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skad [1K]
3 years ago
9

How many solutions over the complex number system does this polynomial have? 7x^5-33x^4-4x^2+3x+52=0

Mathematics
2 answers:
telo118 [61]3 years ago
8 0
Hello : 
in : C
5 solutions
Harrizon [31]3 years ago
7 0

Answer:

Over the complex number there exist two roots.

Step-by-step explanation:

Given : Polynomial 7x^5-33x^4-4x^2+3x+52=0

To find : How many solutions over the complex number system does this polynomial have?                

Solution :

Descartes' rule of sign is used to determine the number of real zeros of a polynomial function.

Let f(x)=7x^5-33x^4-4x^2+3x+52

Now, for positive roots write the sign of f(x) and note the changes,

f(x)=+ - - + + i.e, from + to - and - to + two times sign changes.

So, Two positive roots.

Now, for negative roots write the sign of f(-x) and note the changes,

f(-x)=- - - + + i.e, from - to + one time sign changes.

So, One negative roots.

Since, The polynomial is of 5 degree so five roots exist.

i.e, rest two roots are complex.

Therefore, Over the complex number there exist two roots.

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Substitute the x and y values from the ordered pairs into the equation, one by one, until both sides of the equation are equal.

Step-by-step explanation:

A. (-2,-3)

3 * (-2) - 4 * (-3) = 21

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B. (0,7)

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6 0
2 years ago
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Answer:

z = 7

Step-by-step explanation:

Based on the midsegment theorem of a triangle, the length of the midsegment of the triangle (z) = ½ of the length of the 3rd side of the triangle (14)

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4 0
3 years ago
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Bingel [31]
<span>reducible.

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