Polynomial Problem #1

Algebra Level 3

A system of two polynomials f ( x ) = x 5 − 5 x 3 + 4 x + 1 f(x)= x^5 - 5x^3 + 4x + 1 and Q ( x ) = 2 x 2 + x − 1 Q(x)= 2x^2 + x - 1 is such that there are 5 solutions x 1 , x 2 , x 3 , x 4 , x 5 x_1,x_2,x_3,x_4,x_5 for f ( x ) f(x) .

Calculate the product Q ( x 1 ) × Q ( x 2 ) × Q ( x 3 ) × Q ( x 4 ) × Q ( x 5 ) Q(x_1) \times Q(x_2) \times Q(x_3) \times Q(x_4) \times Q(x_5) .


The answer is 77.

Tin Le by Tin Le , 15, Vietnam

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1 solution

Patrick Corn
Sep 24, 2019

It helps that Q ( x ) = ( 2 x − 1 ) ( x + 1 ) . Q(x) = (2x-1)(x+1). So the product is the product of 2 x i − 1 2x_i-1 times the product of x i + 1. x_i+1.

Now x i + 1 x_i+1 are the roots of the monic polynomial f ( x − 1 ) , f(x-1), whose constant term is f ( − 1 ) = 1 , f(-1) = 1, so their product is − 1. -1.

And 2 x i − 1 2x_i-1 are the roots of the monic polynomial 32 f ( x + 1 2 ) , 32 f\left( \frac{x+1}2 \right), whose constant term is 32 f ( 1 / 2 ) = 77 , 32f(1/2) = 77, so their product is − 77. -77.

Hence the product is − 1 ⋅ − 77 = 77 . -1 \cdot -77 = \fbox{77}.

3 Helpful 1 Interesting 3 Brilliant 0 Confused

Very elegant!

Chris Lewis - 1 year, 8 months ago

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