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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:

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Step-by-step explanation:

(5n - 5)(4n - 4)

= 5n(4n) + 5n(-4) - 5(4n) - 5(-4)

= 20n² - 20n - 20n + 20

= 20n² - 40n + 20

Another way to do this:

(5n - 5)(4n - 4)

= 5(n - 1) * 4(n - 1)

= 20(n - 1)(n - 1)

= 20(n - 1)²

= 20(n² - 2n + 1)

= 20n² - 40n + 20

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Answer:

As i understand this:

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We want to create a two digit number such that:

Not all the cards are used.

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Now, first remember how rounding works.

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if b is less than 5, we round down.

Now in this case we want to have a digit equal 6, if we use this in the units place, then when rounding, we will round up.

Then the other digit can be 1, such that we use the cards:

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then our number is 16:

When rounding, 6 > 5, then we round up to 20.

So this number meets all the conditions.

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PilotLPTM [1.2K]
That is  true good job dude keep up the good work
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