Question

find zeros..will give out medals!

Answer 1

descartes rule of sign for this one

Answer 2

can you help me find the answer?

Answer 3

\[f(x)=-2x^3-5x^2+6x+4\] coefficients are \(-2,-5,6,4\)

Answer 4

there is one "change of sign" for \(f(x)\) from \(-5\) to \(+6\)

Answer 5

ok what is next?

Answer 6

so there is one positive zero
you count down by 2s and there cannot be -1 positive zero

Answer 7

ok

Answer 8

oh btw there are 3 zeros all together because the degree is 3, although some may be complex or repeated

Answer 9

now we compute \(f(-x)\)

Answer 10

ok

Answer 11

\[f(-x)=2x^3-5x^2-6x+4\]

Answer 12

ok

Answer 13

the sign of the coefficients of the terms of odd degree change, and the even degree stays the same
the coefficients are \(2,-5,-6,4\)

Answer 14

now there are two "changes in sign" from 2 to -5 and from -6 to 4
is that more or less clear?

Answer 15

ok is that the final answer?

Answer 16

can you help me with one more? it is easier this time

Answer 17

oh no
two changes in sign of \(f(-x)\) means either 2 or 0 negative zeros

Answer 18

oh ok sorry..please finish the problem

Answer 19

so final answer is this:
1 positive zero for sure
either 2 negative zeros and therefore no complex zzeros
or no negative zeros and two complex zeros

Answer 20

ok thanks..can you do one more? please?

Answer 21

this one is easier
\[\sum_3^53n+2\]

Answer 22

put \(n=3\) get \(3\times 3+2=11\)
then put \(n=4\) get \(3\times 4+2=14\)
finally put \(n=5\) get \(17\)
the "sigma" \(\sum\) means add them up

Answer 23

\[11+14+17\]is what you need to compute, and you are done

Answer 24

thats it?

Answer 25

thanks