Q. 184.5( 4 Votes )

# Show that p – 1 is a factor of p^{10} – 1 and also of p^{11} – 1.

Answer :

let h (p) = 𝑝 ^{10} − 1,and g(p) = 𝑝 − 1

Putting g (p) = 0 ⟹ 𝑝 − 1 = 0 ⟹ 𝑝 = 1

According to factor theorem if g(p) is a factor of h(p) , then h(1) should be zero

h(1) = (1)^{10} − 1 = 1 − 1 = 0

⟹ g (p) is a factor of h(p).

Now, we have h (p) = 𝑝 ^{11} − 1, g (p) = 𝑝 − 1

Putting g (p) = 0 ⟹ 𝑝 − 1 = 0 ⟹ 𝑝 = 1

According to factor theorem if g (p) is a factor of h (p) ,

Then h(1) = 0

⟹ (1)^{11} – 1 = 0 hence g (p) the factor of h (p)

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