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kaylesparks2
24.03.2020 •
Mathematics
Given a polynomial that has zeros of −4, 9i, and −9i and has a value of 492 when x=−1. Write the polynomial in standard form axn+bxn−1+…. Answer using reduced fractions when necessary.
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Ответ:
The polynomial that has zeros of −4, 9i, and −9i, and has a value 492 when x = -1 is:
2x³ + 8x² + 162x + 648 = 0
Step-by-step explanation:
Given that the polynomial that has zeros of −4, 9i, and −9i, we can say that
x = -4 => x + 4 = 0
x = 9i and -9i => x = ±9i => x² = (9i)²
Because i² = -1, x² = -81
=> x² + 81 = 0
So together, we can write the polynomial as
(x + 4)(x² + 81) = 0
Expanding the bracket, we have
x³ + 81x + 4x² + 324 = 0
x³ + 4x² + 81x + 324 = 0
Again, we are told that the polynomial is 492 at x = -1.
Put x = -1 in
x³ + 4x² + 81x + 324 = 0
(-1)³ + 4(-1)² + 81(-1) + 324
-1 + 4 - 81 + 324
= 246 ≠ 492
But 492 = 2×246
Since multiplying the polynomial by 2 doesn't change anything, our result is set.
Now,
2(x³ + 4x² + 81x + 324) = 2(0)
2x³ + 8x² + 162x + 648 = 0
When x = -1
2(-1)³ + 8(-1)² + 162(-1) + 648
-2 + 8 - 162 + 648
= 492
As required.
Ответ: