Question

Using Descartes? rule of signs, how many real roots does f(x) = x^3 + x^2 - 3x + 2 contain?

Answer 1

My teacher didnt explain it well and i only have one example in my notes

Answer 2

no positive real roots, no negative real roots
two or zero positive real roots, one or zero negative real roots
one or zero positive real roots, two or zero negative real roots
two or zero positive real roots, two or zero negative real roots ??????

Answer 3

I think it may be the last one? can someone help me understand this?

Answer 4

Check out my post in your last question for an example of how to use the rule of signs...

Answer 5

We have a highest power of \(x^3\) here, so there are 3 roots. We know that because this polynomial only has real coefficients, any complex roots come in pairs. That means our options are 1 real, 2 complex, or 3 real. Any answer that would require 1 or 3 complex roots isn't going work here.

Answer 6

Hopefully you can wrap your head around why that means the last one cannot be correct.

Answer 7

Okay yes i understnad

Answer 8

doing the procedure here as I try to keep my eyelids propped open:

x^3+x^2-3x+2 has 2 sign changes, so 2 or 0 positive real roots
-x^3+x^2+3x+2 has 1 sign change, so 1 negative real root

our only possible combinations that add to 3 roots are

2 positive, 1 negative, 0 complex
or
0 positive, 1 negative, 2 complex

Answer 9

in fact, the truth of the matter is that we have 1 negative at \( x = -2.51155\) and 2 complex at \( (x = 0.755774 \pm 0.474477i\)\)

Answer 10

little confusing lol i get how to do the positive real roots but not the negative? How do you know which signs to change

Answer 11

What you do for the negative is rewrite f(x) as f(-x), but to do that, all you have to do is go through the equation and change the signs of the terms where x is raised to an odd power (1,3,5...)

Answer 12

That's because \((-x)^{2n} = x^{2n}\) but \((-x)^{2n+1} = -(x^{2n+1})\)

Answer 13

Ohhh so dont change the even powers?

Answer 14

even powers are unchanged when we substitute \(-x\) for \(x\), right?

\[f(x) = x^2\]\[f(-x) = (-x)^2 = (-x)*(-x) = x^2\]

Answer 15

Yeahh it all makes sense now! haha

Answer 16

That has litterally been confusing me for hours i didnt realize how simple it was till you explained it lol

Answer 17

It's one of those things that makes you say "yeah, uh huh, I understand that" as you watch someone do it, but then when you have to do it yourself the first few times, that "I understand" turns into "wait, WTF?" :-)

Answer 18

Hahaha really though!!

Answer 19

So for f(x) = x^4 + 3x^2 + 2
this would have no positive real roots, no negative real roots, correct?

Answer 20

f(x): + + + no sign changes, no positive real roots
f(-x): + + + no sign changes, no negative real roots
must have 4 roots, so we have 2 pairs of complex conjugate roots

Answer 21

Roots for this are \[x = \pm i, \,\,x = \pm i\sqrt{2}\]

Answer 22

Okay kind of following you. so my answer is correct though?

Answer 23

Yes, you were correct, as far as you went.

Answer 24

I just was showing you my shorthand notation:

\[x^4+3x+2\]becomes + + + because that is the sign of each of the coefficients (the actual value doesn't matter, just the sign)
and as we only have even power terms, f(-x) is also + + +

just makes it a bit easier to spot sign changes, at least for me. Your mileage may vary, as they say...

Answer 25

Ohh okay yeah that makes sense. One last quick question if you dont mind then i gotta go to bed!!! :p

Answer 26

Ask quickly :-)

Answer 27

Using Descartes? rule of signs, how many real roots does 5x^6 + 4x^5 + 3x^4 - 2x^3 + 2x^2 - x -1

Answer 28

three or one positive real roots, one or zero negative real roots

Answer 29

?

Answer 30

\[5x^6 + 4x^5 + 3x^4 - 2x^3 + 2x^2 - x -1\]
f(x): + + + - + - -
3 positive real roots, or 3-2=1 positive real root

f(-x): + - + + + + -
3 negative real roots, or 3-2=1 negative real root

total of 6 roots, so we have either

3 positive, 3 negative
or
3 positive, 1 negative, 2 complex
or
1 positive, 3 negative, 2 complex
or
1 positive, 1 negative, 4 complex

Answer 31

turns out to be the final case

Answer 32

Okay i counted wrong, im tired :p lol. So three or one positive real roots, three or one negative real roots

Answer 33

Yes, that's correct.

Answer 34

Okay thank you so much! You've been so much help and helped me understand a lot of stuff that i didnt before :D

Answer 35

Glad I could help! Good night, and hopefully you won't see any equations in your dreams :-)

Answer 36

Hahaha hopefully not :p. Goodnight and thanks again!