Please help!!?

Question

Prove that P(x) = x^n - a^n is divisible by (x-a) for all integer values of n.

Answer 1

Thare are many methods to prove it.
1st method,

You may use mathematical induction,
P(x)= x^n-a^n,
for n=1, P(x)=x-a is divisible by (x-a)
Let the relation holds true for n=k,
hence, For n=k, P(x)=x^k-a^k=Q(x-a) (Q=some integer){say}
we are to examine whether the relation holds true for n=k+1
For n=k+1,
P(x)= x^(k+1)- a^(k+1)
=x*x^k - a*a^k
=x*x^k - x*a^k + x*a^k - a*a^k
= x(x^k - a^k) + a^k(x-a)
=Qx(x-a) + a^k(x-a)
=(x-a)(Qx + a^k)
Which is divisible by (x-a),
hence the relation holds true for n= any integer

2nd way,

you may use Remainder Theorem, which states that " if a polynomial P(x) is divided by (x-a), then the remainder is P(a)"

according to Remainder theorem,if P(x)=x^n - a^n is divided by (x-a), the remainder is =
Remainder= P(a)= a^n - a^n = 0
hence P(x)=x^n - a^n is divisible by (x-a) for n= any integer

3rd way,

You may use synthetic division to show that when P(x)=x^n - a^n is divided by (x-a) the remainder is zero, but synthetic division is a nightmare for me so I will not personally recommend you to employ it.

4th method,

you may use the method of vanish,
P(x)=x^n - a^n,
putting x=a, P(a)=a^n - a^n=0,
hence (x-a) is a factor of P(x)=x^n - a^n.

5th method,

you may also employ binomial theorem
P(x)=x^n - a^n
={(x-a)+a}^n - a^n
=(x-a)^n+ (nc1)(x-a)^(n-1)*a + (nc2)(x-a)^(n-2)*a^2 + ......+a^n-a^n.
=(x-a)^n+ (nc1)(x-a)^(n-1)*a + (nc2)(x-a)^(n-2)*a^2 + ......+(nc(n-1))(x-a)*a^(n-1)
=(x-a)[(x-a)^(n-1)+ (nc1)(x-a)^(n-2)*a + (nc2)(x-a)^(n-3)*a^2 +......+(nc(n-1))*a^(n-1)]
hence (x-a) is a factor of P(x)
Or,P(x) is divisible by (x-a)

Answer 2

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Answer 3

You could use mathematical induction.
X^1-a^1=x-a , so the first case is correct
Suppose X^n-a^n is divisible by(x-a) and we say that X^n-a^n=t .
(x-a)(a^n)is divisible by ,(x-a) and that (x-a)(a^n)=b ; =>
x*t-b=x^(n+1)-a^(n+1) is also divisible by (x-a) =>

"x^n - a^n is divisible by (x-a)" means that "x^(n+1) - a^(n+1) is also divisible by (x-a)" .Problem solved .It's also called the domino principle .
P(a) -> p(a+1) -> p(a+2) -> p(a+3)..............

Answer 4

Gjmb's method is certainly quick and easy. Another way to prove this is to just factor it:

x^n - a^n = (x-a) (x^(n-1) + a*x(n-2) + a^2*x(n-3) + a^3*x(n-4) + ... + a^(n-1)*x + a^n))

When you expand the above, you'll see how everything cancels out except the x^n and the -a^n.

Answer 5

When 'n' is even then it can be easily done by applying a^2-b^2=(a-b)*(a+b) recursively, but when 'n' is odd the problem can be solved easily by applying "Binomial Theorem".
Use the series (1-x)^n.

Answer 6

when n=1 x^n-a^n=x-a divisible by x-a true
let it be true fo some n=k
x^k-a^k is divisible by x-a
let x^k-a^k/(x-a)=c wgere c is element of N
therefore x^k-a^k=c(x-a)
to prove it is true for k+1
x^(k+1)-a^(k+1)=x^k*x-a^k*a
=x[a^k+c(x-a)]-a^k*a
a^k(x-a)+cx(x-a)
(x-a)(a^x+cx)
so p(k+1) is divisible by (x-a)
so true for any n
hence proved

Answer 7

On What

Answer 8

that is easy :
a zero of P(x) is x=a thus x-a is a divisor.