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

##### 1 Answer

Jan 26, 2016

#### Answer:

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 - 25

#x^2) # Inside the bracket there is a 'difference of squares'

Note

# a^2 - b^2 = (a - b )(a + b )# hence

# 64 - 25x^2 = ( 8 - 5x )(8 + 5x ) # therefore

# 128 - 50x^2 = 2 (8 - 5x )( 8 + 5x ) # To solve 2 (8 - 5x )( 8 + 5x ) = 0

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

giving that x =

# 8/5 or x = - 8/5 color(black)("which can be written")# x =

#± 8/5 #