Question

For the polynomial f(x) = x^5 - 2x^4 + 8x^2 - 13x + 6 , answer the following questions

a) How many zeros does the function have over the set of complex numbers?
b) List the possible rational zeros of this function.
c) Factor this polynomial completely over the set of complex numbers

Answer 1

(b), (c), (a) may be a better order to answer this question?

Answer 2

well, (a) is trivial, and you do need to know it to successfully do (b) and (c)

Answer 3

Descartes rule of signs first?

Answer 4

we know there are 5 zeros because the highest term is x^5. then we could do DRoS, or just jump right in with the rational root theorem and start sussing them out...

Answer 5

In a) aren't they asking how many complex roots it has?

Answer 6

yeah, I suppose that could be the desired interpretation, in which case DRoS would be the right first step.

Answer 7

but the set of complex numbers contains the set of real numbers...

Answer 8

Well, then for a) the function has 5 roots (some real, some complex).

Answer 9

we know that there have to be some real zeros, since we have a polynomial with only real coefficients, which implies that any complex zeros must be present in conjugate pairs

Answer 10

and we've got an odd number of zeros

Answer 11

yep, there has to be at least one real root.

Answer 12

as it turns out, I've already helped someone else with this very problem. see it at http://openstudy.com/study#/updates/52d71f53e4b0e63d49ac287c

too bad OpenStudy probably won't format it nicely :-(

Answer 13

f(+1) = 1 -2 + 8 - 13 + 6
four sign changes: possible positive roots: 4, 2, 0
f(-1) = -1 -2 +8 +13 + 6
one sign change: definitely one negative root.

Answer 14

that's a different way of doing it than I learned, but clearly equivalent. I just look at the signs of the coefficients, and for the f(-x) case invert the signs of the odd power terms. either way you have to make sure the terms are in descending order. can you see any advantage to your method that I'm missing? maybe a little easier to keep track of where you are in a lengthy polynomial...

Answer 15

they are equivalent. But putting x = +1 or x = -1 and writing the function (without actually evaluating the function) is fairly easy with the additional advantage you may stumble upon a factor by adding up the numbers.
For f(1) had I added up the numbers: f(+1) = 1 -2 + 8 - 13 + 6 = 0
it would have been zero and therefore (x-1) is a factor.

Answer 16

that's a good point. of course, in this case you'd have to remember to repeat the process if you wanted to discover than x=1 is a zero of multiplicity > 1 :-)

Answer 17

So I use Decartes Rule of signs for b, then the rational root theroem for c?

Answer 18

@whpalmer4 @ranga ?

Answer 19

For a) the polynomial has 5 roots because it is a polynomial of degree 5. Some roots may be complex and some may be real.

For b) Use rational roots theorem to list all possible factors

For c) try the list of factors from b). Every time you find a factor, do synthetic division to find the quotient. Try the factors again (as whpalamer4 says there is at least one root with multiplicity greater than 1).

Eventually you will end up with a quadratic quotient. Apply the quadratic formula to find the complex roots.

Answer 20

For b) Use rational roots theorem to list all possible roots.

Answer 21

okay thanks!

Answer 22

You are welcome.

Answer 23

@ranga could you help me get c? I thought I knew how to do it but I don't.

Answer 24

What is your answer to b?

Answer 25

p = factors of 6 = {±1, ±2, ±3, ±6}
q = factors of 1 = {±1}

So all possible roots are:
{±1, ±2, ±3, ±6}

Answer 26

Yes, Let us try each root.
x = +1
f(x) = x^5 - 2x^4 + 8x^2 - 13x + 6
f(1) = 1 - 2 + 8 - 13 + 6 = 0
Therefore, x = 1 is a root and hence (x-1) is a factor.
Divide f(x) by (x-1) using synthetic division.

Answer 27

x^3 term is missing. Use 0 as its coefficient:

1 | 1 -2 0 8 -13 + 6
| 1 -1 -1 7 - 6
|_______________________________
1 -1 -1 7 -6

Quotient is: x^4 - x^3 - x^2 + 7x - 6

Q(x) = x^4 - x^3 -x^2 + 7x - 6

Try x = 1 again with Q(x)

Answer 28

Q(1) = (1)^4 - (1)^3 - (1)^2 + 7(1) - 6
Q(1) = 0

Answer 29

So there is another (x-1) factor.
Divide x^4 - x^3 -x^2 + 7x - 6 by (x-1) using synthetic division.

Answer 30

-1 | 1 -1 -1 7 -6
| -1 2 -1 6
| 1 -2 1 6 0

Answer 31

You have to divide by the root. So dividing by (x-1) means dividing by +1

Answer 32

Also in the above division, the remainder is not zero because it should be -6 on the second line, last item. Sp the remainder will be -12.

Answer 33

oh so,
1 | 1 -1 -1 7 -6
| 1 0 -1 6
| 1 0 -1 6 0

Answer 34

Yes, the new quotient is: x^3 - x + 6
You can go through the other roots from part b)

But I know x = -2 will work

R(x) = x^3 - x + 6
R(-2) = (-2)^3 - (-2) + 6 = -8 + 2 + 6 = 0
So x = -2 is a root and (x+2) is a factor

Divide x^3 - x + 6 by (x+2). Use synthetic division and divide by +2. Remember to use coefficient 0 for the missing x^2 term.

Answer 35

divide by -2 in synthetic division and not +2

Answer 36

-2 | 1 0 -1 6
| -2 4 -6
| 1 -2 3 0

Answer 37

Yes, the quotient is: x^2 - 2x + 3
This is a quadratic equation you can solve by using the quadratic formula.
You will get two complex roots.
Combine that with the other 3 real roots: 1 (twice) and -2 and you will have a total of 5 roots as is expected for a fifth degree polynomial.

Answer 38

is everything we just went over the answer? or just the roots?