Solve 2x^2-3x-2

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

Factor

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

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.

$$a+b=-3 ab=2\left(-2\right)=-4$$

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.

$$a=-4 b=1$$

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

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

Factor out 2x in 2x^{2}-4x.

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

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

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

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.

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

Square -3.

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

Multiply -4 times 2.

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

Multiply -8 times -2.

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

Add 9 to 16.

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

Take the square root of 25.

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

The opposite of -3 is 3.

$$x=\frac{3±5}{2\times 2}$$

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}.

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

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

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

Add \frac{1}{2} to x by finding a common denominator and adding 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

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