Problem:
Let f be a polynomial function such that, for all real x,
f(x2+1)=x4+5x2+3
For all real x,f(x2−1) is
Answer Choices:
A. x4+5x2+1
B. x4+x2−3
C. x4−5x2+1
D. x4+x2+3
E. none of these
Solution:
Since f(x) is a polynomial and f(x2+1) has degree 4, then f(x) has degree 2. That is, f(x)=ax2+bx+c for some constants a,b,c, and
f(x2+1)=x4+5x2+3​=a(x2+1)2+b(x2+1)+c=ax4+(2a+b)x2+(a+b+c).​
Since two polynomials are equal if and only if the coefficients of corresponding terms are equal, we have a=1,2a+b=5 and a+b+c=3. Solving these gives f(x)=x2+3x−1. Thus f(x2−1)=(x2−1)2+3(x2−1)−1=x4+ x2−3.
Alternate solution. Rewrite x4+5x2+3 in terms of powers of x2+1:
x4+5x2+3​=(x4+2x2+1)+(3x2+3)−1=(x2+1)2+3(x2+1)−1.​
Thus setting w=x2+1, we have f(w)=w2+3w−1.
(Actually, since x is real, w as defined can only take on values equal to or greater than 1. Thus so far we have shown only that f(w)=w2+3w−1 when w≥1. However, since f(w) is given to be a polynomial, and since it is identical to the polynomial w2+3w−1 for infinitely many w, it follows that they are identical for all w.)
Therefore, as before, for all x
f(x2−1)=(x2−1)2+3(x2−1)−1=x4+x2−3.