Solve 2q^2+3q-2

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

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 2q^{2}+aq+bq-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 2q^{2}+3q-2 as \left(2q^{2}-q\right)+\left(4q-2\right).

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

Factor out q in the first and 2 in the second group.

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

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

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

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.

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

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.

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

Square 3.

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

Multiply -4 times 2.

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

Multiply -8 times -2.

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

Add 9 to 16.

$$q=\frac{-3±\sqrt{25}}{2\times 2}$$

Take the square root of 25.

$$q=\frac{-3±5}{2\times 2}$$

Multiply 2 times 2.

$$q=\frac{-3±5}{4}$$

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

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

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

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

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

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

Divide -8 by 4.

$$q=-2$$

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}.

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

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

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

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

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

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

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

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 2

$$x ^ 2 +\frac{3}{2}x -1 = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = -\frac{3}{2} rs = -1$$

Two numbers r and s sum up to -\frac{3}{2} exactly when the average of the two numbers is \frac{1}{2}*-\frac{3}{2} = -\frac{3}{4}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = -\frac{3}{4} - u s = -\frac{3}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -1

$$(-\frac{3}{4} - u) (-\frac{3}{4} + u) = -1$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{9}{16} - u^2 = -1$$

Simplify the expression by subtracting \frac{9}{16} on both sides

$$-u^2 = -1-\frac{9}{16} = -\frac{25}{16}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{25}{16} u = \pm\sqrt{\frac{25}{16}} = \pm \frac{5}{4}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =-\frac{3}{4} - \frac{5}{4} = -2 s = -\frac{3}{4} + \frac{5}{4} = 0.500$$

Evaluate

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