How do you write the polynomial function of least degree with integral coefficients that has the give zeros for 2, 4i?

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Answer 1 · 2016-11-25T04:55:42

$f(x)=x^3-2x^2+16x-32$

If the function has a zero at $4i$ , it also has one at $-4i$ .

If a function has a zero at $a$ , it has a factor of $x-a$ .

So, this function has factors of $(x-2)$ , $(x-4i)$ , and $(x+4i)$ .

The function can be written as

$f(x)=(x-2)(x-4i)(x+4i)$

Mutliplying $(x-4i)(x+4i)$ gives

$f(x)=(x-2)(x^2+4i-4i-16i^2)$

Recall that $i^2=-1$

$f(x)=(x-2)(x^2+16)$

Multiply each term in the first binomial by each term in the second binomial.

$f(x)=x^3+16x-2x^2-32$

Rearranging....

$f(x)=x^3-2x^2+16x-32$