Question

State how many imaginary and real zeros the function has.
x^3 + 5x^2 + x + 5

Answer 1

look at the signs changes, to use Descartes Law of Signs

x^3 + 5x^2 + x + 5
^ ^ ^ ^
+ + + +

Answer 2

that'd give you the "real positive" zeros
now for for the "real negative" zeros
let's get f(-x)

Answer 3

f(-x)

-x^3 + 5x^2 - x + 5
^ ^ ^ ^
- + - +

Answer 4

what do i have to do though to get these zeros?

Answer 5

x^3 + 5x^2 + x + 5
^ ^ ^ ^
+ + + +
no no no

so, no real positive zeros

Answer 6

We can use Descartes' Rule of Signs to find this out.

Take the polynomial written in standard form (decreasing powers of the variable) and count the number of sign changes encountered when scanning left to right:

++++ are the signs of the coefficients of f(x), so there are 0 sign changes. That means we have 0 positive, real roots.

Now rewrite the polynomial as f(-x). In effect, just change the signs of the odd-powered terms:

-x^3+5x^2-x+5

Count again in the same fashion
-+-+
3 sign changes, so we have 3 negative real roots, possibly reduced by a multiple of 2.

Complex roots always come in conjugate pairs in polynomials with only real coefficients (no \(i\)'s lurking in the coefficients). We also know that we always have exactly as many roots as the highest exponent of the variable. The difference between the count of real roots from the first two steps and the total is made up by complex roots, again, always in conjugate pairs.

So, we have 0 positive real roots, and either 3 negative, real roots or 1 negative real root + 2 complex roots.

Answer 7

descartes law of signs, is to see how many times the "sign" changes from term to term

Answer 8

Actually finding the values of the roots or zeros is an entirely different matter!

Answer 9

so there are 3 real zeros and 0 imaginary zeros right?

Answer 10

You can use the Rational Root Theorem to make some intelligent guesses as to possible roots. In the case of this polynomial, 1, -1, 5 and -5 would be trial roots to evaluate. If the polynomial evaluates to 0 at x = 1, -1, 5, or -5, then that value of x is a root. (aka zero).

Answer 11

f(-x)

-x^3 + 5x^2 - x + 5
^ ^ ^ ^
- + - +
yes yes yes

it changed 3 times
from negative to positive
then from positive to negative
then again from negative to positive

3 times means, it has either 3 OR 3-2 real negative zeros, so, 3 or 1

Answer 12

No, actually this polynomial has 1 negative real root and 2 complex roots.

Answer 13

so 1 real zero and 2 imaginary zeros?

Answer 14

right
it can have EITHER 3 negative roots, in which case that'd leave room for no complex ones
OR
if it has 1 negative one, then you'd have room for 2 complex ones

Answer 15

so which is the answer the 1 real zero and 2 imaginary zeros ; or 3 real zeros and 0 imaginary zeros

Answer 16

that IS the answer :), both, either this or that

Answer 17

yes, is kinda ambiguous, but descartes law of signs IS ambiguous, but it gives you a good ballpark figure on what you'd get :)

Answer 18

look at it this way, is better than 0

Answer 19

If you try out those possible roots I suggested (courtesy of the RRT):

\[(1)^3+5(1)^2+(1)+5 = 12\]
\[(-1)^3+5(-1)^2+(-1)+5=8\]
\[(5)^3+5(5)^2+(5)+5=260\]
\[(-5)^3+5(-5)^2+(-5)+5=0\]

So x = -5 is a root. Thanks to the factor theorem, we can divide the polynomial by (x-(-5)) and apply the whole procedure again to get the remaining roots.

\[\frac{x^3+5x^2+x+5}{x+5} = x^2+1\]
Now you can find the roots of that, I'm sure, and you'll see that they are imaginary.

Answer 20

well it is multiple choice so which one do i pick?

Answer 21

what are your choices?

Answer 22

Descartes gives you the possible combinations. You do have other tools which will clarify the situation, such as what I just did.

Answer 23

0 imaginary; 3 real

1 imaginary; 2 real

3 imaginary; 0 real

2 imaginary; 1 real

Answer 24

the first and the last choice are good, so I guess if you had to pick one, could be either
because the function will yield either based on the descartes law of signs

Answer 25

the middle two choices are impossible — with a real coefficient polynomial, the imaginary or complex roots MUST come in pairs.

Answer 26

i picked the last one and it was right thank you :)

Answer 27

yw

Answer 28

the last one is the only one that describes the actual roots, so yes, it should be right :-)

\[x = -5, x = \pm i\]
are the roots \((i = \sqrt{-1})\)

Answer 29

btw, i think we should delete our conversation; it's inappropriate when we occupy other's post to chat. Moreover, ours should be private instead of being here, right?