Question

Use Descartes' Rule of Signs to determine the possible numbers of positive and negative real zeros of f. Then, list the possible rational zeros of f and graph f so that some of the possible zeros in parts A and B can be disregarded and determine all of the real zeros of f.

The function given is f(x)=-3x^3+20x^2-36x+16

Answer 1

do you really have to do all that or can you just cut to the chase and list the three zero?

Answer 2

I have to do all of those steps :( lol

Answer 3

do you think you could still help me out? this is my last problem and I would realllly appreciate it

Answer 4

is there maybe a typo in your question? check carefully, because there are two complex zeros and one irrational zero of this one

Answer 5

oh maybe i did not see the minus sign in front!

Answer 6

It's from a textbook so i dont know :o I'll type exactly what it says!

"In Exercises 61-66, (a) use Descartes's Rule of Signs to determine the possible numbers of positive and negative real zeros of f, (b) list the possible rational zeros of f, (c) use a graphing utility to graph f so that some of the possible zeros in parts (a) and (b) can be disregarded, and (d) determine all the real zeros of f."

Answer 7

and then the one that I was assigned to work on was

62. f(x)=-3x^3+20x^2-36x+16

Answer 8

ok fine
lets do d first
\(-(3 x-2) (x-2) (x-4)\) is what you have so the zeros are
\[2,4,\frac{2}{3}\]

Answer 9

now to A
use Descartes's Rule of Signs to determine the possible numbers of positive and negative real zeros of f,
here
\[f(x)=-3x^3+20x^2-36x+16\] and there are three changes in sign of the coefficients, from \(-3\) to \(+20\) then from \(+20\) to \(-36\) and then again from \(-36\) to \(+16\)

Answer 10

that means there could be 3 positive zeros, and you count down by twos, so 3 positive zeros, or 1 positive zero

Answer 11

for the negative zeros we find the changes in sign of \(f(-x)\) and
\[f(-x)=3x^3+20x^2+36x+16\]
this has no changes in sign
so there are no negative zeros for this function

Answer 12

did i lose you yet?

Answer 13

not yet :p this is starting to make some sense!

Answer 14

i hope it is clear how i computed \(f(-x)\) easily
i just changed the sign of degree 3 and 1 term, left the others alone
so to summarize there are either 3 or 1 positive zero, no negative zeros that means either 3 positive zeros or 1 positive zero and 2 complex zeros

Answer 15

because since it has degree 3, there are three zeros all together, not necessarily all real
now on to part B

Answer 16

, (b) list the possible rational zeros of f,
similar to the last one all numbers that divide 16 or fractions where the numerator divides 16 and the denominator divides the leading coefficient of 3

Answer 17

like last time you get
\[\pm1,\pm2,\pm4,\pm8,\pm16\] but not also
\[\pm\frac{1}{3},\pm\frac{2}{3},\pm\frac{4}{3},\pm\frac{8}{3},\pm\frac{16}{3}\] quite a long list

Answer 18

+/-1, +/-2, +/-4, +/-8, +/-16 and then all of that over 3?

Answer 19

right

Answer 20

now since we know there are no negative zeros, we can cut the list in half and get rid of the negative ones

Answer 21

(c) use a graphing utility to graph f so that some of the possible zeros in parts
ok lets do it, we should have done this at the start
http://www.wolframalpha.com/input/?i=-3x^3%2B20x^2-36x%2B16

Answer 22

so just 1,2,4,8,16,1/3,2/3,4/3,8/3, and 16/3?

Answer 23

yes

Answer 24

now check the link i sent you above
you get the graph in it

Answer 25

oh wow and thats it?

Answer 26

you see there are 3 positive zeros, and they are between 0 and 4, and in fact they are \(\frac{2}{3},2,4\)

Answer 27

which means this one factors as \[-(3 x-2) (x-2) (x-4)\]

Answer 28

and yes, that is it!

Answer 29

wow thank you SO much!!! I really appreciate it!!! :)

Answer 30

if i were to do this same exact problem with all parts but a different function of f(x)=4x^4-17x^2+4, how would I begin this one? I know that there are 2 sign changes but I dont know how to get the rest just yet

Answer 31

so far I have that there are 2 sign changes which means 2 positive zeros or 0 positive zeros
then, I did f(-x) to find negative zeros and got 4x^4+17x^2+4

Answer 32

there are no sign changes which means no negative zeros right?

Answer 33

then i found factors of 4 which were +/-1,+/-2,+/-4 but since there are no negative zeros it is only 1,2,4

Answer 34

am i right so far? @satellite73

Answer 35

hold on a moment

Answer 36

\[f(x)=4x^4-17x^2+4\] two changes in sign
so either 2 positive zeros or no positive zeros

Answer 37

but \(f(-x)\) is not right
all the exponents are even, so \(f(-x)=4x^7-17x^2+4\) same as \(f(x)\) the function is "even"

Answer 38

this means 2 or no negative zeros as well

Answer 39

hm.. i factored the function and got (x-2)(x+2)(2x-1)(2x+1) which gave me x=2 x=-2 x=1/2 x=-1/2

Answer 40

oh man! so did that just mess up all of the math that i did?

Answer 41

no no you are right
it is all right

Answer 42

so would my final answer be x=2 x=-2 x=1/2 x=-1/2 AND 2 positive, 2 negative zeros?

Answer 43

you only wrote one thing wrong
\[f(-x) =4x^4+17x^2+4\] that is NOT correct

Answer 44

it should be \(f(-x)=4x^4-17x^2+4\)
the rest is right

Answer 45

and your zeros are right also

Answer 46

\[-2,2,-\frac{1}{2},\frac{1}{2}\] all that is correct
two positive, two negative

Answer 47

oh i see thats what i meant to type on here but i have it right on my paper :p sorry about that! thank you so much!!! :)

Answer 48

yw, hope this helped and it makes some sense

Answer 49

yes it really did thats how i was able to do this problem! many thanks to you, you are awesome!!! :)

Answer 50

yw, my pleasure