# Question #de4b1

Feb 4, 2018

$9 {x}^{2} + 42 x y + 49 {y}^{2} - 1$

#### Explanation:

Instead of triple-foiling and ending up with 9 terms to condense, use this trick:

The first two terms in both polynomials are the same, so you can group them together and treat them as ONE term:

$\left(3 x + 7 y + 1\right) \left(3 x + 7 y - 1\right)$

$= \left(\left(3 x + 7 y\right) + 1\right) \left(\left(3 x + 7 y\right) - 1\right)$

Now, this is in the form $\left(a + b\right) \left(a - b\right)$, and we know that:

$\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$

Therefore, we can write this expression as:

$= {\left(3 x + 7 y\right)}^{2} - {1}^{2}$

To simplify completely, remember the square formula:

${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$

So, we can expand the expression like this:

$\left({\left(3 x\right)}^{2} + 2 \left(3 x\right) \left(7 y\right) + {\left(7 y\right)}^{2}\right) - 1$

$\left(9 {x}^{2} + 42 x y + 49 {y}^{2}\right) - 1$

$9 {x}^{2} + 42 x y + 49 {y}^{2} - 1$