Reciprocal Eqns............

Q 1 ) is the root x = -1 ?

suppose p(x) is a polynomial function of degree n .

then p(x) is a reciprocal function if ... coefficient of x^n-r= ceff of x ^r

consider a general reciprocal function of degree n .

p(x) = a_{0}+a_{1}x + a_{2}x^{2}......+a_{2}x^{n-2}+a_{1}x^{n-1} + a_{0}x^{n}

so, p(\frac{1}{x}) = a_{0}+a_{1}/x + a_{2}/x^{2}......+a_{2}/x^{n-2}+a_{1}/x^{n-1} + a_{0}/x^{n}

so, p(\frac{1}{x}) =\frac{1}{x^{n}} (a_{0}+a_{1}x + a_{2}x^{2}......+a_{2}x^{n-2}+a_{1}x^{n-1} + a_{0}x^{n})

so, p(\frac{1}{x}) =\frac{1}{x^{n}} (p(x))

p(x)=x^{n}p(\frac{1}{x})

putting\; x=-1,

p(-1)=-p(-1)

p(-1)+p(-1)=0

\Rightarrow p(-1)=0

so x = -1 is a root

Edit: fogot to mention that here n = odd

Q1:

After putting x = -1, we get:

P (-1) = (-1)ⁿ P( -1)

Then we get (-1)ⁿ = 1, which is not possible for an odd value of n.

However, if you try with 1, we get

P (1) = 1ⁿ P(1)

For any defined real value of n, we find this equality exists.

Hence, 1 is a root of the polynomial.

Qsn 2.

Just cross multiply and expand.

You get the equation as :

-78 x⁶ + 133 x⁵ - 133 x + 78 = 0.

Putting x = 1 and x = -1 satisfies this polynomial.

Divide the polynomial by (x-1)(x+1)

We get :

(x+1)(x-1)( -78 x⁴ + 133 x³ -78 x² + 133 x -1) = 0

The remnant polynomial cannot be factorised in my opinion, so the only real roots i have obtained are 1 and -1.

Sum of roots = 0.

Q1: the product of roots is -1. Also if k is a root 1/k also is, thus forming pairs multiplying to 1. Since 1 cannot be a root, the unpaired root is forced to be -1

Hmm...Qwerty, i got ur point... had tht discussion wid shakexpear y'day nite...
Jus got a bit cnfused since u hdnt mentioned tht n = odd.