Question

Use Descartes' Rule of Signs to determine the number of negative real zeros of the following function: f(t) = -5t^7 + 8t^2 − 2

Answer 1

http://ededu.net/Main/Math/175precal/175Q2/175Q24A1.gif <-- use descartes rule of signs then

any ideas on how many real positive ones?

Answer 2

1 ? @jdoe0001

Answer 3

hmmm well.... how did you get 1 though?

Answer 4

Ohh wait the -5 well it would be 2 because the change from - to + is 1 then back to - is 2?

Answer 5

one sec

Answer 6

yeap

checking f( x )
f(t) = -5t^7 + 8t^2 − 2
^ ^ ^
yes yes

so 2
OR
2-2 = 0
real positive solutions or zeros

checking now for f( -x )
f(t) = +5t^7 + 8t^2 − 2
^ ^ ^
nope yes

so 1 real negative zero or solution

Answer 7

Alright i despise that one mistake messes it up

Answer 8

soh for positive you just look at the signs and for negative you switch them?

Answer 9

be right back

Answer 10

well...for negative roots.... what you do is replace the "x" argument with "-x"

so say \(\large {
f(-t) = -5({\color{brown}{ -t}})^7 + 8({\color{brown}{ -t}})^2 - 2\\ \quad \\
\begin{cases}
(-t)^7\to -t\cdot -t\cdot -t\cdot -t\cdot -t\cdot -t\cdot -t\to -t^7\\
(-t)^2\to -t\cdot -t\to +t
\end{cases}
\\ \quad \\
f(-t) = +5{\color{brown}{ -t}}^7 + 8{\color{brown}{ t}}^2 - 2
}\)

thus, usually when the exponent of the variable is EVEN, then the sign remains the same
and when it's ODD, the sign changes

since - * - * - = -
and
- * - = +

Answer 11

back sorry

Answer 12

Hokay i understand can we do a couple more jsut to make sure

Answer 13

sure

Answer 14

Use Descartes' Rule of Signs to determine the number of positive real zeros of the following function. f(x) = -3x^4 + 7x^2 − x + 5

Answer 15

Well this equals 3

Answer 16

well..you're asked for only the real positive ones

so you don't have to check for f( -x )

just for f(x) which is given thus

f(x) = -3x^4 + 7x^2 − x + 5
^ ^ ^ ^
yes yes yes

3 changes....thus real POSITIVE ONES will be
3
OR
3-2 = 1

Answer 17

recall that you'd usually subtract any values greater than 1 by 2

so say if we have had say 6 changes
then it'd be 6 or 6 -2, or 6-2-2, or 6-2-2-2
so 6 or 4 or 2 or 0


if it were 5 then
5 or 5-2, or 5-2-2

or 5 or 3 or 1

Answer 18

soh for this one it would be 1 or 3?

Answer 19

yeap

Answer 20

hokay. Use Descartes' Rule of Signs to determine the number of negative real zeros of the following function. f(x) = -9x^4 + 2x^3 + x^2 + x − 1

soh this one has 2 changes before using -x

Answer 21

yeap
but if you use f( -x )
recall... if the exponent is EVEN, it remains the same sign
if ODD, it changes

Answer 22

hokay soh -9x^4 -2^3 +x^2 - x -1?

Answer 23

2 changes

Answer 24

yeap

so 2 OR 2-2 = 0
real negative roots

Answer 25

Use Descartes' Rule of Signs to determine the number of negative real zeros of the following function. f(x) = 2x^4 − 5x^3 + 8x^2 − 3x + 1

Soh the changes before -x is 3?

Answer 26

I lie its 4

Answer 27

so the real positive ones are
4 or
4-2
or 4-2-2

Answer 28

ok soh 2x^4 +5x^3 +8x^2 + 3x + 1 There is no change for negative

Answer 29

yeap so 0 negative ones... it doesn't have any real negative solutions or roots

Answer 30

Thank you for your help soooo much

Answer 31

yw