Chris Cruz

2021-12-11

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.

3- i, i, and 2

3- i, i, and 2

usaho4w

Beginner2021-12-12Added 39 answers

Step 1

Given:

The zeros of the polynomial 3-i, i and 2

Step 2

The complex roots occur in pair, so conjugate of 3-i and i also be the zeros of the same polynomial

So, the polynomial has zeros: 3-i, 3+i, i , -i and 2

If$a}_{1},{a}_{2},{a}_{3},{a}_{4},{a}_{5$ are the zeros of the polynomial then the polynomial in factored form is written as:

$f\left(x\right)=a(x-{a}_{1})(x-{a}_{2})(x-{a}_{3})(x-{a}_{4})(x-{a}_{5})$

Where a is the leading coefficient since the given polynomial is monic so the leading coefficient is 1.

$\Rightarrow f\left(x\right)=1(x-(3-i))(x-(3+i))(x-\left(i\right))(x-(-i))(x-2)$

$\Rightarrow f\left(x\right)=1(x-3+i)(x-3-i)(x-i)(x+i)(x-2)$

Simplifying we get$(x-3+i)(x-3-i)={x}^{2}-6x+10$

$(x-i)(x+i)={x}^{2}+1$

$\Rightarrow f\left(x\right)=({x}^{2}-6x+10)({x}^{2}+1)(x-2)$

$\Rightarrow f\left(x\right)={x}^{5}-8{x}^{4}+23{x}^{3}-28{x}^{2}+22x-20$

Therefore, the required monic polynomial is$f\left(x\right)={x}^{5}-8{x}^{4}+23{x}^{3}-28{x}^{2}+22x-20$

Given:

The zeros of the polynomial 3-i, i and 2

Step 2

The complex roots occur in pair, so conjugate of 3-i and i also be the zeros of the same polynomial

So, the polynomial has zeros: 3-i, 3+i, i , -i and 2

If

Where a is the leading coefficient since the given polynomial is monic so the leading coefficient is 1.

Simplifying we get

Therefore, the required monic polynomial is

$\frac{20b}{{\left(4{b}^{3}\right)}^{3}}$

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