Question

PRE-CALCULUS QUESTIONS.
Solved Problems: 1, 2,
Since the preview doesn't show equations they are typed below -

Answer 1

1. Is -4 a zero of \[f(x)=x^5-x^3+2x^2-3x+1\]Show two ways to find the answer.
2. Given a function and one of its zeros, find all the zeros of a function.\[f(x)=x^3-7x^2+17x-15\]
3. State the # of positive/negative real zeros and imaginary zeros for this function\[f(x)=-7x^4+2x^3+x^2-16\]

Answer 2

One way is to plug it in and see if you get 0.

Answer 3

another way is to divide by \((x-(-4))\) and see if the result is a polynomial.

Answer 4

@wio the second could be done by either long or synthetic division, right?

Answer 5

yeah

Answer 6

Alright, gotcha.
What about #2?

Answer 7

What's the zero for number 2?

Answer 8

Oh, right I forgot to put it up. It's \[2+i\]

Answer 9

4. Write the function of least degree with integral coefficients that has the following zeros: \[2+/-i\sqrt{3},~-1\]
7. Find all of the zeros of f(x) = 2x^4-9x^3+2x^2+21x-10\]

Answer 10

For #2 I think you divide the function with that factor you get:
x=2+i
x-2-i

And solve the the rest equated to 0. Does that make sense?

Answer 11

I'm trying it with synthetic but I'm stuck...

Answer 12

Hmm try:
(x-2-i)(x+a)(x+b) expand and use comparison? >.<

Answer 13

I understand your reasoning but it's not working when I apply it.
- Hold on, there are only two minutes left on the computer I'm working at (library) and I have to save my work. I'll come back in about ... an hour

Answer 14

Back.
#7 rewrite: \[f(x)=2x^4-9x^3+2x^2+21x-10\]

Answer 15

Too tough for me tonight. Good luck.

Answer 16

LOL. Okay...
@douglaswinslowcooper

Answer 17

Q.2
For any polynomial with real coefficients -
-complex roots exist in conjugate

For any polynomial with rational coefficients -
-irrational roots exist in conjugate

Now can you solve Q2 ??

Answer 18

What. @ _ @

Answer 19

Are you saying that all the roots have "i"?

Answer 20

Nope; what I mean to say is - If (2 + i) is a root of above expression; then (2 - i) would also be one of the roots. (try to guess why? )

Answer 21

Oh.... right.
I forgot the reason... is it because of the complex #?

Answer 22

I am not sure what "complex #" really means :P

BTW, this rule will work as long as all the coefficients are real.

Answer 23

I mean a number with the "i" attached, sorry.
Alright... gimme a sec to figure this out lols

Answer 24

Maybe, after this you can teach me #language :D

Answer 25

XD alrighty

Answer 26

So, can you solve Q2 now ??

Answer 27

Sorry, hold on

Answer 28

So the zeros are 2+i and 2-i... and others...?

Answer 29

I got stuck...

Answer 30

Total number of zeros of a polynomial is equal to the degree of polynomial.
As you already have two roots; only 1 remains (and it would be real)

Answer 31

I can see that but I can't solve :/

Answer 32

Divide f(x) by (x-2-i)(x-2+i) to get a linear equation in f(x).

And my previous post was wrong; the third root may be real or may be repeated.

Answer 33

Okay..... Hold on a sec.

Answer 34

Almost done. And I booted you up a level on your SmartScore. :p

Answer 35

O yeah.. XD

Answer 36

Sorry, I'm lagging... hold on ... maybe 20 minutes

Answer 37

Never mind; take you time :)

Answer 38

(2+i)(2-i)= 4 -2i + 2i - 1 = 3?

Answer 39

Actually -
(2+i)(2-i)= 4 -2i + 2i - i^2 = 4 + 1 = 5

In general; you can write -
(a + ib)(a - ib) = a^2 + b^2

Answer 40

o.o oh, right.
Hmmm... okay

Answer 41

Alright, so then I use synthetic division with the 5?

Answer 42

No;
You need to divide f(x) by (x-2-i)(x-2+i)
or in other words; you need to divide f(x) by (x^2 - 4x + 5)

Answer 43

Oh, okay.

Answer 44

... I have no idea how to divide two polynomials.

Answer 45

Do I simplify into factors and then simplify that?