Neat Fact for Integral Solutions to a polynomial
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02 Jul 2013, 02:09
Hello!
Consider any polynomial \(f(x) = A_1x^n+A_2x^{n-1}+.....A_n\)
Assumption : All the co-efficients for the given polynomial have to be integral,i.e. \(A_1,A_2,A_3....A_n\) are all integers.
Fact:Any integral solution(root) for the above polynomial will always be a factor(positive/negative) of the constant term : \(A_n\)
Example I : \(f(x) = 5x^2-16x+3\). Thus, we know that if the given polynomial has any integral solutions, then it will always be out of one of the following : \(-3,-1,1,3\)
We see that only x=3 is a root for the given polynomial. Also, we know that product of the roots is\(\frac {3}{5}\). Hence, the other root is \(\frac {1}{5}\)
Example II : Find the no of integral solutions for the expression \(f(x) = 3x^4-10x^2+7x+1\)
A. 0
B. 1
C. 2
D. 3
E. 4
For the given expression, instead of finding the possible integral solutions by hit and trial, we can be rest assured that if there is any integral solution, it will be a factor of the constant term ,i.e. 1 or -1. Just plug-in both the values, and we find that f(1) and f(-1) are both not equal to zero. Thus, there is NO integral solution possible for the given expression--> Option A.
Example III : Find the no of integral solutions for the expression \(f(x) = 4x^4-8x^3+9x-3\)
A. 0
B. 1
C. 2
D. 3
E. 4
Just as above, the integral roots of the given expression would be one of the following : -3,-1,1,3. We can easily see that only x = -1 satisfies. Thus, there is only one integral solution for the given polynomial-->Option B.
Hence, keeping this fact in mind might just reduce the range of the hit and trial values we end up considering.