# How do you factor 5x^{2}+18x-35?

Oct 14, 2016

The factorization of $5 {x}^{2} + 18 x - 35$ is $\left(x + 5\right) \left(5 x - 7\right)$.

#### Explanation:

Remember that $5 {x}^{2}$ is the result of "first times first" multiplication, $18 x$ is the sum of "outside times outside" and "inside times inside" multiplications, and $- 35$ is the result of "last times last" multiplication.

To begin, think about the factor pairs whose product is $35$.
$1 \cdot 35$ and $5 \cdot 7$ are the only options.

Now consider the factor pair whose product is $35 {x}^{2}$.
$1 x \cdot 5 x$ is the only one.

We know that the two "first" terms in the factorization will be $1 x$ and $5 x$. We can easily discard $1 \cdot 35$ as the factor pair to obtain the product of $35$, because of the large sums that would result from adding the outside & inside products.

Therefore we must decide on the proper placement of the $5$ and $7$, along with one $+$ and one $-$ & the effect of the $5 x$, to get the outside & inside sum of $18 x$.

$\left(x + 7\right) \left(5 x - 5\right)$ does not work, because the "outside times outside" multiplication results in $- 5 x$ & the "inside times inside" multiplication results in $35 x$. The sum of these two products is $30 x$. Try again.

$\left(x + 5\right) \left(5 x - 7\right)$ works. Do FOIL to verify:

$\left(x + 5\right) \left(5 x - 7\right)$
$5 {x}^{2} - 7 x + 25 x - 35$
$5 {x}^{2} + 18 x - 35$

This verifies that the correct factorization is
$\left(x + 5\right) \left(5 x - 7\right)$.