# How do you factor the expression 49p^2 + 63pq - 36q^2?

Jan 23, 2017

$49 {p}^{2} + 63 p q - 36 {q}^{2} = \left(7 p + 12 q\right) \left(7 p - 3 q\right)$

#### Explanation:

Note that $49 {p}^{2} = {\left(7 p\right)}^{2}$ and $36 {q}^{2} = {\left(6 q\right)}^{2}$

Further note that $63 = {3}^{2} \cdot 7$ is divisible by $7$ and by $3$ but not by $6$.

So let's look at this quadratic in terms of $7 p$ and $3 q$...

$49 {p}^{2} + 63 p q - 36 {q}^{2} = {\left(7 p\right)}^{2} + 3 \left(7 p\right) \left(3 q\right) - 4 {\left(3 q\right)}^{2}$

Note that $4 - 1 = 3$ and $4 \cdot 1 = 4$, so we find:

${\left(7 p\right)}^{2} + 3 \left(7 p\right) \left(3 q\right) - 4 {\left(3 q\right)}^{2} = \left(\left(7 p\right) + 4 \left(3 q\right)\right) \left(\left(7 p\right) - \left(3 q\right)\right)$

$\textcolor{w h i t e}{{\left(7 p\right)}^{2} + 3 \left(7 p\right) \left(3 q\right) - 4 {\left(3 q\right)}^{2}} = \left(7 p + 12 q\right) \left(7 p - 3 q\right)$

Jan 23, 2017

$- 36 \left(q + \frac{7}{12} p\right) \left(q - \frac{7}{3} p\right)$

#### Explanation:

Making $q = \lambda p$ and substituting

$49 {p}^{2} + 63 p q - 36 {q}^{2} = \left(49 + 63 \lambda - 36 {\lambda}^{2}\right) {p}^{2}$

but $\left(49 + 63 \lambda - 36 {\lambda}^{2}\right) = - 36 \left(\lambda + \frac{7}{12}\right) \left(\lambda - \frac{7}{3}\right)$

so

$- 36 \left(\lambda + \frac{7}{12}\right) \left(\lambda - \frac{7}{3}\right) {p}^{2} = - 36 \left(\lambda p + \frac{7}{12} p\right) \left(\lambda p - \frac{7}{3} p\right) = - 36 \left(q + \frac{7}{12} p\right) \left(q - \frac{7}{3} p\right)$