Help to find the real and complex zeroes of: (x - 3)(x + 2)(x+2i)(x - 2i)
you can set the individual factors (x-3) , (x+2), etc . equal to zero
those will give you the 'zeroes' . this is a term i dislike, i prefer to call them roots or solutions
Well, based on the zero product property, when you have for example, \(\large\color{black}{ (x-\color{blue}{a})(x \color{red}{+ b} )(x-\color{green}{c} )}\) then you know when, \(\large\color{black}{ x=\color{blue}{a},~~ \color{red}{- b}, ~~~~~OR~~~~~\color{green}{c} }\) then the expression is zero, right?
Yup. The original question states: Is it possible to create a fourth-degree polynomial with only 2 real zeros? Demonstrate how to do this and explain your steps.
Whatever makes one of the parenthesis, --- one part of the product, to be zero. And you just need to solve for x in your head in 4 cases: \(\large\color{black}{ 1)~~x - 3=0}\) \(\large\color{red}{ 2)~~x + 2=0}\) \(\large\color{black}{ 3)~~x+2i=0}\) \(\large\color{red}{ 4)~~x - 2i=0}\)
And for your original question, you will have a positive answer. Want a quick explanation why?
I mena yes, by a positive answer.
So, you can have this, \(\large\color{black}{ \rm (quadratic~~~~~polynomial) \color{red}{(quadratic~~~~~polynomial)=0}}\) where the black has real solutions, and red has imaginary solutions.
The impossible part would be to have 3 imaginary and 1 real, or 1 imaginary and 3 reals.
The reason for this is, because for at least 1 polynomial out of 2, you will need one imaginary and one real solutions, and you know that in a quadratic polynomial, EITHER; 1) both solutions are real 2) both solutions are imaginary
If anything doesn't make sense, I"ll elaborate...;0
I think i understand :) Is that it?
yes:)
As long as you don't have a situation where you get a quadratic polynomial with 1 imaginary solution, and one real. again, either both solutions are real, or they are both imaginary. (whether they are extraneous or not).
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