Factor the expression by grouping. First, the expression needs to be rewritten as $6x^{2}+ax+bx+1$. To find $a$ and $b$, set up a system to be solved.
$$a+b=5$$ $$ab=6\times 1=6$$
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 $6$.
$$1,6$$ $$2,3$$
Calculate the sum for each pair.
$$1+6=7$$ $$2+3=5$$
The solution is the pair that gives sum $5$.
$$a=2$$ $$b=3$$
Rewrite $6x^{2}+5x+1$ as $\left(6x^{2}+2x\right)+\left(3x+1\right)$.
$$\left(6x^{2}+2x\right)+\left(3x+1\right)$$
Factor out $2x$ in $6x^{2}+2x$.
$$2x\left(3x+1\right)+3x+1$$
Factor out common term $3x+1$ by using distributive property.
$$\left(3x+1\right)\left(2x+1\right)$$
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 $ax^{2}+bx+c=0$.
$$6x^{2}+5x+1=0$$
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.
$$x=\frac{-5±\sqrt{5^{2}-4\times 6}}{2\times 6}$$
Square $5$.
$$x=\frac{-5±\sqrt{25-4\times 6}}{2\times 6}$$
Multiply $-4$ times $6$.
$$x=\frac{-5±\sqrt{25-24}}{2\times 6}$$
Add $25$ to $-24$.
$$x=\frac{-5±\sqrt{1}}{2\times 6}$$
Take the square root of $1$.
$$x=\frac{-5±1}{2\times 6}$$
Multiply $2$ times $6$.
$$x=\frac{-5±1}{12}$$
Now solve the equation $x=\frac{-5±1}{12}$ when $±$ is plus. Add $-5$ to $1$.
$$x=-\frac{4}{12}$$
Reduce the fraction $\frac{-4}{12}$ to lowest terms by extracting and canceling out $4$.
$$x=-\frac{1}{3}$$
Now solve the equation $x=\frac{-5±1}{12}$ when $±$ is minus. Subtract $1$ from $-5$.
$$x=-\frac{6}{12}$$
Reduce the fraction $\frac{-6}{12}$ to lowest terms by extracting and canceling out $6$.
$$x=-\frac{1}{2}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $-\frac{1}{3}$ for $x_{1}$ and $-\frac{1}{2}$ for $x_{2}$.
Multiply $\frac{3x+1}{3}$ times $\frac{2x+1}{2}$ by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
