Princeton Review Precaluclus Question
Princeton Review Precaluclus Question
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.
"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.

 Posts: 5
 Joined: Wed Oct 01, 2008 4:26 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 11 correspondence between the algebraic sets (set of common zeros to every polynomial in the ideal) of nspace 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 nonzero polynomial divides zero, so for any k in the field, k is a root and xk divides the polynomial (zero).
All other definitions are satisfied also, even if they look wierd: every nonzero polynomial divides zero, so for any k in the field, k is a root and xk divides the polynomial (zero).
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
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
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 nonzero 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.

 Posts: 1
 Joined: Thu Sep 12, 2019 5:32 am
Re: Princeton Review Precaluclus Question
Hey, thanks for asking this question. I also thought this problem was weird and stuck on it.
From reading various replies across the interwebs, my understanding is that:
However, the question is still quite strange as it was not covered in the text and other replies stated it usually appears in formal algebraic geometry and theories of degrees topics that are not seen on the GRE subject test.
From reading various replies across the interwebs, my understanding is that:
 If p(x) had ended up being equal to 0, then it would have been the zero polynomial and have no degree or a degree of 1 or infinity, depending on the math system. In all cases, this would make the statement that "p(x) is a real polynomial of degree ≤ 4" incorrect.
 If p(x) had ended up being equal to a nonzero constant, then it would not have been the zero polynomial and have a degree of 0. This would make the statement that "p(x) is a real polynomial of degree ≤ 4" correct.
However, the question is still quite strange as it was not covered in the text and other replies stated it usually appears in formal algebraic geometry and theories of degrees topics that are not seen on the GRE subject test.
Re: Princeton Review Precaluclus Question
It says that you CAN find 5 distinct roots, not that there are 5 distinct roots. Mof you go on looking for roots, you can find 6, 7, etc...
BTW, the degree of the 0:polynomial is not 0
https://en.m.wikipedia.org/wiki/Degree_ ... polynomial
BTW, the degree of the 0:polynomial is not 0
https://en.m.wikipedia.org/wiki/Degree_ ... polynomial