Difference between revisions of "2003 AMC 12A Problems/Problem 21"
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<math>\text{(A)}\ a \qquad \text{(B)}\ b \qquad \text{(C)}\ c \qquad \text{(D)}\ d \qquad \text{(E)}\ e</math> | <math>\text{(A)}\ a \qquad \text{(B)}\ b \qquad \text{(C)}\ c \qquad \text{(D)}\ d \qquad \text{(E)}\ e</math> | ||
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== Solution == | == Solution == | ||
=== Solution 1 === | === Solution 1 === | ||
| − | Let the roots be <math>r_1=0, r_2, r_3, r_4, r_5</math>. According to [[Vieta's formulas]], we have <math>d=r_1r_2r_3r_4 + r_1r_2r_3r_5 + r_1r_2r_4r_5 + r_1r_3r_4r_5 + r_2r_3r_4r_5</math>. The first four terms contain <math>r_1=0</math> and are therefore zero, thus <math>d=r_2r_3r_4r_5</math>. This is a product of four non-zero numbers, therefore <math>d</math> must be non-zero <math>\Longrightarrow \mathrm{(D)}</math>. | + | Let the roots be <math>r_1=0, r_2, r_3, r_4, r_5</math>. According to [[Vieta's formulas | Viète's formulae]], we have <math>d=r_1r_2r_3r_4 + r_1r_2r_3r_5 + r_1r_2r_4r_5 + r_1r_3r_4r_5 + r_2r_3r_4r_5</math>. The first four terms contain <math>r_1=0</math> and are therefore zero, thus <math>d=r_2r_3r_4r_5</math>. This is a product of four non-zero numbers, therefore <math>d</math> must be non-zero <math>\Longrightarrow \mathrm{(D)}</math>. |
=== Solution 2 === | === Solution 2 === | ||
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[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 20:12, 29 September 2021
Problem
The graph of the polynomial
![\[P(x) = x^5 + ax^4 + bx^3 + cx^2 + dx + e\]](http://latex.artofproblemsolving.com/8/5/a/85a8f024c29a92d2b5ebd8359ba595c26a38a65a.png)
has five distinct 
-intercepts, one of which is at 
. Which of the following coefficients cannot be zero?
Solution
Solution 1
Let the roots be 
. According to Viète's formulae, we have 
. The first four terms contain 
and are therefore zero, thus 
. This is a product of four non-zero numbers, therefore 
must be non-zero 
.
Solution 2
Clearly, since is an intercept, 
must be 
. But if was , 
would divide the polynomial, which means it would have a double root at , which is impossible, since all five roots are distinct.
See Also
| 2003 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 20 |
Followed by Problem 22 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
