Question

Using the given zero, find all other zeros of f(x).

-2i is a zero of f(x) = x^4 - 21x^2 - 100

Answer 1

We have a polynomial with only real coefficients, so our complex roots must come in conjugate pairs. What is the conjugate of \(-2i\)?

Answer 2

+2i?

Answer 3

imaginary roots always come in pairs right? :)

Answer 4

yea

Answer 5

yes, you are correct. what is (x-2i)(x+2i)? We can divide that out of our polynomial to get a simpler polynomial to find the remaining roots.

Answer 6

all i know so far is

(x+2i)*(2-2i)*(x+a)*(x+b) =0 :P could be wrong tho.

Answer 7

change that to
(x+2i)(x-2i)(x+a)(x+b) = 0
and I'll agree with you

Answer 8

\((x+2i)(x-2i) \implies x^2-(2i)^2 \implies x^2-4i^2 \implies x^2+4\)

Answer 9

so, if we have
\[x^2 + a^2 = 0\]\[x^2=-a^2\] \[x=\pm ai\]
So that means we can skip multiplying and just go directly to \[(x-2i)(x+2i) = x^2+4\]
Now let's divide our original polynomial by \(x^2+4\)

Answer 10

a quick look suggests that it's going to be \(x^2-25\) because we can take out \(x^2(x^2+4) = x^4+4x^2\) and that leaves us with \(-25x^2-100\) which is just \(-25(x^2+4)\).

What are the zeros of \(x^2-25\)?

Answer 11

You should recognize that as a difference of squares...

Answer 12

\[a^2-b^2 = (a-b)(a+b)\]

Answer 13

oh!

Answer 14

(x-25)(x+25)

Answer 15

sorry!

Answer 16

(x-5)(x+5)

Answer 17

right. so the final two zeros are?

Answer 18

+/- 5?

Answer 19

yes. so the entire list of zeros of the polynomial is
\[x = \{-2i,2i,-5,5\}\]