Solve (2x^2+3x-2)

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

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.

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

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 -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.

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

Square 3.

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

Multiply -4 times 2.

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

Multiply -8 times -2.

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

Add 9 to 16.

$$x=\frac{-3±\sqrt{25}}{2\times 2}$$

Take the square root of 25.

$$x=\frac{-3±5}{2\times 2}$$

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.

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

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

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

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

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

Divide -8 by 4.

$$x=-2$$

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

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

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

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

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

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

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

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

Evaluate

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