 Lorraine Harvey

2021-12-12

Write a polynomial f(x) that satisfies the given conditions.
Degree 3 polynomial with integer coefficients with zeros −3i and 9/5. vicki331g8

Step 1
In this case, a polynomial function of 3 with the zeros 3i and 9/5 is written.
Step 2
Let f(x) be the ploynomial function
Given that 3i are the zeros of f(x), +3i are the zeros of f. (x)
[because complex conjugate each other]
Since −3i be the zeros of f(x) then (x+3i) be the factor of f(x).
Since +3i be the zeros of f(x) then (x−3i) be tge factor of f(x)
Since 9/5 be the zeros of f(x) , then (x−9/5) be the factor of f(x).
Step 3
Therefore f(x)=(x+3i)(x−3i)(x−9/6)
$=\left({x}^{2}-{\left(3i\right)}^{2}\right)\left(x-\frac{9}{5}\right)$
$={x}^{2}\left(x-\frac{9}{5}\right)+9\left(x-\frac{9}{5}\right)$
$={x}^{3}-\frac{9{x}^{2}}{5}+9x-\frac{81}{5}$
$={x}^{3}-\frac{9{x}^{2}}{5}+9x-\frac{81}{5}$
Hence $f\left(x\right)={x}^{3}-\frac{9{x}^{2}}{5}+9x-\frac{81}{5}$ Zeros -3i and $\frac{9}{5}$
Complex zero must have conjugate

$f\left(x\right)=\left(x-3i\right)\left(x+3i\right)\left(x-\frac{9}{5}\right)$
$=\left({x}^{2}-{\left(3i\right)}^{2}\right)\left(x-\frac{9}{5}\right)$
$=\left({x}^{2}+9\right)\left(x-\frac{9}{5}\right)$
$={x}^{3}+9x-\frac{9}{5}{x}^{2}-\frac{81}{5}$
$={x}^{3}-\frac{9}{5}{x}^{2}+9x-\frac{81}{5}$

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