Question

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

Answer 1

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`