show that ( x+a) is a factor of xn+an for any odd positive integer n
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Answered by
16
(xn+an)
Taking n common...
n(x+a)
hence proved///
Taking n common...
n(x+a)
hence proved///
Answered by
50
Hi friend,
Given,
(x+a) is a factor of xn+an
x+a = 0
x = -a
Put x = -a
xn + an
→ (-a)n + an
→ -an + an
→ 0
Therefore, (x+a) is a factor of xn + an
(OR)
xn + an
→n(x+a)
Therefore, n and (x+a) are factors.
hope it helps..
Given,
(x+a) is a factor of xn+an
x+a = 0
x = -a
Put x = -a
xn + an
→ (-a)n + an
→ -an + an
→ 0
Therefore, (x+a) is a factor of xn + an
(OR)
xn + an
→n(x+a)
Therefore, n and (x+a) are factors.
hope it helps..
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