Factor the expression by grouping. First, the expression needs to be rewritten as $2x^{2}+ax+bx-18$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-5$$ $$ab=2\left(-18\right)=-36$$
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 $-36$.
Rewrite $2x^{2}-5x-18$ as $\left(2x^{2}-9x\right)+\left(4x-18\right)$.
$$\left(2x^{2}-9x\right)+\left(4x-18\right)$$
Factor out $x$ in the first and $2$ in the second group.
$$x\left(2x-9\right)+2\left(2x-9\right)$$
Factor out common term $2x-9$ by using distributive property.
$$\left(2x-9\right)\left(x+2\right)$$
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 $ax^{2}+bx+c=0$.
$$2x^{2}-5x-18=0$$
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 $x=\frac{5±13}{4}$ when $±$ is plus. Add $5$ to $13$.
$$x=\frac{18}{4}$$
Reduce the fraction $\frac{18}{4}$ to lowest terms by extracting and canceling out $2$.
$$x=\frac{9}{2}$$
Now solve the equation $x=\frac{5±13}{4}$ when $±$ is minus. Subtract $13$ from $5$.
$$x=-\frac{8}{4}$$
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 $\frac{9}{2}$ for $x_{1}$ and $-2$ for $x_{2}$.
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 $2$
$$x ^ 2 -\frac{5}{2}x -9 = 0$$
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 $rs = C$
$$r + s = \frac{5}{2} $$ $$ rs = -9$$
Two numbers $r$ and $s$ sum up to $\frac{5}{2}$ exactly when the average of the two numbers is $\frac{1}{2}*\frac{5}{2} = \frac{5}{4}$. 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 $u$.
$$r = \frac{5}{4} - u$$ $$s = \frac{5}{4} + u$$
To solve for unknown quantity $u$, substitute these in the product equation $rs = -9$
$$(\frac{5}{4} - u) (\frac{5}{4} + u) = -9$$
Simplify by expanding $(a -b) (a + b) = a^2 – b^2$
$$\frac{25}{16} - u^2 = -9$$
Simplify the expression by subtracting $\frac{25}{16}$ on both sides
$$-u^2 = -9-\frac{25}{16} = -\frac{169}{16}$$
Simplify the expression by multiplying $-1$ on both sides and take the square root to obtain the value of unknown variable $u$
