Question

show that (p-1)is a factor of both p10-1 and p11-1

Answer 1

Answer:

Show that P(1) is a factor of p(10-1) and p(11-1)

Step-by-step explanation:

Well, this question here may be a little bit difficult to make you understand out here as we only have access to MS word doc. A lot of other functions needs to be accessed so white board or a chalk board is the best option, still will try my best.

You asked to prove that (p^-1) is a factor of both p^10-1 and p^11-1

Lets see,

Assuming here that  g(p) = p^10-1.

  H(p) = p^11-1, we try and plug in values for p -1 in equation g(p) = p^10-1, we get

  g(1) = 1^10-1 = 1 – 1 (as 1^10 = 1)

  g(1) = 1-1 = 0. So, p-1 is a factor of g(p).

Now again, we plug in p = 1 in second equation, and we get

  h(1) = (1) 11-1 = (1)^10-1 = 1-1 = 0. Hence p-1 is a factor of h(p).

Answer 2

Proved that (p-1) is a factor of both $\bold{\mathrm{p}^{10}-1 \text { and } \mathrm{p}^{11}-1}$  

To find:

Show that (p-1) is a factor of both$\mathrm{p}^{10}-1 \text { and } \mathrm{p}^{11}-1$

Solution:

Given: We are to prove that (p-1) is a factor of both $\mathrm{p}^{10}-1 \text { and } \mathrm{p}^{11}-1$

As it is given that (p - 1) is a factor of both, we can say that

p - 1 = 0

Hence, p = 1  

Thus, we got the value of p = 1.  

Now, let us substitute the above value found i.e. p = 1, in the given equations $\mathrm{p}^{10}-1 \text { and } \mathrm{p}^{11}-1$

$f(1)=(1)^{10}-1\quad (\text{Substituting}\quad \mathrm{p}=1 \text { in } \mathrm{p}^{10}-1)$

$\begin{aligned} &=1-1 \\ &=0 \end{aligned}$

We very well know that any number of times 1 is multiplied by 1 itself, we get answer as 1 only.

$f(1)=(1)^{11}-1\quad (\text{Substituting}\quad \mathrm{p}=1 \text { in } \mathrm{p}^{11}-1)$

$\begin{aligned} &=1-1 \\ &=0 \end{aligned}$

Hence, this shows that p-1 is the factor of both.