Question

What are the possible number of positive, negative, and complex zeros of f(x) = -2x3 + 5x2 + 6x - 4 ?

Positive: 2 or 0; Negative: 1; Complex: 2 or 1
Positive: 1; Negative: 2 or 0; Complex 2 or 0
Positive: 2 or 0; Negative: 1; Complex: 2 or 0
Positive: 1; Negative: 2 or 0; Complex: 0

Answer 1

@aum @ganeshie8

Answer 2

Descartes' Rule, anyone? ; )

Answer 3

I am not that familiar with it? could you please explain it? :)

Answer 4

For the positive real roots, you count the number of sign changes as your expression is right now. It starts with a -2 then goes into a +5, so that's one sign change. It remains a positive with the 6x, but changes to a negative with the 4. So that means that there is a total of 2 possible positive real roots or 0. You always go down by two. so 2 or 0 positive real roots.

Answer 5

Now for the negative roots, you fill in your x's with -x's to see if that changes the signs anywhere in your expression.

Answer 6

\[-2(-x)^{3}+5(-x)^{2}+6(-x)-4\]

Answer 7

negative x cubed leaves the x as a negative, but multiply that by the -2 that is already there and your sign is now positive. The negative x squared leaves the x a positive, but since the sign in front of the 5 is already positive, there is no sign change there. the sign will change for the positive 6 x when you change the x to a negative x. So that's the first change in the expression. and the sign with the 4 is negative so no change there.

Answer 8

So here's what we have:

Answer 9

it makes sense so far :)

Answer 10

That table means that you can have 2 positive roots and 1 negative root and no complex roots. You have to have a total of 3 because this is a third degree polynomial. You can also have 0 real roots, 1 negative root, and 2 complex roots. You cannot have 1 negative roots because you have 1 negative, not enough to take 2 away. So the negative root is 1 no matter what. You will find the complex roots AFTER you find the possible combinations of the positive and negative roots; what's needed to make up your degree of your polynomial after you find the negative and positive roots is complex. You can't find complex directly.

Answer 11

So that's your lesson on Descartes' Rule! And to think...it took my Algebra 2 class a whole hour of listening to me crab at them til they got it!

Answer 12

I see...lol so you're a teacher? thank you for explaining you make sense!

Answer 13

I meant to say up there that you cannot have 0 negative roots. I said you cannot have 1 negative roots. That was a typo...sorry.

Answer 14

it's fine

Answer 15

Do you sort of understand?

Answer 16

TY for the medal, btw.

Answer 17

np, I understand for the positive but so for the negative do we always need to plug in -x in here .... −2(−x)3+5(−x)2+6(−x)−4

Answer 18

yes for the negative you always have to replace the x with -x. Because the -x is negative.

Answer 19

positive signs use the (x), negative signs use the (-x). Always.

Answer 20

ok thx for the clarification :) we end with 2 positive roots and 1 negative root and no complex roots right?

Answer 21

Look at the table. You can have 2 positive, 1 negative and no complex...or you can have 0 positive, 1 negative, and 2 complex.

Answer 22

Your problem does ask for complex as well as positive and negative.

Answer 23

Those are all the possibilities for this third degree polynomial.

Answer 24

alright thanks! can you help me with another one that focuses on negative?

Answer 25

You would have to actually factor it to find out the right combination of positives and negatives and complex.

Answer 26

I can sure try to help!

Answer 27

thx! :)

What are the possible number of negative zeros of f(x) = 2x7 + 2x6 + 7x5 + 7x4 - 4x3 + 4x2 ?

2 or 0
3 or 1
4, 2, or 0
7, 5, 3, or 1

Answer 28

BTW that last answer is your third choice down.

Answer 29

Now let me look at the one you just posted...

Answer 30

thx again! :)

Answer 31

The rule of thumb here is that if the exponent is an odd number, when you have -x to that odd power, you're going to end up with a negative. Like with 2x^7, originally the 2 is positive, but (-x)^7 is negative, so a negative times a positive is negative. (-x)^6 is a positive but since the 2 is already a positive, there is no sign change there. The next term, (-x)^5 is a negative, and since the sign on the 7 is a positive, a negative times a positive is negative. So that is another sign change.So far we have gone from negative to positive to negative. Two changes so far.

Answer 32

Now for the (-x)^4, that is a positive number and since the sign on that 7 is positive, there is no sign change there. On the (-x)^3, that is a negative number, times the negative on the 4 and you get a positive. So another change. Lastly we have (-x)^2, which is positive times the positive on the 4 and no sign change. SOOOOOOO.......

Answer 33

wait so only a sign change happens when we go from negative to positive to negative? cuz i thought in the last problem it was different?

Answer 34

It looks like we have a total of 3 sign changes here which makes it a total of 3 negative roots or 1 negative root.

Answer 35

Nope, there is only a sign change when it goes from a positive to a negative and then back to a positive.

Answer 36

even like when there's two either positive or negative?

Answer 37

like pos neg neg pos