Factor the expression by grouping. First, the expression needs to be rewritten as $5x^{2}+ax+bx+6$. To find $a$ and $b$, set up a system to be solved.
$$a+b=13$$ $$ab=5\times 6=30$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is positive, $a$ and $b$ are both positive. List all such integer pairs that give product $30$.
$$1,30$$ $$2,15$$ $$3,10$$ $$5,6$$
Calculate the sum for each pair.
$$1+30=31$$ $$2+15=17$$ $$3+10=13$$ $$5+6=11$$
The solution is the pair that gives sum $13$.
$$a=3$$ $$b=10$$
Rewrite $5x^{2}+13x+6$ as $\left(5x^{2}+3x\right)+\left(10x+6\right)$.
$$\left(5x^{2}+3x\right)+\left(10x+6\right)$$
Factor out $x$ in the first and $2$ in the second group.
$$x\left(5x+3\right)+2\left(5x+3\right)$$
Factor out common term $5x+3$ by using distributive property.
$$\left(5x+3\right)\left(x+2\right)$$
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 $ax^{2}+bx+c=0$.
$$5x^{2}+13x+6=0$$
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 $x=\frac{-13±7}{10}$ when $±$ is plus. Add $-13$ to $7$.
$$x=-\frac{6}{10}$$
Reduce the fraction $\frac{-6}{10}$ to lowest terms by extracting and canceling out $2$.
$$x=-\frac{3}{5}$$
Now solve the equation $x=\frac{-13±7}{10}$ when $±$ is minus. Subtract $7$ from $-13$.
$$x=-\frac{20}{10}$$
Divide $-20$ by $10$.
$$x=-2$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $-\frac{3}{5}$ 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 $5$
$$x ^ 2 +\frac{13}{5}x +\frac{6}{5} = 0$$
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 $rs = C$
$$r + s = -\frac{13}{5} $$ $$ rs = \frac{6}{5}$$
Two numbers $r$ and $s$ sum up to $-\frac{13}{5}$ exactly when the average of the two numbers is $\frac{1}{2}*-\frac{13}{5} = -\frac{13}{10}$. 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 $u$.
