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#