 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 usaho4w

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:
$f\left(x\right)=a\left(x-{a}_{1}\right)\left(x-{a}_{2}\right)\left(x-{a}_{3}\right)\left(x-{a}_{4}\right)\left(x-{a}_{5}\right)$
Where a is the leading coefficient since the given polynomial is monic so the leading coefficient is 1.
$⇒f\left(x\right)=1\left(x-\left(3-i\right)\right)\left(x-\left(3+i\right)\right)\left(x-\left(i\right)\right)\left(x-\left(-i\right)\right)\left(x-2\right)$
$⇒f\left(x\right)=1\left(x-3+i\right)\left(x-3-i\right)\left(x-i\right)\left(x+i\right)\left(x-2\right)$
Simplifying we get $\left(x-3+i\right)\left(x-3-i\right)={x}^{2}-6x+10$
$\left(x-i\right)\left(x+i\right)={x}^{2}+1$
$⇒f\left(x\right)=\left({x}^{2}-6x+10\right)\left({x}^{2}+1\right)\left(x-2\right)$
$⇒f\left(x\right)={x}^{5}-8{x}^{4}+23{x}^{3}-28{x}^{2}+22x-20$
Therefore, the required monic polynomial is $f\left(x\right)={x}^{5}-8{x}^{4}+23{x}^{3}-28{x}^{2}+22x-20$

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