Factor the expression by grouping. First, the expression needs to be rewritten as $6x^{2}+ax+bx+10$. To find $a$ and $b$, set up a system to be solved.
$$a+b=19$$ $$ab=6\times 10=60$$
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 $60$.
Rewrite $6x^{2}+19x+10$ as $\left(6x^{2}+4x\right)+\left(15x+10\right)$.
$$\left(6x^{2}+4x\right)+\left(15x+10\right)$$
Factor out $2x$ in the first and $5$ in the second group.
$$2x\left(3x+2\right)+5\left(3x+2\right)$$
Factor out common term $3x+2$ by using distributive property.
$$\left(3x+2\right)\left(2x+5\right)$$
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 $ax^{2}+bx+c=0$.
$$6x^{2}+19x+10=0$$
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 $x=\frac{-19±11}{12}$ when $±$ is plus. Add $-19$ to $11$.
$$x=-\frac{8}{12}$$
Reduce the fraction $\frac{-8}{12}$ to lowest terms by extracting and canceling out $4$.
$$x=-\frac{2}{3}$$
Now solve the equation $x=\frac{-19±11}{12}$ when $±$ is minus. Subtract $11$ from $-19$.
$$x=-\frac{30}{12}$$
Reduce the fraction $\frac{-30}{12}$ to lowest terms by extracting and canceling out $6$.
$$x=-\frac{5}{2}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $-\frac{2}{3}$ for $x_{1}$ and $-\frac{5}{2}$ for $x_{2}$.
Multiply $\frac{3x+2}{3}$ times $\frac{2x+5}{2}$ by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
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 $6$
$$x ^ 2 +\frac{19}{6}x +\frac{5}{3} = 0$$
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 $rs = C$
$$r + s = -\frac{19}{6} $$ $$ rs = \frac{5}{3}$$
Two numbers $r$ and $s$ sum up to $-\frac{19}{6}$ exactly when the average of the two numbers is $\frac{1}{2}*-\frac{19}{6} = -\frac{19}{12}$. 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 $u$.
