Question

Find the complex zeros of the polynomial function. Write f in factored form.
f(x)=x^3-5x^2+16x-30
Use the complex zeros to write f in factored form.

Answer 1

using the "rational root test", you'd find that 3, is a root of it, a real root
so dividing it for (x-3), that is
$$ \bf
\frac{x^3-5x^2+16x-30}{x-3} \implies x^2-2x+10\\
\text{so you have then}\\
f(x)=(x-3)(x^2-2x+10)
$$

to get the 2 complex roots, I gather I'd use the quadratic formula on the trinomial

Answer 2

The example shows f(x)=(x-2)(x-4-5i)(x-4+5i) for what the examples answer looks like

Answer 3

hmm well

Answer 4

$$\bf
\frac{x^3-5x^2+16x-30}{x-3} \implies x^2-2x+10\\
\text{so you have then}\\
f(x)=(x-3)(x^2-2x+10)\\
x= \cfrac{ 2 \pm \sqrt { 4-40}}{2} \implies x= \cfrac{ 2 \pm \sqrt {-36}}{2}\\
x= \cfrac{ 2 \pm 6i}{2} \implies 1\pm 3i
$$

Answer 5

\(\large f(x)=x^3-5x^2+16x-30 \implies \bf (x-3)(x-1-3i)(x-1+3i)\)

Answer 6

woops, got truncated :/

Answer 7

I wish there was away I could take a picture of the exampes and post them

Answer 8

\(\large f(x)=x^3-5x^2+16x-30 \\ \implies \bf (x-3)(x-1-3i)(x-1+3i)\)

Answer 9

just take a screenshot, in paste it in Paint, and save, and here click on [Attach File]

Answer 10

Find the complex zeros of the polynomial function. Write f in factored form.
f(x)=x^3+8

Answer 11

I don't know how to take a screen shot on a mac

Answer 12

well, is usually easy to post the question in the channel, so we can all see it and help, and revise each other

Answer 13

taking screenshots on a mac.... I think is about the same

Answer 14

http://guides.macrumors.com/Taking_Screenshots_in_Mac_OS_X

Answer 15

ok. I did post the next question on this feed

Answer 16

well, I guess on the main channel :)

either way, this one is pretty simple

Answer 17

$$ \bf
a^3+b^3 = (a+b)(a^2-ab+b^2)\\
x^3+8 \implies x^3+2^3\\
\implies (x+2)(x^2-2x+2^2)
$$

Answer 18

so, there's right off, the real positive root, (x+2)
the rest, you can again, use the quadratic formula to squeeze out the complex ones

Answer 19

First, find the real zero of the function by taking a real cubic root. Then divide f(x) by the linear corresponding to the real zero. Finally, find the complex zero of the resulting quotient using the quadratic formula.

Answer 20

well, right, the thing is to find a real cubic root, usually you'd need to use the \(\bf \text{rational root test}\)

Answer 21

the one you just posted is quite simple because there's an identity already for that form \(\bf a^3+b^3 = (a+b)(a^2-ab+b^2)\)

Answer 22

it's f(x)=x^3+8

Answer 23

\(\bf \color{blue}{a^3+b^3 = (a+b)(a^2-ab+b^2)}\\
x^3+8 \implies x^3+2^3\\
\implies (x+2)(x^2-2x+2^2) \)

Answer 24

that is what I typed in

Answer 25

well, yes, that's a factored form, from the trinomial, you can get the complex ones by using the quadratic formula like for the previous one

Answer 26

when I typed that answer in it said it was wrong then it told me
First, find the real zero of the function by taking a real cubic root. Then divide f(x) by the linear corresponding to the real zero. Finally, find the complex zero of the resulting quotient using the quadratic formula.

Answer 27

yes, because you haven't use the quadratic formula to get the complex ones yes

Answer 28

yet rather

Answer 29

I don't know how to get that answer.

Answer 30

\((x+2)(x^2-2x+2^2)\)
so this is one root, a real positive one => (x+2)
so the other 2 will come out of \(\bf (x^2-2x+2^2)\)
use the quadratic formula on that one, solve for "x"

Answer 31

I have to use websites to help me answer problems and the site I use when I type this in won't give me an option to solve for x

Answer 32

so, just use the quadratic formula, you pretty much have the values to plug right in

Answer 33

i don't know how to do it at all that is why I have to ask for help

Answer 34

$$
\text{let's use the quadratic formula on}\\
\large {
(x^2-2x+2^2) \implies (\color{red}{1}x^2\color{blue}{-2}x+\color{green}{2}^2)\\
x= \cfrac{ - \color{blue}{b} \pm \sqrt { \color{blue}{b}^2 -4\color{red}{a}\color{green}{c}}}{2\color{red}{a}}
}
$$

Answer 35

so who goes where?

Answer 36

see, rather :/
see who goes where? just plug them in and you'd get the values

Answer 37

I've been using this site for my math
http://www.mathway.com/answer.aspx?p=calg?p=(6)SMB10(x)+(7)SMB10(2x)SMB011?p=1?p=?p=?p=?p=?p=0?p=?p=0?p=?p=?p=SolveSMB15theSMB15EquationSMB15forSMB15x