Question

Find a polynomial function P(x) of least possible degree, having real coefficients, with the zeros 2 and 3i. Let the leading coefficient be 1.

Answer 1

Can the zeroes only be 2 and 3i, can there be more zeroes?

Answer 2

The problem only listed 2 and 3i :/ Here is how I started the problem: 1(x-2)((x-(x-3i))((x-(x+3i))

Answer 3

now just foil those.

Answer 4

so 9x-18???

Answer 5

i didnt do the math. but i dont see how you come out with a linear term with 3 x's in there. you should have a cubic function

Answer 6

1(x-2)((x-(x-3i))((x-(x+3i)) -- In this expression you have subtracted x - (x - 3i) so the two x's cancel out. But it should be just x - 3i in place of x - (x - 3i) and the same for x - (x + 3i).

Answer 7

yep, i didnt pay enough attention to catch that. good eye

Answer 8

here is simpler way to deal with imaginary zeros
\[x = 3i\]
\[x^2 = 9i^2\]
\[x^2 = -9\]
\[x^2 +9 = 0\]
polynomial is:
--> (x-2)(x^2 +9)

Answer 9

thats interesting. it looks so simple. why isnt it taught that way? and what makes that legal to do out of curiosity?

Answer 10

i guess. it just makes sense. because it is a zero of the polynomial.

Answer 11

right, it works because you are just working backwards
start with solution then work back until right side is zero, then you have the factor

x = 2

x-2 = 0

Answer 12

i wish they taught it that way. makes doing the math a lot quicker

Answer 13

yep some teachers do

Answer 14

x3−8x2+22x−20 is what I got for the polynomial :/ I broke it down to (x-2)(x-3-i)(x-3+i) and then multiplied.

Answer 15

Multiply each term in the first polynomial by each term in the second polynomial

Answer 16

no try my method

(x-2)(x^2 +9) = x^3 -2x^2 +9x -18

Answer 17

okie. However I'm still a little unclear to where the x^2+9 came from...

Answer 18

can you not see my earlier post above??

i show steps

Answer 19

so you put: x=3i because 3i is one of the zeros. and you squared it. X^2+9=3i Okay, I see how you got there. Thank you! i'm going to try your way on similar problems till I get it down :)

Answer 20

:)

it works for when the zero is a sqrt too
anytime the zero is an imaginary or sqrt they come in pairs so the factor is always a quadratic

if have any other questions concerning this tag me

Answer 21

awesome! made things much simpler