Factor the expression by grouping. First, the expression needs to be rewritten as $14x^{2}+ax+bx-1$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-5$$ $$ab=14\left(-1\right)=-14$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is negative, the negative number has greater absolute value than the positive. 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 $-5$.
$$a=-7$$ $$b=2$$
Rewrite $14x^{2}-5x-1$ as $\left(14x^{2}-7x\right)+\left(2x-1\right)$.
$$\left(14x^{2}-7x\right)+\left(2x-1\right)$$
Factor out $7x$ in $14x^{2}-7x$.
$$7x\left(2x-1\right)+2x-1$$
Factor out common term $2x-1$ by using distributive property.
