Solve -2x^2-3x+2

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

Factor

$$-\left(2x-1\right)\left(x+2\right)$$

Solution Steps

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

$$a+b=-3 ab=-2\times 2=-4$$

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 -4.

$$1,-4 2,-2$$

Calculate the sum for each pair.

$$1-4=-3 2-2=0$$

The solution is the pair that gives sum -3.

$$a=1 b=-4$$

Rewrite -2x^{2}-3x+2 as \left(-2x^{2}+x\right)+\left(-4x+2\right).

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

Factor out -x in the first and -2 in the second group.

$$-x\left(2x-1\right)-2\left(2x-1\right)$$

Factor out common term 2x-1 by using distributive property.

$$\left(2x-1\right)\left(-x-2\right)$$

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.

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

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.

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

Square -3.

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

Multiply -4 times -2.

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

Multiply 8 times 2.

$$x=\frac{-\left(-3\right)±\sqrt{9+16}}{2\left(-2\right)}$$

Add 9 to 16.

$$x=\frac{-\left(-3\right)±\sqrt{25}}{2\left(-2\right)}$$

Take the square root of 25.

$$x=\frac{-\left(-3\right)±5}{2\left(-2\right)}$$

The opposite of -3 is 3.

$$x=\frac{3±5}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{3±5}{-4}$$

Now solve the equation x=\frac{3±5}{-4} when ± is plus. Add 3 to 5.

$$x=\frac{8}{-4}$$

Divide 8 by -4.

$$x=-2$$

Now solve the equation x=\frac{3±5}{-4} when ± is minus. Subtract 5 from 3.

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

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

$$x=\frac{1}{2}$$

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

$$-2x^{2}-3x+2=-2\left(x-\left(-2\right)\right)\left(x-\frac{1}{2}\right)$$

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

$$-2x^{2}-3x+2=-2\left(x+2\right)\left(x-\frac{1}{2}\right)$$

Subtract \frac{1}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-2x^{2}-3x+2=-2\left(x+2\right)\times \frac{-2x+1}{-2}$$

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

$$-2x^{2}-3x+2=\left(x+2\right)\left(-2x+1\right)$$

Evaluate

$$-\left(2x-1\right)\left(x+2\right)$$