Princeton Review Precaluclus Question

Forum for the GRE subject test in mathematics.
amateur
Posts: 42
Joined: Wed Sep 10, 2008 9:41 am

Princeton Review Precaluclus Question

Postby amateur » Wed Oct 01, 2008 11:39 am

Q. 14 Page 28:

"Given that p(x) is a real polynomial of degree <= 4 such that one can find five distinct solutions to the equation p(x) = 5, what is the value of p(5)?" Answer Choices A. 0, B. 1, C. 4, D. 5, E. Cannot be determined.

While solving this question, I was puzzled as to when a polynomial of degree 4 or less can have five DISTINCT roots? The question was absurd to me, so no answer made sense.

The answer at the back of the book states the polynomial must be zero.
So for example p(x) = a*x^4 + b*x^3 + c*x^2 + d*x + e - 5, where all a, b, c, d, e - 5 = 0.

Therefore
p(1) - 5 = 0 => x = 1 is a root;
p(2) - 5 = 0 => x = 2 is a root...so on.

Therefore, p(5) - 5 = 0, So p(5) = 5.

Isn't it plain nonsense? I mean does a zero polynomial having 5 distinct roots make sense? How about 25,000 distinct roots, 36437 repeated roots, 23123 complex roots,...

Amateur.

Wonderacle
Posts: 5
Joined: Wed Oct 01, 2008 4:26 pm

Postby Wonderacle » Wed Oct 01, 2008 4:34 pm

as you have suspected
a non-zero polynomial of degree n can have at most n distinct zeros
therefore, p(x)-5 must be zero because it has degree of 4 and at least 5 distinct zeros

CoCoA
Posts: 42
Joined: Wed Sep 03, 2008 5:39 pm

Postby CoCoA » Wed Oct 01, 2008 6:42 pm

Everything is a root of the zero polynomial. This is required when we get to algebraic geometry, and (when the field is algebraically closed) have a 1-1 correspondence between the algebraic sets (set of common zeros to every polynomial in the ideal) of n-space and the radical ideals of polynomials in n variables over the field: the ideal generated by the zero polynomial corresponds to the entire space, which is an algebraic set.

All other definitions are satisfied also, even if they look wierd: every non-zero polynomial divides zero, so for any k in the field, k is a root and x-k divides the polynomial (zero).

rookie
Posts: 2
Joined: Thu Oct 02, 2008 5:49 pm

Postby rookie » Sun Oct 05, 2008 6:05 pm

If q(x)=p(x)-5 of degree \le 4 has five distinct zeroes it has means p(x) is a constant polynomial: p(x)=5 so p(5)=5

amateur
Posts: 42
Joined: Wed Sep 10, 2008 9:41 am

Postby amateur » Mon Oct 06, 2008 4:29 am

First of all, I am really thankful to all participants of this forum. It has been especially very helpful in preparing for the GRE and made my preparation relevant, lively and exciting.

Coming back to the question, I checked on several websites and found that the degree of a zero polynomial is undefined. Some authors define it to be equal to -infinity, Under this situation, I don't think it makes sense to talk of the degree of a zero polynomial to be <= 4, or at best, this statement is controversial. Check:

http://en.wikipedia.org/wiki/Degree_of_ ... polynomial
http://planetmath.org/encyclopedia/ZeroPolynomial2.html
http://mathworld.wolfram.com/ZeroPolynomial.html

CoCoA
Posts: 42
Joined: Wed Sep 03, 2008 5:39 pm

Postby CoCoA » Mon Oct 06, 2008 4:06 pm

You are correct that many (most?) sources define the zero polynomial as having no degree. Some do give it a degree, either -1 or -infinity, for the purpose of a complete theory of degrees. Nevertheless, your original question had the answer p(x)=5, which is a non-zero constant polynomial, which does have degree 0. The wording in one answer or another may obscure the point, but the point is that q(x)=p(x)-5=0 is used to determine the zeros, but the solutions of p(x)=5 are the zeros of q(x)=0, and p(x)=5 is used to determine the degree in this question.




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