Question

Form a polynomial f(x) with real coefficients having given degree and zeros. degree 5; zeros -8;-i;4+i

Answer 1

Well a polynomial, by defintion of the linear factorization theorm states that a degree of n>1, then it can be written in linear factors with its zeros.

It is written in the form (x-a)(x-b) where this is a 2nd degree polynomial with zeros a and b

If x = a is a zero, then (x - a) is a factor. If a + bi is a zero, then a - bi is a zero as well.

So see how you have 3 zeros given but this is a 5th degree polynomial? That neans it has 5 factors, but where are the other two? Well they are the cnjugates of -i and 4+i, do you know what those are?

Answer 2

umm I really don't know.... I have spent over 2hours trying to figure this problem out

Answer 3

Please Help!!!!!

Answer 4

Lol, ok we can work through this and I'll stay until you get it, do you know how you have real numbers, numbers you can write like 1, 2, 3.14159..., 8.99579, 22/7, and then you have imaginary numbers such as \[\Large i=\sqrt{-1}\]

Answer 5

yes

Answer 6

well you can sometimes write them together to make what is called a "complex number" it it written in the form \[\Large a+bi\] where a and b are real numbers and i is the imaginary number

Answer 7

really any number can be written as a complex number, like \[\Large 1+2i\]
is a complex number but so is 2, 2 is a complex number because it can be written like this \[\Large 2+0i\] becasue 0 times anything is still 0, and -3i can be written as a complex number because its the same thing as \[\Large 0-3i\]

Answer 8

OK, sorry im back anyways, a complex number ahs a conjugate and the conjugate can be described as saying if you have a complex number (rmemeber any number real or imaginary or a combo of both can be written as this way) in the form \[\Large a+bi\]
then its conjuagte can be denoted as \[\Huge a-bi\]

Answer 9

Now, theres a theorem called the fundamental theormrem of algebra that states that if you have a polynomial with n-degree, it has n-roots (zeros) so in this case its a 5 degree, so it has 5 zeros. Also it says that if a complex number and its conjugate is also a zero so if they say -i is a zero then its conjugate ( this can be written as 0-i=0+i=i) is i. so i and -i are zeros tot his function.

Also 4+i is zero meaning that 4-i is also a zero so now that accounts for two more zeros so now you can wirte it in linear factorization form

\[\Large (x-(-8)\times(x-(i))\times(x-(-i))\times(x-(4+i))\times(x-(4-i))\]

Answer 10

wow, is the last number (x-6)

Answer 11

no sorry it flew off the page \[ (x-(-8)\times(x-(i))\times(x-(-i))\times(x-(4+i))\times(x-(4-i))\]

Answer 12

:) thanks sooooo much... your the best of the class this year!!!!lol

Answer 13

Did you understand my explanaation? Also all you have to do now is multiply the two last ones together then the two ones aftert hat and then multiply by (x-(-8))=(x+8) btw

Answer 14

yes I did, and plus I took notes.. thanx!!!

Answer 15

np, thats what we're here for :)