Question

Show that p-1 is a factor of p^10-1 and also of p^11-1

Answer 1

For 1st factor
p(x)=p^10-1
q(x)=p-1
According to question it is a factor
therefore,q(x)=p-1
=1-1
=0{we have to make the factor 0 that's why we have to make it 0 ..For that we have to subtract it with 1}
p(x)=p^10-1
=1^10-1{by putting the value of x}
= 1-1
=0{we get the answer 0 therefore it is a factor}
For 2nd factor
p(x)=p^11-1
q(x)=p-1
According to question it is a factor
therefore ,q(x)=p-1
=1-1
=0
p(x) =p^11-1
=1^11-1
=1-1
=0
so they are the factors .....plz give this a brainliest answer

Answer 2

Step-by-step explanation:

Step-by-step explanation:Let p−1=0

Step-by-step explanation:Let p−1=0p=1

Step-by-step explanation:Let p−1=0p=1For p−1 to be factor, the equation on substituting p=1 should equate to 0

Step-by-step explanation:Let p−1=0p=1For p−1 to be factor, the equation on substituting p=1 should equate to 0i) f(p)=p

Step-by-step explanation:Let p−1=0p=1For p−1 to be factor, the equation on substituting p=1 should equate to 0i) f(p)=p 10

Step-by-step explanation:Let p−1=0p=1For p−1 to be factor, the equation on substituting p=1 should equate to 0i) f(p)=p 10 −1

Step-by-step explanation:Let p−1=0p=1For p−1 to be factor, the equation on substituting p=1 should equate to 0i) f(p)=p 10 −1f(1)=1−1=0

Step-by-step explanation:Let p−1=0p=1For p−1 to be factor, the equation on substituting p=1 should equate to 0i) f(p)=p 10 −1f(1)=1−1=0ii) f(p)=p

Step-by-step explanation:Let p−1=0p=1For p−1 to be factor, the equation on substituting p=1 should equate to 0i) f(p)=p 10 −1f(1)=1−1=0ii) f(p)=p 11

Step-by-step explanation:Let p−1=0p=1For p−1 to be factor, the equation on substituting p=1 should equate to 0i) f(p)=p 10 −1f(1)=1−1=0ii) f(p)=p 11 −1

Step-by-step explanation:Let p−1=0p=1For p−1 to be factor, the equation on substituting p=1 should equate to 0i) f(p)=p 10 −1f(1)=1−1=0ii) f(p)=p 11 −1f(1)=1−1=0

Step-by-step explanation:Let p−1=0p=1For p−1 to be factor, the equation on substituting p=1 should equate to 0i) f(p)=p 10 −1f(1)=1−1=0ii) f(p)=p 11 −1f(1)=1−1=0∴ Hence p−1 is a factor of both

Step-by-step explanation:Let p−1=0p=1For p−1 to be factor, the equation on substituting p=1 should equate to 0i) f(p)=p 10 −1f(1)=1−1=0ii) f(p)=p 11 −1f(1)=1−1=0∴ Hence p−1 is a factor of bothp

Step-by-step explanation:Let p−1=0p=1For p−1 to be factor, the equation on substituting p=1 should equate to 0i) f(p)=p 10 −1f(1)=1−1=0ii) f(p)=p 11 −1f(1)=1−1=0∴ Hence p−1 is a factor of bothp 10

Step-by-step explanation:Let p−1=0p=1For p−1 to be factor, the equation on substituting p=1 should equate to 0i) f(p)=p 10 −1f(1)=1−1=0ii) f(p)=p 11 −1f(1)=1−1=0∴ Hence p−1 is a factor of bothp 10 −1 & p

Step-by-step explanation:Let p−1=0p=1For p−1 to be factor, the equation on substituting p=1 should equate to 0i) f(p)=p 10 −1f(1)=1−1=0ii) f(p)=p 11 −1f(1)=1−1=0∴ Hence p−1 is a factor of bothp 10 −1 & p 11

Step-by-step explanation:Let p−1=0p=1For p−1 to be factor, the equation on substituting p=1 should equate to 0i) f(p)=p 10 −1f(1)=1−1=0ii) f(p)=p 11 −1f(1)=1−1=0∴ Hence p−1 is a factor of bothp 10 −1 & p 11 −1