Solve -2x^2+2x+40

$$-2 { x }^{ 2 } +2x+40$$

Factor

$$-2\left(x-5\right)\left(x+4\right)$$

Solution Steps

Factor out 2.

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

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

$$a+b=1 ab=-20=-20$$

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 -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 1.

$$a=5 b=-4$$

Rewrite -x^{2}+x+20 as \left(-x^{2}+5x\right)+\left(-4x+20\right).

$$\left(-x^{2}+5x\right)+\left(-4x+20\right)$$

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

$$-x\left(x-5\right)-4\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(-x-4\right)$$

Rewrite the complete factored expression.

$$2\left(x-5\right)\left(-x-4\right)$$

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.

$$-2x^{2}+2x+40=0$$

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.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-2\right)\times 40}}{2\left(-2\right)}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\left(-2\right)\times 40}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-2±\sqrt{4+8\times 40}}{2\left(-2\right)}$$

Multiply 8 times 40.

$$x=\frac{-2±\sqrt{4+320}}{2\left(-2\right)}$$

Add 4 to 320.

$$x=\frac{-2±\sqrt{324}}{2\left(-2\right)}$$

Take the square root of 324.

$$x=\frac{-2±18}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-2±18}{-4}$$

Now solve the equation x=\frac{-2±18}{-4} when ± is plus. Add -2 to 18.

$$x=\frac{16}{-4}$$

Divide 16 by -4.

$$x=-4$$

Now solve the equation x=\frac{-2±18}{-4} when ± is minus. Subtract 18 from -2.

$$x=-\frac{20}{-4}$$

Divide -20 by -4.

$$x=5$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -4 for x_{1} and 5 for x_{2}.

$$-2x^{2}+2x+40=-2\left(x-\left(-4\right)\right)\left(x-5\right)$$

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

$$-2x^{2}+2x+40=-2\left(x+4\right)\left(x-5\right)$$

Evaluate

$$-2\left(x-5\right)\left(x+4\right)$$