# How do you factor completely 49x^2-9?

Jun 29, 2016

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

#### Explanation:

We know that, ${A}^{2} - {B}^{2} = \left(A - B\right) \left(A + B\right) .$

Given Exp.$= {\left(7 x\right)}^{2} - {3}^{2} = \left(7 x - 3\right) \left(7 x + 3\right) .$

Jun 29, 2016

$49 {x}^{2} - 9 = \left(7 x + 3\right) \left(7 x - 3\right)$

#### Explanation:

As both the monomials in the binomial are squares, with a negative sign between them, we can the identity ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$

Here we have $49 {x}^{2} - 9$

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

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