Factor the expression by grouping. First, the expression needs to be rewritten as $7a^{2}+pa+qa-35$. To find $p$ and $q$, set up a system to be solved.
$$p+q=-44$$ $$pq=7\left(-35\right)=-245$$
Since $pq$ is negative, $p$ and $q$ have the opposite signs. Since $p+q$ is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product $-245$.
$$1,-245$$ $$5,-49$$ $$7,-35$$
Calculate the sum for each pair.
$$1-245=-244$$ $$5-49=-44$$ $$7-35=-28$$
The solution is the pair that gives sum $-44$.
$$p=-49$$ $$q=5$$
Rewrite $7a^{2}-44a-35$ as $\left(7a^{2}-49a\right)+\left(5a-35\right)$.
$$\left(7a^{2}-49a\right)+\left(5a-35\right)$$
Factor out $7a$ in the first and $5$ in the second group.
$$7a\left(a-7\right)+5\left(a-7\right)$$
Factor out common term $a-7$ by using distributive property.
$$\left(a-7\right)\left(7a+5\right)$$
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 $ax^{2}+bx+c=0$.
$$7a^{2}-44a-35=0$$
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 $a=\frac{44±54}{14}$ when $±$ is plus. Add $44$ to $54$.
$$a=\frac{98}{14}$$
Divide $98$ by $14$.
$$a=7$$
Now solve the equation $a=\frac{44±54}{14}$ when $±$ is minus. Subtract $54$ from $44$.
$$a=-\frac{10}{14}$$
Reduce the fraction $\frac{-10}{14}$ to lowest terms by extracting and canceling out $2$.
$$a=-\frac{5}{7}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $7$ for $x_{1}$ and $-\frac{5}{7}$ for $x_{2}$.
