Find a monic polynomial f(x) of least degree over C that has the given numbers as zeros, and a monic polynomial g(x) of least degree with real coefficients that has the given numbers as zeros.

2i, 3

2i, 3

Gregory Emery
2022-01-16
Answered

Find a monic polynomial f(x) of least degree over C that has the given numbers as zeros, and a monic polynomial g(x) of least degree with real coefficients that has the given numbers as zeros.

2i, 3

2i, 3

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jgardner33v4

Answered 2022-01-17
Author has **35** answers

Step 1: Introduction

Monic Polynomial: Mono stands for Single Therefore it is a type of polynomial having single variable or single constraints used in the function. The degree of the polynomial or the highest power of the given polynomial is 1.

Step 2: To Find

A monic polynomial f(x) of least degree over C that has the given numbers as zeros, and

a monic polynomial g(x) of least degree with real coefficients that has the given numbers as zeros.

Step 3: Solution to the question

Both 2i and 3 are the zeroes of the functions, therefore,

We have,

f(x)=(x-2i)(x-3)

=x(x-3)-2i(x-3)

$={x}^{2}-3x-2ix+6i$

$={x}^{2}-(3x+2i)x+6i$

g(x)=(x-2i)(x+2i)(x-3)

$=({x}^{2}-{\left(2i\right)}^{2})(x-3)$

$=({x}^{2}+4)(x+3)$

$={x}^{3}+3{x}^{2}+4x+12$

Monic Polynomial: Mono stands for Single Therefore it is a type of polynomial having single variable or single constraints used in the function. The degree of the polynomial or the highest power of the given polynomial is 1.

Step 2: To Find

A monic polynomial f(x) of least degree over C that has the given numbers as zeros, and

a monic polynomial g(x) of least degree with real coefficients that has the given numbers as zeros.

Step 3: Solution to the question

Both 2i and 3 are the zeroes of the functions, therefore,

We have,

f(x)=(x-2i)(x-3)

=x(x-3)-2i(x-3)

g(x)=(x-2i)(x+2i)(x-3)

Medicim6

Answered 2022-01-18
Author has **33** answers

To find a monic polynomial f(x) of least degree over C that has the given numbers as zeroes and a monic polynomials g(x) of least degree with real coefficient that has given numbers as zeroes.

2i, 3

f(x)=(x-2i)(x-3)

$={x}^{2}-(3+2i)x+6i$

g(x)=(x-2i)(x+2i)(x-3)

$=({x}^{2}+4)(x-3)$

$={x}^{3}-3{x}^{2}+4x-12$ .

2i, 3

f(x)=(x-2i)(x-3)

g(x)=(x-2i)(x+2i)(x-3)

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To be fair I wasn't able to make any progress. I tried using substitution for the logarithms, but it didn't help at all.

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Any help? Clue?

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What I did is:

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