Question

one factor of p⁶+p³-p²+p-1​

Answer 1

Answer:

p+3= -a×b = 123

Step-by-step explanation:

hope it will help you

Answer 2

Answer:

Let p−1=0

p=1

For p−1 to be factor, the equation on substituting p=1 should equate to 0

i) f(p)=p

10

−1

f(1)=1−1=0

ii) f(p)=p

11

−1

f(1)=1−1=0

∴ Hence p−1 is a factor of both

p

10

−1 & p

11

−1