# 16y^2-x^2-6x-9. How can I factorise this? I know that -(x^2+6x+9) is equal to -(X+3)^2, and 16y^2 can be written as (4y)^2 but I can't quite get the solution (4y-x-3)(4y+X+3)

Feb 28, 2018

$16 {y}^{2} - {x}^{2} - 6 x - 9 = \left(4 y - x - 3\right) \left(4 y + x + 3\right)$

#### Explanation:

First observe that $16 {y}^{2} - {x}^{2} - 6 x - 9$

= $16 {y}^{2} - \left({x}^{2} + 6 x + 9\right)$

= ${\left(4 y\right)}^{2} - {\left(x + 3\right)}^{2}$

Now observe that this is of type ${A}^{2} - {B}^{2}$, where $A = 4 y$ and $B = \left(x + 3\right)$. Now factors of ${A}^{2} - {B}^{2} = \left(A - B\right) \left(A + B\right)$

Hence we can write ${\left(4 y\right)}^{2} - {\left(x + 3\right)}^{2}$ as

$\left(4 y - \left(x + 3\right)\right) \left(4 y + \left(x + 3\right)\right)$

= $\left(4 y - x - 3\right) \left(4 y + x + 3\right)$