Astudent claims that -4i is the only imaginary root of a quadratic polynomial equation that has real coefficients.

1. what is the student’s mistake?
2. write one possible polynomial that has the correct roots from part a in standard form.

explain your answer. you!

1. There will be two roots not only one

2.a( +16) =0

Step-by-step explanation:

Simply assume a quadratic equation

Here a,b,c are real independent variables.

As given by student -4i is one of it's root it must satisfy the equation.

-16a +4ib+c=0

 b must be zero otherwise either a or c must be imaginary number which is not possible.

So b =0

then, c= 16a

a+c=a+16a

Quadratic is a( +16) =0 here a is any real number.

4th option is correct xy=-10

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all the angles opposite to the angle 150 would be 150 and all the angles that are next to 150 are 30 because they equal 180 it you add it up.

step-by-step explanation: