if-p-x-is-a-polynomial-and-p-x-p-1-x-p-x-p-1-x-p-3-28-then-p-x-p-4-

Question Number 806 by 123456 last updated on 16/Mar/15

i f p (x) i s a p o l y n o m i a l a n d

p (x) p (\frac{1}{x}) = p (x) + p (\frac{1}{x})

p (3) = 28

t h e n

p (x) = ?

p (4) = ?

Commented by 123456 last updated on 16/Mar/15

x = \pm 1 \Rightarrow \frac{1}{x} = \pm 1

p (\pm 1) p (\pm 1) = p (\pm 1) + p (\pm 1)

{[p (\pm 1)]}^{2} - 2 p (\pm 1) = 0

p (\pm 1) [p (\pm 1) - 2] = 0

p (\pm 1) = 0 \lor p (\pm 1) = 2

Commented by prakash jain last updated on 16/Mar/15

p (x) = \underset{i = 0}{\sum^{n}} a_{i} x^{i}

p (x) p (\frac{1}{x}) = p (x) + p (\frac{1}{x})

Equating coefficients for x^{- n}

a_{0} a_{n} = a_{n} \Rightarrow a_{0} = 1

Equating coefficients for x^{0}

a_{0}^{2} + a_{1}^{2} + \dots + a_{n}^{2} = 2 a_{0}

a_{1}^{2} + \dots + a_{n}^{2} = 1 \dots . (i)

Equating coefficients for x^{1}

a_{0} a_{1} + a_{1} a_{2} + . . + a_{n - 1} a_{n} = a_{1}

a_{1} a_{2} + a_{2} a_{3} + . . + a_{n - 1} a_{n} = 0

Equating coefficients for x^{2}

a_{1} a_{3} + a_{2} a_{4} + . . + a_{n - 2} a_{n} = 0

Equating coefficent for x^{n - 1}

a_{0} a_{n - 1} + a_{1} a_{n} = a_{n - 1}

a_{0} = 1 hence a_{1} a_{n} = 0

a_{n} \neq 0 (poynomial of degree n) hence a_{1} = 0

So all coefficient (a_{1} \dots a_{n - 1}) = 0 .

From (i) a_{n} = 1

p (x) = 1 + x^{n}

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

= p (x) + p (\frac{1}{x})

p (3) = 28 = 1 + 3^{n} \Rightarrow n = 3

p (4) = 1 + 4^{3} = 65

Answered by prakash jain last updated on 16/Mar/15

See comments

p (x) = 1 + x^{3}

p (1) = 2

p (- 1) = 0

p (4) = 65