(B^-1)B = 1
B^(-1xB) = 1/(B^B)
Chat with our AI personalities
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)
-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.
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