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.

Degree 3 polynomial with integer coefficients with zeros −3i and 9/5.

vicki331g8

Beginner2021-12-13Added 37 answers

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)

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

$={x}^{2}(x-\frac{9}{5})+9(x-\frac{9}{5})$

$={x}^{3}-\frac{9{x}^{2}}{5}+9x-\frac{81}{5}$

$={x}^{3}-\frac{9{x}^{2}}{5}+9x-\frac{81}{5}$

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

Nadine Salcido

Beginner2021-12-14Added 34 answers

Zeros -3i and $\frac{9}{5}$

Complex zero must have conjugate

$x=\pm 3i\text{}x=\frac{9}{5}$

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

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

$=({x}^{2}+9)(x-\frac{9}{5})$

$={x}^{3}+9x-\frac{9}{5}{x}^{2}-\frac{81}{5}$

$={x}^{3}-\frac{9}{5}{x}^{2}+9x-\frac{81}{5}$

Complex zero must have conjugate

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