Question

Can anyone help me with these two cause im really confused?
1. Find a third-degree polynomial equation with rational coefficients that has roots -4 and 6+i
A. x^3-8x^2-11x+148=0
B. x^3-8x^2-12x+37=0
C. x^3-12x^2+37x=0
D. x^3-8x^2-11 =0

2. what does Descartes's rule of signs tell you about the real roots of the polynomial?
-2x^3+3x^2-5x-2=0

A. there is one positive root and either 2 or 0 negative roots
B. there are either 2 or 0 positive roots and there are either 2 or 0 negative roots
C. there is one positive root and one negative root.
D. there are either 2 or 0 positive

Answer 1

D^and one negative root

Answer 2

you can either construct an equation from the given roots, or test the roots to see which option they fit into

Answer 3

the rule of sign refers to a property that descartes noticed; the number of roots has a relationship to the number of sign changes in a setup

Answer 4

rule of sign doesnt tell us what the roots are, it just helps us determine a possible number

-2x^3 + 3x^2 - 5x - 2 = 0
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1 1

there are 2 sign changes as is, so there is at most 2 positive roots, and since complex roots come in pairs, there is also a chance of 0 positive roots being taken up by 2 complex roots

Answer 5

letting x be a negative term, we can assess the negative roots

-2(-x)^3 + 3(-x)^2 - 5(-x) - 2 = 0

2x^3 + 3x^2 + 5x - 2 = 0
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1

there has to be 1 negative root, which cannot be removed in pairs

Answer 6

so for #2 it would be B or D?

Answer 7

B suggests that it would have 2 or 0 negative roots; and we can rest assured that it has 1 negative root. So B is out

D has no mention of negative roots, but it retains the idea of 2 or 0 positives which is true

Answer 8

so how would you do #1 im really confused o.O

Answer 9

#1 can be constructed by recalling the complex roots come in conjugate pairs, so we have 3 roots stated. Then the function has to be a product of the stated roots-x

for example: spose we are given the roots (or zeroes) to be: a,b,c

the function can then be defined as: f(x) = (a-x)(b-x)(c-x)

Answer 10

okay so how would i do the problem :P

Answer 11

what are the stated roots?

Answer 12

it is -4 and 6+i?

Answer 13

thats only 2 of them, you need to remember what a conjugate pair is ...

Answer 14

well thats the only one it shows on the problem :P

Answer 15

it gives you 1 real root, and 1 complex (imaginary) root.

complex roots come in conjugate pairs ... so they have actually given you all 3 roots to play with, you simply have to recall what it means to be a conjugate pair

Answer 16

so far we can construct this thing as:\[f(x)=(-4-x)(6+i-x)(??-x)\]
where ?? is the conjugate of 6+i

Answer 17

or, we can see if the rule of signs can help us narrow down the odds to 1 negative and 2 complexes

Answer 18

so would it be 6-i-x?

Answer 19

yes

Answer 20

okay so f(x)=(-4-x)(6+i-x)(6-i-x)
then what do you do after that

Answer 21

multiply it out of course ....

Answer 22

we only need the constant products since all the options have different constant values

Answer 23

what is:\[-4(6+i)(6-i)\]??

Answer 24

do you have multiply it ?

Answer 25

yes

Answer 26

did i do this right -24-24i

Answer 27

-24i-24i

Answer 28

not quite :) conjugate pairs produce a difference of squares:

(a+b)(a-b) = a^2-b^2

(6+i)(6-i) = 6^2 - i^2
= 36 - (-1)
= 37

what is 4(37) ?

Answer 29

148 right?

Answer 30

yes, and which option has a constant of 148?

Answer 31

option A

Answer 32

then option A is the correct choice

Answer 33

thank you for the help :)