Factor the expression by grouping. First, the expression needs to be rewritten as $7x^{2}+ax+bx-2$. To find $a$ and $b$, set up a system to be solved.
$$a+b=13$$ $$ab=7\left(-2\right)=-14$$
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 $-14$.
$$-1,14$$ $$-2,7$$
Calculate the sum for each pair.
$$-1+14=13$$ $$-2+7=5$$
The solution is the pair that gives sum $13$.
$$a=-1$$ $$b=14$$
Rewrite $7x^{2}+13x-2$ as $\left(7x^{2}-x\right)+\left(14x-2\right)$.
$$\left(7x^{2}-x\right)+\left(14x-2\right)$$
Factor out $x$ in the first and $2$ in the second group.
$$x\left(7x-1\right)+2\left(7x-1\right)$$
Factor out common term $7x-1$ by using distributive property.
$$\left(7x-1\right)\left(x+2\right)$$
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 $ax^{2}+bx+c=0$.
$$7x^{2}+13x-2=0$$
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.
Now solve the equation $x=\frac{-13±15}{14}$ when $±$ is plus. Add $-13$ to $15$.
$$x=\frac{2}{14}$$
Reduce the fraction $\frac{2}{14}$ to lowest terms by extracting and canceling out $2$.
$$x=\frac{1}{7}$$
Now solve the equation $x=\frac{-13±15}{14}$ when $±$ is minus. Subtract $15$ from $-13$.
$$x=-\frac{28}{14}$$
Divide $-28$ by $14$.
$$x=-2$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{1}{7}$ 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 $7$
$$x ^ 2 +\frac{13}{7}x -\frac{2}{7} = 0$$
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 $rs = C$
$$r + s = -\frac{13}{7} $$ $$ rs = -\frac{2}{7}$$
Two numbers $r$ and $s$ sum up to $-\frac{13}{7}$ exactly when the average of the two numbers is $\frac{1}{2}*-\frac{13}{7} = -\frac{13}{14}$. 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 $u$.
