During a mission near an asteroid, Emek and Tara prepared to launch a probe through a narrow observation zone. The flight computer showed that possible launch times satisfied the polynomial equation t^3 - 9t^2 + 14t + 24 = 0, where t represented hours relative to the current time. Emek tested an integer value and found that t = 4 made the polynomial zero, so he factored out (t - 4). The remaining quadratic was t^2 - 5t - 6, which Tara factored as (t - 6)(t + 1). The three candidate times were therefore t = 4, t = 6, and t = -1. They rejected -1 because it represented an hour that had already passed. Mission data also showed that after 5 hours, the asteroid would rotate into a safer orientation, so t = 4 was too early. That left t = 6 as the only usable solution. They programmed the launch for six hours later, and the probe passed safely through the observation zone, sending back detailed images of the asteroid.

Challenges

Challenge 1 of 5

Challenges

Challenge 1 of 4
Hint
Substitute 4 for every t and evaluate each term carefully before adding the results.
A student claims that t = 4 cannot be a zero of t^3 - 9t^2 + 14t + 24 because all four terms are nonzero when t = 4. Which calculation correctly checks the student's claim?
4^3 - 9(4^2) + 14(4) + 24 = 64 - 144 + 56 + 24 = 0, so t = 4 is a zero.4^3 - 9(4^2) + 14(4) + 24 = 64 - 36 + 56 + 24 = 108, so t = 4 is not a zero.4^3 - 9(4) + 14(4) + 24 = 64 - 36 + 56 + 24 = 108, so t = 4 is not a zero.4^3 - 9(4^2) + 14(4) + 24 = 64 - 144 + 14 + 24 = -42, so t = 4 is not a zero.
Correct! The terms combine to give 0, so t = 4 is a zero even though the individual terms are nonzero.
Not quite. Substitute t = 4 into every term and evaluate the powers and products carefully.
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Hint
Compare each candidate time with 5.5 hours and remember that negative time represents the past.
Suppose the asteroid became safe only after 5.5 hours instead of after 5 hours. Using the same candidate times t = 4, t = 6, and t = -1, how many candidate times would still be usable if a launch must occur after the asteroid becomes safe?
Correct! Only t = 6 occurs after 5.5 hours.
Check each of the three candidate times and count only those that occur after 5.5 hours.
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Hint
A mathematically valid solution may still fail a restriction from the real-world situation.
Another student says that because t = 4 and t = 6 are both positive solutions of the polynomial equation, either one should be safe for the launch. Which statement best explains the error in this reasoning?
Only the smallest positive solution of a polynomial can be used in a real situation.The solution t = 4 should be changed to t = -4 because launch times must have opposite signs.A solution must satisfy both the polynomial equation and the mission condition; t = 4 satisfies the equation but occurs before the asteroid becomes safe after 5 hours.Positive solutions cannot represent future times when a polynomial has a negative solution.
Correct! Both 4 and 6 are zeros, but only 6 also satisfies the mission requirement that the launch occur after 5 hours.
Not quite. Separate the question of whether a value solves the equation from whether it satisfies the mission's timing restriction.
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Asteroit Yakın Geçişini Planlamak
TR
ARC_G11_PA_02_STORY_03
Asteroit Yakın Geçişini Planlamak
Planning the Asteroid Flyby
EN
ARC_G11_PA_02_STORY_03
Planning the Asteroid Flyby