# How do you factor 128-50x^2=0 ?

Jan 26, 2016

2(8 - 5x )(8 + 5x )

#### Explanation:

Firstly , consider the factors of 128 and 50. Both have a common

factor of 2 . Factorising will give: 2 (64 - 25x^2)

Inside the bracket there is a 'difference of squares'

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

hence $64 - 25 {x}^{2} = \left(8 - 5 x\right) \left(8 + 5 x\right)$

therefore $128 - 50 {x}^{2} = 2 \left(8 - 5 x\right) \left(8 + 5 x\right)$

To solve 2 (8 - 5x )( 8 + 5x ) = 0

then 8 - 5x = 0 or 8 + 5x = 0

giving that x =$\frac{8}{5} \mathmr{and} x = - \frac{8}{5} \textcolor{b l a c k}{\text{which can be written}}$

x = ± 8/5