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boyakko [2]
3 years ago
12

What is the leading coefficient of a cubic polynomial that has a value of −208 when x=1, and has zeros of 5, 5i, and −5i?

Mathematics
1 answer:
Marianna [84]3 years ago
6 0

Answer:

2

Step-by-step explanation:

We already have the zeros, so we can write the cubic polynomial in this general form:

y = a(x - x_1)(x - x_2)(x - x_3)

Where:

x_1 = 5

x_2 = 5i

x_3 = -5i

So we have that:

y = a(x -5)(x - 5i)(x + 5i)

y = a(x -5)(x^2 + 25)

To find the value of the leading coefficient 'a', we can use the point (1, -208) given:

-208 = a(1 -5)(1 + 25)

-208 = a(-4)(26)

a = -208 / (-104) = 2

So the leading coefficient is 2.

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Answer:

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3 years ago
Can anyone help me with trig identities?
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