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3 years ago

Find a polynomial function f in standard form with zeros 3i, -2, 1

1 answer:

3 0

The first thing to remember is that when you have a zero of 3i, you will also have a zero of -3i. This is because of the way imaginary solutions are created (taking the square root of a negative number). When you take the sqrt you get TWO answers.
The factors for these zeros are (x - 3i)(x+3i)(x+2)(x-1)
When you multiply the factors, the polynomial will be:
x^4 + x^3 + 7x^2 + 9x - 18

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a 1 =− 3 1 a, start subscript, 1, end subscript, equals, minus, start fraction, 1, divided by, 3, end fraction a i = a i − 1

MrRa [10]

Answer:

$S_{75}=-506.888989761$ X 10^{32}$

Step-by-step explanation:

Given the sequence

$a_1=-\dfrac{1}{3}\\ a_i=a_{i-1}\cdot (-3)$

$a_2=a_{2-1}\cdot (-3)=a_1 \cdot (-3) = -\dfrac{1}{3}\cdot (-3)=1\\a_3=a_{3-1}\cdot (-3)=a_2 \cdot (-3) = 1 \cdot (-3)=-3\\a_4=a_{4-1}\cdot (-3)=a_3 \cdot (-3) = -3 \cdot (-3)=9\\$

Therefore the sequence is:

$-\dfrac{1}{3}, 1, -3, 9, ...$

This is a geometric sequence where the:

First Term, $a_1=-\dfrac{1}{3}$

Common ratio, r =-3

We want to determine the sum of the first 75 terms.

For a geometric sequence, the sum:

$S_n=\dfrac{a_1(1-r^n)}{1-r} \\$

Therefore:

$S_{75}=\dfrac{-\dfrac{1}{3}(1-(-3)^{75})}{1-(-3)} \\\\=\dfrac{-(1-(-3)^{75})}{4*3}\\\\S_{75}=\dfrac{(-3)^{75}-1}{12}\\\\S_{75}=-506.888989761$ X 10^{32}$

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Answer:.

Step-by-step explanation:

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