Question

How do you solve `8x^3 - 32x=0`?

Answer 1

There is no need to use 'Comparing method' as suggested.
`x=0,-2,2`

Explanation:

Given
`8x^3-32x=0`
Inspection reveals that `8x` is a factor of right hand side cubic.
`:. 8x(x^2-4)=0`
The quadratic term can be rewritten using the formula, since `4=2^2`
`a^2-b^2=(a+b)(a-b)`
Hence the factors are
`8x(x+2)(x-2)=0`
Now `8!=0`, setting remaining factors equal to zero, we obtain
`x=0, (x+2)=0 and (x-2)=0`
We obtain three values of `x=0,-2,2`