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?
@satellite73
@phi
@amistre64
your initial product form was: f(x) = (x-1) (x-1) (x+2i) (x-2i) correct?
Yes.
and, when you are multiply number together; and one of them is equal to zero; what does that do for the answer?
for example: 8*6*23*n = ? ; when n=0
It makes it zero.
thats the key to this problem then: (x-1) (x-1) (x+2i) (x-2i) = 0 when any of the (...) parts equal zero
So I plug in numbers for x, and the zeros are whichever numbers, when plugged into the problem, make the equation = 0?
exactly, therefore (x-1) = 0 when x=? (x-1) = 0 when x=? (x+2i) = 0 when x=? (x-2i) = 0 when x=?
so my zeros are... 1, and + or - 2i?
yes
Oh, thank you.
youre welcome, and good luck ;)
:)
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