Question

Without actual division prove X + a is a a factor of(x)=x^5+a^5

Answer 1

Answer:

✏Given that (x+a) is a factor of p(x)=x^5+a^5

✏If (x+a) is factor of p(x) it means that -a is a zero of p(x) ...i.e.x=-a

✏Now,

$= > p(x) = x {}^{5} + a {}^{5} \\ = > p( - a) = ( - a) {}^{5} + a {}^{5} \\ = > p( - a) = - a {}^{5} + a {}^{5} \\ = > p( - a) = 0$

⭐Hence Proved⭐

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Answer 2

Answer :

Step-by-step explanation:

f(x) = x^5 + a^5 & let p(x) = X+a

To prove , x + a is a factor of f(x) without performing actual division.

By using remainder theorem,

zero of p(x) = -a

i.e p(-a)=0

Now putting this value of x in f(x) and checking whether this vale of x is zero of f(x) or not :

f(-a) = (-a)^5+(a)^5=0

Hence -a is zero of f(x) also.

So, p(x) is factor of g(x)

Hence Proved.