Question

Find the polynomial function f(x) of lowest degree which has zeros 2i (of multiplicity of 2) and 3 (of mult. of 1) and for which f(1)=5.

Answer 1

with real coefficients?

Answer 2

whatever that means

Answer 3

do you know a simple quadratic that has one zero of \(2i\)>

Answer 4

oh actually (x-2i)^2

Answer 5

lol

Answer 6

wait is that question to me @satellite73 or to the person

Answer 7

lol

Answer 8

but if he wants real coefficients which I'm assume he does
he will need both (x-2i)^2 and (x+2i)^2 to be factors of the given poly

Answer 9

that is why the i'm asking about the real coefficients thing...

Answer 10

Yes, using real coef. so that would make the lowest degree 5?

Answer 11

of course they are supposed to be real
the quadratic with zero of \(2i\) is \(x^2+4\) since if \[x=2i\\x^2=-4\\
x^2+4=0\]

Answer 12

yes, degree 5

Answer 13

polys can have complex coefficients

Answer 14

but yeah if you want real coefficients
you can say you have x=2i twice
which means you have the factor x-2i twice
what I mean is (x-2i)^2 will be a factor of your poly
and since we want real coefficents you will neeed also x=-2i twice
or the factor x+2i twice
so you will have (x-2i)^2(x+2i)^2 which you can expand to see you will have all real coefficients
then if x=3 you know what else is a factor of your poly?
once you found that I will on and say how to use the f(1)=5 thing

Answer 15

or you can use what @satellite73 said above

Answer 16

x^2+4=0 has solutions x=2i or x=-2i
so to get those zeros twice how many (x^2+4) 's do you need in your poly?

Answer 17

I'm at (x-2i)^2(x+2i)^2(x-3) and setting f(1) would make it (1-2i)^2(1+2i)^2(-2)

Answer 18

whoa nelly too much work !

Answer 19

one sec though
...so yeah you do have all of that (x-2i)^2(x+2i)^2(x-3)
but you need one more thing since you have the condition f(1)=5
you need a constant in which to solve for so that when you replace x with 1 you get output 5
so you have
\[f(x)=a(x-2i)^2(x+2i)^2(x-3)\]

Answer 20

multiply out and deal with real numbers

Answer 21

a would be 5

Answer 22

no one wants to compute what you wrote \[f(x)=a(x^2+4)^2(x-3)\] then put \(f(1)=5\) and solve for \(a\)

Answer 23

\[f(1)=a(1-2i)^2(1+2i)^2(1-3) \\ 5=a[(1-2i)(1+2i)]^2(-2) \\ 5=a[1-4i^2]^2(-2) \\ 5=a[1+4]^2(-2)\]

Answer 24

so a isn't 5

Answer 25

Ok, I see. So 5=-50a?

Answer 26

right and then divide both sides by -50

Answer 27

and reduce fraction

Answer 28

couple of things I want to mention if they weren't clear above
I used law of exponents \[a^2 \cdot b^2 =(a \cdot b)^2\]
and I also used that easy formula for multiplying conjugates \[(a+bi)(a-bi)=a^2-b^2 \cdot i^2=a^2+b^2\]
or you might know it better as
\[(a-b)(a+b)=a^2-b^2\]

Answer 29

Yes, it makes sense where it came from.