Question

How do I write a polynomial function with a degree of 4 and real coefficients in standard form that has 3, -1, and 1+5i as its zeros?

Answer 1

@Vocaloid

Answer 2

Can you help me with this problem?

Answer 3

if 1 + 5i is a zero then 1 - 5i must also be a zero [forgot what the rule is called]

Answer 4

okay...

Answer 5

so for every zero you are given, write an expression for x that gives you that zero

for example, for "3" let (x-3) be part of your final function (since x - 3 = 0 gives you 3 as a zero)

Answer 6

how do i write and expression for x that gives me the complex numbers?

Answer 7

Oh Welcome to Questioncove

Answer 8

then you have

(x - 3)
(x + 1)
(x - [1+5i])
(x + [1-5i])

multiply these together and you should have a polynomial with degree 4

Answer 9

okay.

Answer 10

thank you.

Answer 11

wait

Answer 12

(x-[1+5i]) gives me what result?

Answer 13

made a typo on that last one

(x - 3)
(x + 1)
(x - [1+5i])
(x - [1-5i])

Answer 14

anyway, if you multiply these 4 things together you should get the answer

Answer 15

(x-[1+5i]) means that x = 1 + 5i is a zero [as stated by the original problem]

Answer 16

how did you get the last two?

Answer 17

ok

Answer 18

if x = 1 + 5i = 0

then x - (1+5i) = 0
and x - (1 - 5i) = 0 using that same logic

Answer 19

so (x-[1-5i]) means that...?

Answer 20

1 - 5i is one of the zeros, with me so far?

Answer 21

yes, so (x-[1-5i]) is x=1-5i?

Answer 22

yes

Answer 23

ok let me solve it then

Answer 24

you don't need to solve it

Answer 25

you just need to write down all of your zero expressions and then multiply them together

Answer 26

that what i meant (misunderstanding)

Answer 27

but wait why when you typed (x-[1+5i]) and (x-[1-5i]) you used x- and not x+?

Answer 28

if 3 is our zero, then we have x - 3 = 0 as our zero expression, right?

same logic applies for [1-5i]

Answer 29

x - the zero = 0

Answer 30

ok good

Answer 31

can you please help me how do i multiply (x-[1+5i]) and (x-[1-5i])?

Answer 32

use foil

Answer 33

ok

Answer 34

how do i multiply -[1+5i] and -[1-5i]?

Answer 35

(A-B)(C-D) = AC - AD - BC + BD

treat [1+5i] and [1-5i] as your B and D values

Answer 36

ok let me try it will probably last about two minutes for my brain to process please be patient with me.

Answer 37

it's perfectly fine, take your time and be careful with your signs ^^

Answer 38

(x^2-2x-3)(x^2-x[1-5i]-x[1+5i] is this correct so far?

Answer 39

i still need the last part the L from FOIL

Answer 40

yeah that's what I was going to say^^

Answer 41

but i am confused on that one

Answer 42

the last term is just

[1 + 5i] * [1-5i]

which can be solved using foil or difference of squares

Answer 43

(A+B)(A-B) = A^2 - B^2

Answer 44

but what about the negatives that where in front of them?

Answer 45

they are positive

Answer 46

the negative signs cancel out

Answer 47

so if it would of had a negative and a positive then it would be a negative right?

Answer 48

sure

Answer 49

when dealing with complex numbers?

Answer 50

(-[1 + 5i]) * (-[1-5i]) = (-1)(-1)[1 + 5i] * [1-5i] so the signs cancel out

Answer 51

the fact that they're complex numbers have nothing to do with this, this is just basic foil rules

Answer 52

we're not using the signs inside the complex terms yet

Answer 53

so the result of multiplying them is -9?

Answer 54

that's not what I get ^^ check your algebra again

Answer 55

[1 + 5i] * [1-5i] = 1^2 - (5i)^2 = ?

Answer 56

ok let me mind process it about 2 mins

Answer 57

remember that i^2 = -1

Answer 58

what does -(5i)^2 represent?

Answer 59

the quantity 5i squared, then multiplied by -1

Answer 60

so that means you got -5i and -5i for the OI part of the FOIL method?

Answer 61

one of those should be + 5i

Answer 62

because i got -5i and + 5i

Answer 63

good, those cancel out so we only have the first and last terms

Answer 64

I got 1-5i+5i-10i^2

Answer 65

check your last term again

Answer 66

5i * 5i = ?

Answer 67

(1+5i)(1-5i), for the L part of the FOIL method I got -5i * 5i = -10i^2

Answer 68

hint: 5*5 = ?

Answer 69

10

Answer 70

* means multiply

what is 5 times 5?

Answer 71

alright

Answer 72

its 25

Answer 73

bruh -_- let me try it then

Answer 74

good, so

[1 + 5i] * [1-5i] = 1^2 - (5i)^2 = 1 - 25i^2 = ?

Answer 75

26

Answer 76

good

so let's add that to what you already wrote earlier.

Answer 77

(x^2-2x-3)(x^2-x[1-5i]-x[1+5i] + 26)

Answer 78

(x^2-2x-3)(x^2-x[1-5i]-x[1+5i]+26)

Answer 79

now how do i multiply all of this?

Answer 80

distributive property

Answer 81

however

Answer 82

(x^2-x[1-5i]-x[1+5i]+26)

^ you can simplify this to get rid of all i terms

Answer 83

how can i do that?

Answer 84

start by distributing inside the parentheses

Answer 85

ok so like multiply x(1-5i)?

Answer 86

yes (be careful with signs)

Answer 87

ok

Answer 88

do i cancel the brackets ([]) or leave them

Answer 89

after you distribute you should remove the appropriate brackets

Answer 90

so for the first one is it -x+5ix?

Answer 91

yup keep going

Answer 92

ok let me keep going

Answer 93

(x^2-x+5ix-x-5ix+26)

Answer 94

good, and do you see anything you can cross/cancel out?

Answer 95

ok let me try it

Answer 96

so i got (x^2-2x+26)

Answer 97

good, now you put that back together with what we had before

Answer 98

(x^2 - 2x - 3)*(x^2 - 2x + 26)

then distribute

Answer 99

ok let me try it

Answer 100

x^2(x^2-2x+26)-2x(x^2-2x+26)-3(x^2-2x+26)?

Answer 101

yeah, but keep distributing

Answer 102

I think it's better if you go term by term instead of lumping (x^2-2x+26) together

Answer 103

how?

Answer 104

start by distributing according to these arrows

Answer 105

ok but its basically the same thing right?

Answer 106

yeah i guess

Answer 107

alright i like your strategy because this will not fit on my test paper

Answer 108

i am still here i am just slow for this because i am learning something new ok be patient please

Answer 109

no need to apologize ^^

Answer 110

so is it x^4-4x^3+37x^2-46x-78?

Answer 111

that 37 should be a 27 but other than that good job ^^

Answer 112

wait a 27?

Answer 113

oh yes

Answer 114

i know why 30 -3

Answer 115

ok thank you very much that is the final result right?

Answer 116

yup

Answer 117

how do i gratify you is there a way to give you points or something?

Answer 118

you can click "best response" if you'd like c;

Answer 119

where on your last comment?

Answer 120

it doesn't matter, any comment will work

Answer 121

ok thank you very much.