x2-81=(x-9)(x+9) Factors: x-9, x+9
5x2 + 9x + 13 has no real factors.
I'm presuming its factors for a monic quadratic trinomial--of form : X squared + BX + C . You need the numbers that MULTIPLY to give C and ADD up to give B Eg X squared + 8X + 12 factors to ( X+2) (X + 6 ) BUT X squared + 7X + 12 factors to (X +3 )(X + 4) Hope this helps. Phil.
7(x + 3)(x - 3)
(x - 5)(x - 3)
3x2 - 11x - 4 = (x-4)(3 x + 1)
There are no rational 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)
Assuming that is x squared minus 16: x2 - 16 = (x + 4)(x - 4) (The difference of two squares.)
(x - 5)(x + 4)
(x+4)(x-7)
(X-5)(X+4)=0 so X-5=0 -----X=5 and X=4=0 so X=-4 it factors to (X-5)(X+4) so X=5,-4
This expression factors as x -1 quantity squared.
(x + 2)(x - 4)
4
x(x + 3)(x - 4)
x2-10x-24 = (x+2)(x-12) when factored