Solve -2x^2-3x+20

$$-2 { x }^{ 2 } -3x+20$$

Factor

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

Solution Steps

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

$$a+b=-3 ab=-2\times 20=-40$$

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 -40.

$$1,-40 2,-20 4,-10 5,-8$$

Calculate the sum for each pair.

$$1-40=-39 2-20=-18 4-10=-6 5-8=-3$$

The solution is the pair that gives sum -3.

$$a=5 b=-8$$

Rewrite -2x^{2}-3x+20 as \left(-2x^{2}+5x\right)+\left(-8x+20\right).

$$\left(-2x^{2}+5x\right)+\left(-8x+20\right)$$

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

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

Factor out common term 2x-5 by using distributive property.

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

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.

$$-2x^{2}-3x+20=0$$

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.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-2\right)\times 20}}{2\left(-2\right)}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-2\right)\times 20}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-\left(-3\right)±\sqrt{9+8\times 20}}{2\left(-2\right)}$$

Multiply 8 times 20.

$$x=\frac{-\left(-3\right)±\sqrt{9+160}}{2\left(-2\right)}$$

Add 9 to 160.

$$x=\frac{-\left(-3\right)±\sqrt{169}}{2\left(-2\right)}$$

Take the square root of 169.

$$x=\frac{-\left(-3\right)±13}{2\left(-2\right)}$$

The opposite of -3 is 3.

$$x=\frac{3±13}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{3±13}{-4}$$

Now solve the equation x=\frac{3±13}{-4} when ± is plus. Add 3 to 13.

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

Divide 16 by -4.

$$x=-4$$

Now solve the equation x=\frac{3±13}{-4} when ± is minus. Subtract 13 from 3.

$$x=-\frac{10}{-4}$$

Reduce the fraction \frac{-10}{-4} to lowest terms by extracting and canceling out 2.

$$x=\frac{5}{2}$$

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 \frac{5}{2} for x_{2}.

$$-2x^{2}-3x+20=-2\left(x-\left(-4\right)\right)\left(x-\frac{5}{2}\right)$$

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

$$-2x^{2}-3x+20=-2\left(x+4\right)\left(x-\frac{5}{2}\right)$$

Subtract \frac{5}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-2x^{2}-3x+20=-2\left(x+4\right)\times \frac{-2x+5}{-2}$$

Cancel out 2, the greatest common factor in -2 and 2.

$$-2x^{2}-3x+20=\left(x+4\right)\left(-2x+5\right)$$

Evaluate

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