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Write the polynomial in standard form and identify the zeros of the function and the x-intercepts of its graph. f(x)=(x-1)(x-1)(x+2i)(x-2i)


So far I have foiled and got the polynomial in standard form as :

f(x)=x^4-2x^3-5x^2-8x+4


How do I find the zeros?

Verified Answer :

\(f(x)=(x-1)(x-1)(x+2i)(x-2i) \\f(x)=(x^2-2x+1)(x^2-(2i)^2) \\f(x)=(x^2-2x+1)(x^2-4i^2) \\f(x)=(x^2-2x+1)(x^2-4(-1)) \\f(x)=f(x)=(x^2-2x+1)(x^2+4) \\f(x)=x^4-2x^3+x^2+4x^2-8x+4 \\f(x)=x^4-2x^3+5x^2-8x+4\)

zeros of the function

\(f(x)=(x-1)(x-1)(x+2i)(x-2i) \\x-1=0 \Rightarrow x=1 \\x+2i=0 \Rightarrow x=-2i \\x-2i=0 \Rightarrow x=2i\)

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