Question

Use the factor theorem to prove that (x+a) is a factor of (x^n+a^n) for any odd positive integer n

Answer 1

We know that if (x - α) is a factor of a polynomial p(x), then p(α) will be 0.

Here we have to prove that (x + a) is a factor of (x^n + a^n) for any odd positive integer n.

Let p(x) = (x^n + a^n).

If (x + a) has to be a factor of p(x), then p(-a) shall be equal to 0. So,

p(-a) = (-a)^n + a^n

Since n is an odd positive integer, it leaves remainder 1 on division by 2. So let n = 2k + 1, for any whole number k.

p(-a) = (-a)^(2k + 1) + a^(2k + 1)

p(-a) = (-a)^(2k) · (-a) + a^(2k + 1)

p(-a) = ((-a)^2)^k · (-a) + a^(2k + 1)

Since the square of any real number is non - negative,

p(-a) = (a^2)^k · (-a) + a^(2k + 1)

p(-a) = a^(2k) · a · (-1) + a^(2k + 1)

p(-a) = a^(2k + 1) · (-1) + a^(2k + 1)

p(-a) = - a^(2k + 1) + a^(2k + 1)

p(-a) = - a^n + a^n

p(-a) = 0

Hence Proved!

Answer 2

Answer:

Let p(x) = x^n + a^n , where n is odd positive integer.

Take (x+a)= 0

=> x = -a

Consider:

p(-a) = (-a) ^n + (a) ^n

= -a^n + a^n

= 0

Since, n is odd.

By Factor theorem,

(x+a) is a factor of p(x) when n is odd positive integer.

Step-by-step explanation: