x^(2) - 100 =>
x^(20 - 10^(2) =>
( x - 10)(x + 10) Done!!!!!
9 - x2 will factor into (3 - x)(3 + x)
x2(x3 + 1) is the best you can do there.
x2-9x = x(x-9)
If you mean "factors", the two monomials have the common factor x2. Divide each factor by x2 to get the other factor.
x2-49 = (x-7)(x+7)
(x + 10)(x - 10)
2 to the third power x2=highest common factor of 100 and 96=16
If by "what is the answer to" you mean "how do you factor", then the answer is as follows: x2 + 52x + 100 = x2 + 50x + 2x + 100 = x(x + 50) + 2(x + 50) = (x + 2)(x + 50)
(x - 4)(x - 25)
(x - 10)(x - 10)
You're looking for factors of 100 that differ by 20. (x + 5)(x - 20)
(x+20)(x-5) x=5 or -20
x2 - 4
9 - x2 will factor into (3 - x)(3 + x)
x^2+25x-100 is not factorable because there are not numbers that when multiplied equal -100 and when added equal 25.
x2-3x-28
x2-14 does factor but not pretty; it factors to: (x+sqrt14)(x-sqrt14).