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Polynomial phobia abandoned! (Posted on 2019-04-14) Difficulty: 4 of 5
What is the minimum value of p(2) if the following 4 conditions are followed?

1) p(x) is a polynomial of degree 17.

2) All roots of p(x) are real.

3) All coefficients are positive.

4) The coefficient of x is 1.

5) The product of roots of p(x) is -1.

No Solution Yet Submitted by Danish Ahmed Khan    
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Some work, guess and test | Comment 1 of 2
I'm not able give a unconfutable solution. What I will do is to build the simplest polynomial following the conditions and to just test if this polynomial could be the solution. 

p(x)= a0x17+a1x16+a2x15+...+a16x+a17.

p(x)=a0*q(x)
=a0*(x17+(a1/a0)x16+(a2/a0)x15+...+(a16/a0)x+(a17/a0).

The roots of p(x) will be the same as the roots of q(x)

Consider q(x). If it has 17 real roots all are negative (Descartes). Then q(x)=(x+r1)*(x+r2)*...*(x+r17) and 

r1*r2*...*r17= - a17/a0. As for condition 5) we know it is = -1 we get: 

a17=a0

For condition 4 we know a16=1 

But a16/a0 = 1/a0 = sum of all sixteen products of r1 to r17 (excluding a root by turn in any factor of the sum). We call it S

So S=1/a0 = -(1/r17) -(1/r16) - ... - (1/r1) applying also condition five.

The simplest polynomial with the required conditions will be that with all the seventeen roots have the same value =-1

In this case: 1/a0= 17 and a0=1/17 Then,

p(x)=a0 (q(x) =(1/17)*(x+1)17 

p(x) is a polynomial grade 17 all coefficients are positive, all roots are real, the coefficient of x is 1 and the product of all the roots is -1

p(2)= 7596480.17 for this polynomial.

Now we are going to test what happen when we change the value of two roots, in a way that all the conditions are still followed

So we take now r1=-4/5  then r2= -5/4 to maintain the product of all the roots in -1

The new a0 is given by : 1/a0= -(1/r17) -(1/r16) - ... - (1/r1) =15+(4/5)+(5/4) = 341/20   Then a0= 20/341

The new p(x)=(20/341)*(x+4/5)(x+5/4)(x+1)15

p(2) = 7658360.92

It seems then that when we change two roots under the conditions the tendency in the value of p(2) is to grow. 

So the proposed solution is for p(x)=(1/17)*(x+1)17 

See that anyway this is not what happen with a polynomial of grade three. There the value of p(2) is lower when there is one root =-1 and the two other roots as separated as possible. 

Edited on April 19, 2019, 1:25 pm
  Posted by armando on 2019-04-19 08:10:30

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