Question

What are the roots of the following polynomial equation?
(x + 5)(x + 2i)(x – 5i) = 0

Answer 1

They're staring you right in the face!

Answer 2

Set each piece equal to zero, via the zero product principle. Then, according to the complex conjugates theorem, you know that if -2i is a solution then...

Answer 3

i got -2i , and 5i

Answer 4

There's one more.

Answer 5

is that right?

Answer 6

actually... there are a total of 5 zeros

Answer 7

Do enlighten us!

Answer 8

see my earlier post... complex conjugates theorem tells us that if -2i is a root of a polynomial equation, then +2i is ALSO a root.

Answer 9

Example 1. Let P(x) = 5x³ − 4x² + 7x − 8. Then a root of that polynomial is 1, because


P(1) = 5· 1³ − 4· 1² + 7· 1 − 8
=5 − 4 + 7 − 8
=0


It is traditional to speak of a root of a polynomial. Of a function in general, we speak of a zero.

This is just an example I swiped off a website but maybe it will help.
Hope it does.

Answer 10

\((x + 5)(x + 2i)(x – 5i) = 0\) is a degree-3 polynomial. The fundamental theorem of algebra, basically, states that a polynomial of degree \(n\) has \(n\) roots. Which one is it?

Answer 11

It's not... it's a 5th degree polynomial with 1 real root and 4 imaginary roots. This means it has only one x-intercept.

Answer 12

How it it a degree-5 polynomial?

Answer 13

there are 3

Answer 14

if you expand there is no way to get a higher than x^3 term.

Answer 15

Look up "Complex Conjugates Theorem." It states that if 3 + i is a root of an equation, so is 3 - i. So, if you're only solving the equation listed, you're correct (there are only 3 terms listed in the product so it seems that there are only 3 solutions). However, as the question asked what are the roots of the equation, you must also consider the conjugates of any complex root.

Answer 16

imaginary roots always come in pairs and i think 2i is the pair of 5i

Answer 17

i believe the answer is more simple then where over looking it, -5,-2i,5i i think we just have to wright the inverse of the in the ( )

Answer 18

You got confused.

Answer 19

No, that's not right, timo86m. 2i must pair with -2i and 5i must pair with -5i.

Answer 20

The Complex Conjugate Theorem requires that the polynomial has real coefficients. If you expand this thing out, you'll have complex ones.

Answer 21

This is an elaborate debate.

Answer 22

Guys, this is why I always recommend you take everything you get told in here with a huge grain of salt.

Answer 23

I am with across on this one

Answer 24

here is what you get when you expand it

x^3-(3*I)*x^2+10*x+5*x^2-(15*I)*x+50

Answer 25

Usually when a question is asked about roots in a HS class where some are imaginary, they are testing an understanding of Complex Conjugates theory... And this is coming from a veteran math teacher of 10 years. I guess it would be helpful to know if Complex Conjugates were discussed in this chapter or unit.

Answer 26

ok even maple tells me

{x = -5}, {x = -2*I}, {x = 5*I}

Answer 27

Clearly, that is not being tested here today.

Answer 28

I didn't think it was clear and I'm sorry if I confused the discussion in any way. I was trying to help and nothing I've explained is wrong. If this is an Algebra 2 class, then I'm probably over-explaining. If this is precalc, then this would be part of the discussion... at least it is in the school where I teach.

Answer 29

it doesn't matter the class the point is it has 3 zeros. Do you finally get that?

Answer 30

I know where you're coming from, and I totally agree with your bringing of the Complex Conjugate Theorem. I just don't want @Tea-Is-God thinking that a degree-3 polynomial has 5 roots. :)

Answer 31

i think by his logic

(x+2*I)*(x-5*I) would have 4 roots.

Answer 32

thank you for the spirited debate, @across. @timo86m, there is more to my point than you're realizing. your comment and tone with "do you finally get that" was unnecessary given that i was only trying to help.

Answer 33

You're welcome. And @timo86m, that wasn't necessary at all.

Answer 34

No i mean it in a way do you get it? it is literal question. Sorry it sounded bad.