Problem:
The graph of the polynomial
P(x)=x5+ax4+bx3+cx2+dx+e
has five distinct x-intercepts, one of which is at (0,0). Which of the following coefficients cannot be zero?
Answer Choices:
A. a
B. b
C. c
D. d
E. e
Solution:
Since P(0)=0, we have e=0 and P(x)=x(x4+ax3+bx2+cx+d). Suppose that the four remaining x-intercepts are at p,q,r, and s. Then
x4+ax3+bx2+cx+d=(xβp)(xβq)(xβr)(xβs),
and d=pqrsξ =0.
Any of the other constants could be zero. For example, consider
P1β(x)=x5β5x3+4x=x(x+2)(x+1)(xβ1)(xβ2)
and
P2β(x)=x5β5x4+20x2β16x=x(x+2)(xβ1)(xβ2)(xβ4).
\section*{OR}
Since P(0)=0, we must have e=0, so
P(x)=x(x4+ax3+bx2+cx+d)
If d=0, then
P(x)=x(x4+ax3+bx2+cx)=x2(x3+ax2+bx+c)
which has a double root at x=0. Hence dξ =0.
\mathrm
There is also a calculus-based solution. Since P(x) has five distinct zeros and x=0 is one of the zeros, it must be a zero of multiplicity one. This is equivalent to having P(0)=0, but Pβ²(0)ξ =0. Since
Pβ²(x)=5x4+4ax3+3bx2+2cx+d, we must have 0ξ =Pβ²(0)=d
The problems on this page are the property of the MAA's American Mathematics Competitions