How do I find the complete factored form of a polynomial with a degree of #3#, having a leading coefficient of #2# with some zeros #i# and #1#?

1 Answer
George C.
Mar 30, 2018

#2(x-i)(x+i)(x-1)#

Explanation:

Assuming we want the standard form of the cubic to have real coefficients, any zeros occur in complex conjugate pairs. So if #i# is a zero then so is #-i#.

Also note that #x=a# is a zero if and only if #(x-a)# is a factor.

Hence we can write our cubic polynomial in factored form as:

#f(x) = 2(x-i)(x+i)(x-1)#

We can multiply this out to find standard form:

#2(x-i)(x+i)(x-1) = 2(x^2-i^2)(x-1)#

#color(white)(2(x-i)(x+i)(x-1)) = 2(x^2+1)(x-1)#

#color(white)(2(x-i)(x+i)(x-1)) = 2(x^3-x^2+x-1)#

#color(white)(2(x-i)(x+i)(x-1)) = 2x^3-2x^2+2x-2#

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