Polynomial distinction

Algebra Level 3

If P ( x ) P(x) , Q ( x ) Q(x) , R ( x ) R(x) and S ( x ) S(x) are the polynomials such that

P ( x 5 ) + x Q ( x 5 ) + x 2 R ( x 5 ) = ( x 4 + x 3 + x 2 + x + 1 ) S ( x ) P(x^5) + xQ(x^5) + x^2R(x^5) = (x^4 + x^3 + x^2 + x + 1)S(x)

which of the following is always true?

( x − 1 ) (x-1) is a factor of P ( x ) P(x) ( x + 1 ) (x+1) is a factor of P ( x ) P(x) ( x − 1 ) (x-1) is a factor of P ( x + 1 ) P(x + 1) ( x + 1 ) (x+1) is a factor of P ( x − 1 ) P(x - 1)

Danish Ahmed by Danish Ahmed , 24, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Mark Hennings
Nov 9, 2020

Let ζ = e 2 π i 5 \zeta = e^{\frac{2 \pi i}{5}} be a primitive fifth root of unity. Subsituting x = ζ k x =\zeta^k into the polynomial identity gives R ( 1 ) + ζ k Q ( 1 ) + ζ 2 k R ( 1 ) = 0 1 ≤ k ≤ 4 R(1) + \zeta^kQ(1) + \zeta^{2k}R(1) \; = \; 0 \hspace{2cm} 1 \le k \le 4 Since the Vandermonde matrix ( 1 ζ ζ 2 1 ζ 2 ζ 4 1 ζ 3 ζ 6 ) \left(\begin{array}{ccc} 1 & \zeta & \zeta^2 \\ 1 & \zeta^2 & \zeta^4 \\ 1 & \zeta^3 & \zeta^6 \end{array}\right) is nonsingular we deduce that P ( 1 ) = Q ( 1 ) = R ( 1 ) = 0 P(1) = Q(1) = R(1) =0 . Thus x − 1 x-1 divides P ( x ) P(x) , Q ( x ) Q(x) and R ( x ) R(x) .

1 Helpful 2 Interesting 3 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...