Question

Using the given zero, find one other zero of f(x). Explain the process you used to find your solution.

2 - 3i is a zero of f(x) = x4 - 4x3 + 14x2 - 4x + 13.

Answer 1

do you know how to do long division?

Answer 2

ahemm. I meant, long division of polynomials that is

Answer 3

kind of.. not really sure how they even got the i in 2-3i though.

Answer 4

well.. is a complex root, that is, an imaginary expression, and in this case, it's given

Answer 5

oh okay. so how do i find the other zero?

Answer 6

ok... well, the complex root comes from using the the square root from say the quadratic formula, like \(\bf x= \cfrac{ - b \pm \sqrt { b^2 -4ac}}{2a}\)

notice the \(\boldsymbol{\pm}\) which means it has a "negative version" of the root, and a "positive version" of it

so if we see a complex one, we can tell that it's not there all by her lonesome, it has a companion, her conjugate

Answer 7

so... gimme a sec

Answer 8

\(\bf 2 - 3i \quad conjugate \implies 2+3i\\
\textit{so we can say that our roots are}\\
x = 2 - 3i\implies x - 2 + 3i=0\implies (x - 2 + 3i)=0\\ \quad \\
x = 2 + 3i\implies x - 2 - 3i=0\implies (x - 2 - 3i)=0\\ \quad \\
\textit{thus}\\ \quad \\
(x - 2 + 3i)(x - 2 - 3i)=0\\ \quad \\
\textit{recall that }\quad (a-b)(a+b) = (a^2-b^2)\\ \quad \\
(x - 2 + 3i)(x - 2 - 3i)=0\implies (x - 2) + (3i)(x - 2) - (3i)=0\\ \quad \\
(x - 2)^2 - (3i)^2\implies x^2-4x+2^2-3^2i^2\implies x^2-4x+4-9(-1)\\ \quad \\
x^2-4x+9\implies x^2-4x+13\)

Answer 9

so the quotient of dividing our polynomial, for THAT root, will give us a quadratic, that has our other 2 roots

Answer 10

hmm... I'm not getting 0 as remainder

Answer 11

Wait a sec.... if you have one complex root, then its complex conjugate has to be another root.

So "find one other zero".... just give the complex conjugate as another zero.

Answer 12

hmmm done that

Answer 13

ohhh... I see shoot

Answer 14

they don't want ALL ROOTS, just some OTHER ROOT

Answer 15

lol yes, I see that... just wans't sure why we were worrying about doing long division of the polynomial. :)

Answer 16

har har har.. well, yes I was =)

Answer 17

Still always a valuable exercise. :)

Answer 18

hmmm,, it has not real roots, no wonder the polynomial wasn't giving me a 0 remainder heheh

Answer 19

well, that's what I get for not doing a rule of signs check first =)

Answer 20

i don't get it. so x^2-4x+13 is the answer?

Answer 21

hmmm no, see what they're asking is

"Using the given zero, find \(\bf \text{one other zero}\) of f(x). Explain the process you used to find your solution."

Answer 22

so, as I explained before, complex roots don't come all by their lonesome, they have their conjugate companion

so the \(\bf \text{one other zero}\) will be its conjugate :)

Answer 23

so 2 + 3i?

Answer 24

yeap