# The polynom p(x) = a_n*x^n + a_n-1*x^(n-1) + ... + a_0 has one positive root given that all a_i real and a_i >0 but a_0 <0. Can you show this?This problem must deal with the fundamental...

The polynom p(x) = a_n*x^n + a_n-1*x^(n-1) + ... + a_0 has one positive root given that all a_i real and a_i >0 but a_0 <0. Can you show this?

This problem must deal with the fundamental Theorem of Algebra, but I can not imagine that we are asked to proof this theorem, so there must be straightforward approach. Can you help?

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### 2 Answers

Given the polynomial `p(x)=a_nx^(n)+a_(n-1)x^(n-1)+...+a_1x+a_0` with `a_0<0` and all other `a_i>0` with real coefficients: show that there is one positive root.

Note that `p(0)<0` and `lim_(x->oo)p(x)=oo` , so by the intermediate value theorem there exists at least one zero on `(0,oo)` .(All polynomials are everywhere continuous)

To show that there is only one positive zero, consider the first derivative of the function. All coefficients of the derivative are positive, so for `x>0,p'(x)>0` . Thus the function is monotonically increasing on `(0,oo)` . This implies that if a>b, then p(a)>p(b).

Thus there exists a positive root, and only one positive root for p(x).

Apply Lagrange's theorem for a polynomial `P(x) = a_n*x^d + a_n-1*x^(d-1) + ... + a_0.`

The upper bound for the positive roots of P(x) is the number `1+(A/a_n)^(1/(m+1)).`

`` `A = max{a_i ; coefficient x^(d-i)lt0}`

To prove the existence of one positive root, use the proof of the theorem for P(x).

`|P(x)| gt= a_n*x^n + .... + a_m*x^(n-m) - A(x^(n-m-1) + ... + 1)`

Notice that the terms of the sum denotes the terms of a geometrical progression.

Addition of these terms yields `(x^(n-m)-1)/(x-1), ` therefore:

**`|P(x)| gt= a_n*x^n + .... + a_m*x^(n-m) - A*(x^(n-m)-1)/(x-1)` **