Solve 2x^2-x-6

$$2 x ^ { 2 } - x - 6$$

Factor

$$\left(x-2\right)\left(2x+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=-1 ab=2\left(-6\right)=-12$$

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 -12.

$$1,-12 2,-6 3,-4$$

Calculate the sum for each pair.

$$1-12=-11 2-6=-4 3-4=-1$$

The solution is the pair that gives sum -1.

$$a=-4 b=3$$

Rewrite 2x^{2}-x-6 as \left(2x^{2}-4x\right)+\left(3x-6\right).

$$\left(2x^{2}-4x\right)+\left(3x-6\right)$$

Factor out 2x in the first and 3 in the second group.

$$2x\left(x-2\right)+3\left(x-2\right)$$

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

$$\left(x-2\right)\left(2x+3\right)$$

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.

$$2x^{2}-x-6=0$$

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.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 2\left(-6\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-1\right)±\sqrt{1-8\left(-6\right)}}{2\times 2}$$

Multiply -8 times -6.

$$x=\frac{-\left(-1\right)±\sqrt{1+48}}{2\times 2}$$

Add 1 to 48.

$$x=\frac{-\left(-1\right)±\sqrt{49}}{2\times 2}$$

Take the square root of 49.

$$x=\frac{-\left(-1\right)±7}{2\times 2}$$

The opposite of -1 is 1.

$$x=\frac{1±7}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{1±7}{4}$$

Now solve the equation x=\frac{1±7}{4} when ± is plus. Add 1 to 7.

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

Divide 8 by 4.

$$x=2$$

Now solve the equation x=\frac{1±7}{4} when ± is minus. Subtract 7 from 1.

$$x=-\frac{6}{4}$$

Reduce the fraction \frac{-6}{4} to lowest terms by extracting and canceling out 2.

$$x=-\frac{3}{2}$$

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{3}{2} for x_{2}.

$$2x^{2}-x-6=2\left(x-2\right)\left(x-\left(-\frac{3}{2}\right)\right)$$

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

$$2x^{2}-x-6=2\left(x-2\right)\left(x+\frac{3}{2}\right)$$

Add \frac{3}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$2x^{2}-x-6=2\left(x-2\right)\times \frac{2x+3}{2}$$

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

$$2x^{2}-x-6=\left(x-2\right)\left(2x+3\right)$$

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

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

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

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

Two numbers r and s sum up to \frac{1}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{1}{2} = \frac{1}{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>

$$r = \frac{1}{4} - u s = \frac{1}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -3

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

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

$$\frac{1}{16} - u^2 = -3$$

Simplify the expression by subtracting \frac{1}{16} on both sides

$$-u^2 = -3-\frac{1}{16} = -\frac{49}{16}$$

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

$$u^2 = \frac{49}{16} u = \pm\sqrt{\frac{49}{16}} = \pm \frac{7}{4}$$

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

$$r =\frac{1}{4} - \frac{7}{4} = -1.500 s = \frac{1}{4} + \frac{7}{4} = 2$$

Evaluate

$$\left(x-2\right)\left(2x+3\right)$$