Question

Write the polynomial of the smallest degree with roots 6i, -6i, and 5.

Answer 1

if you could show your work I would really appreciate it!

Answer 2

any polynomial P(x) with a set of roots r1, r2, ... rn can be written in factored form as
P(x) = (x-r1)(x-r2)...(x-rn)

Smallest degree implies that none of the roots are repeated, so we'll have 3 factors.

One other thing to remember is that complex roots in a polynomial with only real coefficients (no i appearing in the polynomial) come in conjugate pairs. If one root is a+bi, then a-bi will also be a root (i = sqrt(-1)). Here, we have both roots in the conjugate pair listed, but often these problems will only list one to check your understanding of this concept.

Then having written it in factored form as above, you just multiply the whole thing out to get something like

P(x) = x^3 - x^2r1 -x^2r2 + xr1r2...

Answer 3

Right, I understand that part, but I don't understand how to factor the numbers if that makes sense. So I know I will have x^3 - x^2 + x in that pattern but I don't know what numbers to stick with the x^2 and x because I am having a hard time factoring...

Answer 4

I'll do an example:

find poly of minimum degree with roots 2 and 3i
Well, we need to add -3i as a root to get a complex conjugate pair, so our roots are
2, 3i, -3i
our polynomial is P(x) = (x-2)(x-3i)(x-(-3i)) = (x-2)(x-3i)(x+3i)

to expand that, it generally is easier to multiply the conjugate pair terms together first, so we get rid of the pesky i's:

(x-3i)(x+3i) = x^2 +3ix - 3ix -9i^2 = x^2 - 9i^2
but i^2 = -1, so that becomes x^2 - 9(-1) = x^2 + 9

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

Answer 5

you don't need to do any factoring, you need to do multiplication...

Answer 6

Can you please do what you just did to the example I originally posted? I would REALLY appreciate it!

Answer 7

we can check our work by plugging in the various roots and making sure the polynomial evaluates to 0 as it does.

let's do x = 2:
P(2) = x^3 - 2x^2 + 9x - 18 = (2)^3 - 2(2)^2 + 9(2) - 18 = 8 - 2*4 + 18 - 18 = 0
good so far!

x = 3i:

P(3i) = (3i)^3 - 2(3i)^2 + 9(3i) - 18 = 27i^3 - 2*9i^2 + 27i - 18
remember i^2 = -1 and i^3 = -1*i
=-27i -18(-1) + 27i - 18
= -27i + 18 + 27i - 18
= 0
again, success. x = -3i works out very much the same, so I won't bother to do it.

Answer 8

let's work through your problem together. I know how to do this, you're the one the needs to practice it :-)

desired roots are 6i, -6i, and 5. the polynomial is constructed by multiplying (x-root) together for each root. what will the factored polynomial be?
P(x) = (x-root1)(x-root2)(x-root3) = ?

Answer 9

(x-6i)(x+6i)(x-5)=0

Answer 10

well, the =0 isn't desired here — it won't always be = 0, only when x = one of the roots

otherwise, that's correct. now can you expand
P(x) = (x-6i)(x+6i)(x-5)
into something that will have an x^3, x^2, x, etc?

Answer 11

ok

Answer 12

This is where I'm stuck....I don't know how to expand this

Answer 13

I know x times x times x and x^3 but what next?

Answer 14

pick two of the terms and multiply them together. Multiply the result with the third term.

(x+a)(x+b)(x+c) = [(x+a)(x+b)]*(x+c) = (x^2 + ax + bx + ab)(x+c) right?

Answer 15

just like 2*3*4 can be found by multiplying 2*3 = 6 and 6*4 = 24

Answer 16

ok so (x-6i)(x+6i)

Answer 17

yes, I would start with that, even if they didn't appear first in the list

Answer 18

equals x^2+6Ix-6ix-36i^2

Answer 19

yes, but you can simplify that...

Answer 20

sorry... x^2+6ix-6ix-36i^2

Answer 21

I know I can but I need help with this part.... the "i" confuses me

Answer 22

well, the middle terms combine to what?

Answer 23

well what does the "i" stand for? 1?

Answer 24

doesn't matter! 6ix - 6ix = ?

Answer 25

0

Answer 26

so I'm left with x^2-36?

Answer 27

right! now it does matter what i = for the last term. i is the symbol for the square root of -1. so the square root of -4, for example, is the same as the square root of (4 * -1) which is the same as the square root of -1 * the square root of 4, so the answer is 2i.

what you need to remember is that i^2 = -1. so we had x^2 - 36i^2. if we replace i^2 with -1, we get x^2 - 36(-1) = x^2 + 36

agreed?

Answer 28

yes

Answer 29

so now I multiply (x^2-36)(x-5) right?

Answer 30

it's good to know how to do this, but here's a shortcut that will save you some time:

the complex conjugate pairs are (a+bi) and (a-bi), as I mentioned earlier. a may be 0, as it is in this problem (we could have written the roots as 0+6i and 0-6i without doing anything except causing more wear and tear on our pencil)

when you multiply (x-a)(x+a) you get x^2 - ax + ax -a^2 = x^2 - a^2. this is called a difference of squares, hopefully the reason for that name is obvious :-)

when we multiply the conjugate pair root factors together, we are doing just such a multiplication. here's the shortcut: you can just write down the first part of each root factor (here, "x"), square it, then add the square of the second part of each root factor (here, "6"), and because of the effect of the i^2, we write x^2 + a^2 instead of x^2 - a^2.

so (x+6i)(x-6i) becomes x^2 + 6*6 = x^2 + 36. bam! just write the answer, no need to go through all the steps.

Answer 31

to answer your last question, no, you multiply (x^2 PLUS 36)(x-5)

Answer 32

remember, the (x^2+36) comes from the two complex roots. complex roots arise when we have something like x^2 + 36 = 0

if we try to solve that,
x^2 + 36 - 36 = 0 - 36
x^2 = -36
x = sqrt root of -36
x = 6i or -6i

if we accidentally change it to x^2 - 36 as you've done, then our roots would be
x^2 - 36 = 0
x^2 = 36
x = sqrt root of 36
x = 6 or -6

which is an entirely different kettle of fish...important that you not accidentally change a - to a + or vice versa, if you want to get the right answer(s) :-)

Answer 33

OH I see...

Answer 34

so remember that the multiplication of a pair of complex root factors should always give you x^2 + some number, not x^2 - some number...

Answer 35

So I got x^3-5x^2+36x-180

Answer 36

Ok good to know

Answer 37

and that is exactly correct!!!

Answer 38

YEA!!! Thank you so much for all your help!

Answer 39

You bet. It gets easier with practice. Just work carefully and watch those pesky + and - signs :-)

Answer 40

and when you are done, if in any doubt, you can plug in the roots in the finished polynomial and see if you get 0. you should, if you didn't make any mistakes on either the expansion, or the evaluation. of course, with a complicated enough polynomial, the chances that you might make a mistake in either place goes up...

Answer 41

ok, thanks for the info :)

Answer 42

I'll check this one for you:
x = 5
P(x) = x^3-5x^2 + 36x - 180
P(5) = (5)^3 - 5(5)^2 + 36(5) - 180
P(5) = 125 - 5*25 + 180 - 180
P(5) = 125 - 125 + 180 - 180 = 0
good so far

x = 6i
P(6i) = (6i)^3 - 5(6i)^2 + 36(6i) - 180
= 216i^3 - 5*36i^2 + 216i - 180
= -216i - 180(-1) + 216i - 180
= -216i + 180 + 216i - 180 = 0

again x = -6i will be similar

Answer 43

x = -6i
P(-6i) = (-6i)^3 - 5(-6i)^2 + 36(-6i) - 180
= -216i^3 - 5(36i^2) -216i - 180
= -216(i^2)i -180(i^2) - 216i - 180
= -216(-1)i - 180(-1) - 216i - 180
= 216i + 180 - 216i - 180 = 0