Difference between revisions of "2003 AMC 12A Problems/Problem 21"
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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 === |
Latest revision as of 20:12, 29 September 2021
Contents
[hide]Problem
The graph of the polynomial
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 |
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