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tia_tia [17]
2 years ago
5

Find the polynomial equation of least degree with roots -1, 3, and (+/-)3i

Mathematics
1 answer:
jasenka [17]2 years ago
7 0
Each of these roots can be expressed as a binomial:

(x+1)=0, which solves to -1
(x-3)=0, which solves to 3
(x-3i)=0 which solves to 3i
(x+3i)=0, which solves to -3i
There are four roots, so our final equation will have x^4 as the least degree

Multiply them together. I'll multiply the i binomials first:
(x-3i)(x+3i) = x²+3ix-3ix-9i²
x²-9i²
x²+9  [since i²=-1]

Now I'll multiply the first two binomials together:
(x+1)(x-3) = x²-3x+x-3
x²-2x-3
Lastly, we'll multiply the two derived terms together:

(x²+9)(x²-2x-3)   [from the binomial, I'll distribute the first term, then the second term, and I'll stack them so we can simply add like terms together]

x^4 -2x³-3x²
 <u>           +9x²-18x-27</u>
x^4-2x³+6x²-18x-27

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Answer:

the fourth one

Step-by-step explanation:

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2 years ago
What is the solution to the inequality 2 + 4/9x _&gt; 4 + x
Vsevolod [243]
Multiply each side by 9:
18 + 4x > 36 + 9x
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Hope this helps!

8 0
2 years ago
You put $350 per month in an investment plan that pays an APR of 3.5% compounded monthly. How much will you have after
melamori03 [73]

Answer: B.

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8 0
2 years ago
The number of pages that Allen, Kevin, Roneisha, and Ben can read in a day is shown below. Allen read 15% of his 56-page book. K
grandymaker [24]
It would be “Allen” since his percentage is far more grater then the rest, even Roneisha because even though her percentage is 1% greater, she has less pages, Ben has more pages but has 1% less.


Basically it’s like 2 bikers racing, 1 biker is faster but can’t go 5 miles, but the other is not that fast but can go 5+ miles.

Now who can go further, it would be the slow one.
3 0
1 year ago
Read 2 more answers
If 8 identical blackboards are to be divided among 4 schools,how many divisions are possible? How many, if each school mustrecei
MAXImum [283]

Answer:

There are 165 ways to distribute the blackboards between the schools. If at least 1 blackboard goes to each school, then we only have 35 ways.

Step-by-step explanation:

Essentially, this is a problem of balls and sticks. The 8 identical blackboards can be represented as 8 balls, and you assign them to each school by using 3 sticks. Basically each school receives an amount of blackboards equivalent to the amount of balls between 2 sticks: The first school gets all the balls before the first stick, the second school gets all the balls between stick 1 and stick 2, the third school gets the balls between sticks 2 and 3 and the last school gets all remaining balls.

 The problem reduces to take 11 consecutive spots which we will use to localize the balls and the sticks and select 3 places to put the sticks. The amount of ways to do this is {11 \choose 3} = 165 . As a result, we have 165 ways to distribute the blackboards.

If each school needs at least 1 blackboard you can give 1 blackbooard to each of them first and distribute the remaining 4 the same way we did before. This time there will be 4 balls and 3 sticks, so we have to put 3 sticks in 7 spaces (if a school takes what it is between 2 sticks that doesnt have balls between, then that school only gets the first blackboard we assigned to it previously). The amount of ways to localize the sticks is {7 \choose 3} = 35. Thus, there are only 35 ways to distribute the blackboards in this case.

4 0
2 years ago
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