(B^-1)B = 1
B^(-1xB) = 1/(B^B)
x4 - x3 - x - 1 rewriting: = x4 - 1 - x3 - x factorising pair of terms: =(x2 + 1)*(x2 - 1) - x*(x2 + 1) = (x2 + 1)*(x2 - 1 - x) or (x2 + 1)*(x2 - x - 1) which cannot be factorised further.
(x2 + 1)(x2 - 2)
x(-b)=m(x-c)
Let y= ab+(- a)(b) +(-a)(-b) factor out -a y= ab+(-a){b+(-b)} y=ab+(-a)(0) y =ab -------------------(1) now factor out b y= b{a+(-a)}+(-a)(-b) y= b(0) +(-a)(-b) y= (-a)(-b)-----------------(2) equate (1) and (2) (-a)(-b)=ab minus x minus = positive
x2-x-30 = (x-1/2)2 - 1/4 -30 as (a+b)2=a2+2ab+b2 with a=x then b=-1/2 = (x-1/2)2 - 121/4 = [(x-1/2)-sqrt(121/4)] [ [(x-1/2)+sqrt(121/4)] as a2-b2=(a-b)(a+b) then = (x-1/2-11/2)(x-1/2+11/2) = (x-6)(x+5)
x(x + 1)(x - 1)
x(x - 1)
x4 - x3 - x - 1 rewriting: = x4 - 1 - x3 - x factorising pair of terms: =(x2 + 1)*(x2 - 1) - x*(x2 + 1) = (x2 + 1)*(x2 - 1 - x) or (x2 + 1)*(x2 - x - 1) which cannot be factorised further.
if you mean x to the second power minus 1 then the answer is (x+1)(x-1)
It's hard to write this on a typewriter keyboard. Do you mean xb + (1-x)b , or xb + 1 - xb , or maybe even x(b+1) - xb ? The first cannot be factored. The second equals 1, so it is already prime (cannot be factored). The third could be written as xb(x) - xb(1), so there is a common factor of xb, and the expression can be factored as xb(x-1). This is prime.
(x + 1)(x - 1)
use this strategy: integral of (b^x) dx = (b^x)/ln(b) + K [K is integration constant, b is not a variable]rewrite (1/c)^(1-x) = ((1/c)^1)*((1/c)^(-x)) = (1/c)*(c^x). (1/c) is a constant, so bring outside the integral, then let b = c in the formula above, and you have (1/c)*(c^x)/ln(c) + K
(x2 + 1)(x2 - 2)
x(-b)=m(x-c)
3-2 x 3-3 = 3-5 or 1/243
Let y= ab+(- a)(b) +(-a)(-b) factor out -a y= ab+(-a){b+(-b)} y=ab+(-a)(0) y =ab -------------------(1) now factor out b y= b{a+(-a)}+(-a)(-b) y= b(0) +(-a)(-b) y= (-a)(-b)-----------------(2) equate (1) and (2) (-a)(-b)=ab minus x minus = positive
-1 is the answer. With an odd number of -1s, we end up back where we started as each multiplication only switches from positive to negative.