 -4+80=76 -5+64=59 -8+40=32 -10+32=22 -16+20=4$$

The solution is the pair that gives sum 4.

$$a=-16 b=20$$

Rewrite y^{2}+4y-320 as \left(y^{2}-16y\right)+\left(20y-320\right).

$$\left(y^{2}-16y\right)+\left(20y-320\right)$$

Factor out y in the first and 20 in the second group.

$$y\left(y-16\right)+20\left(y-16\right)$$

Factor out common term y-16 by using distributive property.

$$\left(y-16\right)\left(y+20\right)$$

Rewrite the complete factored expression.

$$2\left(y-16\right)\left(y+20\right)$$

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.

$$2y^{2}+8y-640=0$$

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.

$$y=\frac{-8±\sqrt{8^{2}-4\times 2\left(-640\right)}}{2\times 2}$$

Square 8.

$$y=\frac{-8±\sqrt{64-4\times 2\left(-640\right)}}{2\times 2}$$

Multiply -4 times 2.

$$y=\frac{-8±\sqrt{64-8\left(-640\right)}}{2\times 2}$$

Multiply -8 times -640.

$$y=\frac{-8±\sqrt{64+5120}}{2\times 2}$$

Add 64 to 5120.

$$y=\frac{-8±\sqrt{5184}}{2\times 2}$$

Take the square root of 5184.

$$y=\frac{-8±72}{2\times 2}$$

Multiply 2 times 2.

$$y=\frac{-8±72}{4}$$

Now solve the equation y=\frac{-8±72}{4} when ± is plus. Add -8 to 72.

$$y=\frac{64}{4}$$

Divide 64 by 4.

$$y=16$$

Now solve the equation y=\frac{-8±72}{4} when ± is minus. Subtract 72 from -8.

$$y=-\frac{80}{4}$$

Divide -80 by 4.

$$y=-20$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 16 for x_{1} and -20 for x_{2}.

$$2y^{2}+8y-640=2\left(y-16\right)\left(y-\left(-20\right)\right)$$

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

$$2y^{2}+8y-640=2\left(y-16\right)\left(y+20\right)$$

Evaluate

$$2\left(y-16\right)\left(y+20\right)$$