How do you write a polynomial with zeros 8, -i, i and leading coefficient 1?

1 Answer
Alan P.
Mar 2, 2017

#color(green)(x^3-8x^2+x-8)#

Explanation:

If the polynomial has zeros #8, -i, and i#
then it has factors #cxx(x-8)(x+i)(x-i)# where #c# is a constant.

Since we are asked for a polynomial with a leading coefficient of #1#,
#c=1# and we can ignore it from here on.

#color(blue)((x+i)(x-i))=x^2-i^2=x^2-(-1)=color(blue)(x^2+1)#
and
#(x-8)color(blue)((x+i)(x-i))=(x-8)xxcolor(blue)(""(x^2+1))=x^3-8x^2+x-8#

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