the factor of (x^2-2) could be looked at if we write (x^2-2) as (x^2-(sqrt(2)^2 now we have the difference of squares so we can factor is... The key point here is where can your factors come from? Most often we factor over the Integers and (X^2-2) cannot be factored over the integers If we allow our field, or set of numbers over which we factor to be the real numbers, then the method I suggested works. If we allow our field to be the complex numbers then we cal always factor an expression. This is an important theorem in algebra.
This expression factors as x -1 quantity squared.
That factors to (x - 5)(x - 1)
The answer to x2 - 2x - 4y2 - 4y =(x - 2y)(x - 2y - 2)
x = ? 42 = x squared minus x
x2 - x - 2 = 0(x - 2)(x + 1) = 0x = 2 and -1
(x - 16)(x + 2) x = 16 or -2
x2-10x-24 = (x+2)(x-12) when factored
(2x - 3)(x + 2)
There are no rational factors.
"minus negative x" means "plus x" (2x - 5)(x + 2) or (2x + 5)(x - 2) depending on what you really meant...
5x2 + 9x + 13 has no real factors.
Factors of -8 are -1 & 8, -2 & 4, -4 & 2 and -8 & 1. Only -4 & 2 add to -2 so factors are (x + 2)(x - 4)
Find factors of -99 which sum to -2. Solution appears to be 9 and -11 (x + 9)(x - 11)
This expression factors as x -1 quantity squared.
That factors to (x - 5)(x - 1)
That factors to (x - 3)(x + 6)
The answer to x2 - 2x - 4y2 - 4y =(x - 2y)(x - 2y - 2)