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.
$$a+b=1$$ $$ab=6\left(-1\right)=-6$$
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 $-6$.
$$-1,6$$ $$-2,3$$
Calculate the sum for each pair.
$$-1+6=5$$ $$-2+3=1$$
The solution is the pair that gives sum $1$.
$$a=-2$$ $$b=3$$
Rewrite $6x^{2}+x-1$ 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 $ax^{2}+bx+c=0$.
$$6x^{2}+x-1=0$$
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.
Now solve the equation $x=\frac{-1±5}{12}$ when $±$ is plus. Add $-1$ to $5$.
$$x=\frac{4}{12}$$
Reduce the fraction $\frac{4}{12}$ to lowest terms by extracting and canceling out $4$.
$$x=\frac{1}{3}$$
Now solve the equation $x=\frac{-1±5}{12}$ when $±$ is minus. Subtract $5$ from $-1$.
$$x=-\frac{6}{12}$$
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.
