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beks73 [17]
3 years ago
8

How many complex roots does the polynomial equation have? 3x5−2x+1=0

Mathematics
2 answers:
dmitriy555 [2]3 years ago
3 0

Answer : The given polynomial equation can have 0,2 or 4 complex roots.

Explanation:-

Given polynomial equation is 3x^5-2x+1=0 which is a polynomial of degree 5.

We know that the complex roots always occur in pair ,therefore the number of complex roots in any polynomial must be an even number but less than equal  to the degree of polynomial.

Thus, the given polynomial equation has 4 complex roots.

Nana76 [90]3 years ago
3 0

Answer:

The polynomial equation have 4 complex roots.

Step-by-step explanation:

We are given a polynomial equation as:

3x^5-2x+1=0

clearly the polynomial equation is a equation of degree 5 hence, the polynomial equation has atmost 5 roots.

Now we find the roots of the equation by it's graph.

we could clearly see that the graph ofthe function passes through the point (-1,0). hence -1 is the root of the polynomial equation, other than this point the graph does not touches or pass through the x-axis.

Hence, the other four roots of the polynomial equation are complex.

Hence, the polynomial equation has 4 complex roots.



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The table represents the start of the division of 3x^4+22x^3+37x^2-7x+10<br>by the indicated divisor
andreyandreev [35.5K]
The quotient is x^3 + 4x^2 -x + 1.

Solution:
By polynomial grid division, we start by the divisor 3x + 10 placed on the column headings.
               3x      10
     x^3    3x^4   

We know that 3x^4 must be in the top left which means that the first row entry must be x^3. So the row and column multiply to 3x^4. We use this to fill in all of the first row, multiplying x^3 by the terms of the column entries.
                 3x       10
     x^3      3x^4   10x^3 
     4x^2

We now got 10x^3 though we want 22x^3. The next cubic entry must then be 12x^3 so that the overall sum is 22x^3.
                  3x         10
     x^3      3x^4     10x^3 
     4x^2    12x^3

Now we have 40x^2, so the next quadratic entry must be -3x^2 so that the overall sum is 37x^2.
                 3x          10
     x^3      3x^4      10x^3 
     4x^2    12x^3    40x^2
     -x       -3x^2      -10x

This time we have -10x, so the next linear entry must be 3x so that the overall sum is 7x.
                 3x          10
     x^3      3x^4     10x^3 
     4x^2    12x^3    40x^2
     -x         -3x^2    -10x
     1           3x         10

The bottom and final term is 10, which is our desired answer. Therefore, we can now read the quotient off the first column:
     3x^4+22x^3+37x^2-7x+10 / 3x + 10 = x^3 + 4x^2 -x + 1
8 0
3 years ago
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