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 #