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.
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 $x_{2}$.
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 $u$.
$$r = \frac{1}{4} - u$$ $$s = \frac{1}{4} + u$$
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 $u$
