Question

Need help with a few questions please! {MEDAL & FAN}

Answer 1

HI!!

Answer 2

which one you want to do first?

Answer 3

Lets Go From Top To bottom please & I've been wanting your help lol but you haven't been responding

Answer 4

ok you want to do it the way it says, or you want to do it the easy way, or both?

Answer 5

Both

Answer 6

lets do it the easy way first
to solve \[x^2+5x+6=0\] is not too hard, because it factors as \[(x+2)(x+3)=0\] making the zeros \(-2\) and \(-3\) so descartes rule of sign, even though we did not use it, better say "no positive zeros, two negative ones"

Answer 7

now lets go ahead and use descartes rule of sign
\[x^2+5x+6\] has three coefficients, respectively \(\{1,5,6\}\)

Answer 8

they are all positive, no changes in sign from \(+\) to \(-1\) so NO positive zeros

Answer 9

Okay I see. The first way you did is easier than the way the teacher wants

Answer 10

now replace \(x\) by \(-x\) and get \[(-x)^2+5(-x)+6=x^2-5x+6\]

Answer 11

@misty can you help me with my question after this please you seem to be good at math

Answer 12

here the coefficents are \(\{1,-5,6\}\) which has two changes in sign from \(+\) to \(-\) and from \(-\) to \(+\)

Answer 13

that means either 2 or 0 negative zeros, you count down by 2

Answer 14

Ok I see, shall we move on?

Answer 15

ok you try the next one
how many changes in sign ?

Answer 16

Two changes in sign

Answer 17

Wait One change!

Answer 18

yes so either 2 positive zeros or no positive zeros, you count down by twos

Answer 19

no 2

Answer 20

Ok So I was correct before lol Sorry about that

Answer 21

from 9 to -4, from -4 to +10

Answer 22

Yeah lol I shouldn't have changed my answer

Answer 23

ok so now we know there are 2 or 0 positive zeros
now turn x in to -x

Answer 24

Ok

Answer 25

you should get \[-9x^2-4x^2+10\] if you do it right
how many changes in sign here ?

Answer 26

Two still

Answer 27

try again

Answer 28

One?

Answer 29

yea -9, -4, +10 only from -4 to +10

Answer 30

that means there is one negative zero

Answer 31

because you count down by twos, so there has to be one negative zero

Answer 32

Two or no positive real roots; One negative real root

Answer 33

yes

Answer 34

Lets move onto the last one? I appreciate your help as well Misty!

Answer 35

your welcome, i hope the idea is getting clearer

Answer 36

how many changes in sign of the last one ?

Answer 37

Three changes in Sign

Answer 38

8, 2, -14, 5

Answer 39

Yeah.

Answer 40

so what are the changes in sign? or maybe that part is not clear

Answer 41

From 8 to 2, 2 to -14, -14 to 5

Answer 42

but 8 and 2 are both positive right?

Answer 43

Yes they are

Answer 44

\[\{+8,+2,-14,+5\}\]

Answer 45

yeah.

Answer 46

so what are the sign changes, from what number to what number?

Answer 47

They change from +2 to -14 then from -14 to +5

Answer 48

right that is two changes

Answer 49

not three
so 2 or no positive zeros

Answer 50

Oh yeah Ive been thinking you were talking about changes in sign as in +,-,+

Answer 51

But now I understand. So i now see Its two changes

Answer 52

oh i see
yeah two changes from plus to minus, from minus to plus
now replace x by \(-x\)

Answer 53

Ok After changing X to -X i'd get

Answer 54

-8x^3 + -2x^3 - -14x + 5

Answer 55

Wait its a few errors in that I think lol but you get the point^

Answer 56

lets clean it up a bit

Answer 57

should be \[-8x^2+2x^2+14x+5\] right?

Answer 58

Yes

Answer 59

how many changes in sign here?

Answer 60

one

Answer 61

yup so one negative zero

Answer 62

one negative real root

Answer 63

xactly

Answer 64

Thanks so much, can you help with this final question?

Answer 65

lol you got a lot of time you can check each one to see which works, will be very annoying and i would not bother

Answer 66

lets just cheat

Answer 67

lol The Algebrea Calc won't work on this

Answer 68

Algebra *

Answer 69

watch this
http://www.wolframalpha.com/input/?i=6x%5E4-13x%5E3%2B13x%5E2-39x-15

Answer 70

Wolfram lol I see. Thanks misty

Answer 71

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