Question

Demonstrate a fourth-degree polynomial with only two real zeros.

Answer 1

Can someone please give me an example of one, i know how to solve them but i am very bad at creating fourth-degree polynomial problems that could be solved with two real zeros.

Answer 2

@amistre64 @Hero , can either one of you guys help me please? i am very sorry to bother you but i really need this one.

Answer 3

@zepdrix can you help me please?

Answer 4

Creating one? Ummmm
Let's start from the ground up.
Let's make up some easy roots that a 4th degree polynomial could have.
4 roots:

x=1
x=-1
x=i
x=-i

2 of those roots are real, the other two are complex ( and are conjugates of one another which is important, complex roots always come in pairs like that ).

So for the first root we'll subtract 1 from each side,
(x-1)=0
Do the same with all of the roots,
(x+1)=0
(x-i)=0
(x+i)=0

We can form a 4th degree polynomial by multiplying all of these factors together.\[\Large (x+1)(x-1)(x-i)(x+i)\quad=\quad 0\]

Answer 5

Ahhh i understand. so next i would just multiply them together right?

Answer 6

Yes :) Understand how to multiply the brackets containing the i's?

Answer 7

but could you help me in explaining how this is done? and a program tells me this equation equals to x^4 - 1, is that correct?

Answer 8

Explain how what is done? :o

Answer 9

Program told you? :x lol
Yes that is the correct equation.

Answer 10

really? but where did the i's go? and how would i take this equation (x^4 -1) and find its zeros?

Answer 11

\[\Large x^4-1\quad=\quad (x^2)^2-(1^2)^2\]We have the difference of squares,
Remember how to factor the difference of squares?\[\Large \color{royalblue}{a^2-b^2=(a-b)(a+b)}\]

Answer 12

i remembered that the i's are equal to -1 so i understand how the equation came to be. and i remember the difference of square, it would be (x^2 -1)(x^2 +1) right?

Answer 13

yes good so far

Answer 14

\[\Large i=\sqrt{-1}\]

Answer 15

\[\Large x^2-1\quad=\quad x^2-1^2\]Again we have difference of squares here.

Answer 16

wouldn't they be (x-1)(x+1)(x+1)(x+1) ? and if i factor this equation (x^4 -1) wouldn't i get (x^2 +1)(x-1)(x+1)

Answer 17

\[\Large x^4-1\quad=\quad \color{orangered}{(x^2-1)}(x^2+1)\]\[\Large =\quad \color{orangered}{(x-1)(x+1)}(x^2+1)\]

The other part is the `sum of squares` which results from multiplying complex conjugates.
It might be easier to understand if we look at it from the factored direction.

\[\Large (x-i)(x+i) \quad=\quad x^2-ix+ix-i^2\]I `FOIL`ed out the brackets ok?

Answer 18

\[\Large =\quad x^2-i^2\]

Answer 19

\[\Large \color{royalblue}{i^2=-1}\]

Answer 20

ahhhhhhhhhhhhhh i^2 = -1 i remember that! that makes more sense

Answer 21

so would this be all the steps needed to prove that this is an equation is a fourth-degree polynomial with two real zeros right? or is there more steps that are important?

Answer 22

This is one of those identities that doesn't come up as often. But it's still worth remembering I think :P

The product of complex conjugates gives you the `sum of squares` similar to how
the product of conjugates gives you the `difference of squares`.

\[\Large (x-\color{orangered}{1}i)(x+\color{orangered}{1}i)\quad=\quad x^2+\color{orangered}{1}^2\]

Answer 23

ummm

Answer 24

Hmm, I think that's all that's needed +_+
You started with 2 real zeros and 2 imaginary zeroes and multiplied them together to form a 4th degree polynomial.

If you wanted to make the problem even simpler you could let the real roots be x=0 and x=0. That would simplify the multiplication a little bit :)

\[\Large (x)(x)(x-i)(x+i)=0\]

Answer 25

Oh snap the time! :O I gotta get to class.
Hope that helps!

Answer 26

ok thank you very very much! :D