Question

how do i find the roots of this equation: f(x) = 2x5 - 9x4 + 12x3 - 12x2 + 10x - 3 = 0

Answer 1

I'd recommend a combination of the following:
Descartes' rule of signs
Rational root theorem
Plotting points and graphing
Synthetic division

Answer 2

thanks :)

Answer 3

Want to give it a shot and see if we can find some of them?

Answer 4

im not sure how

Answer 5

Do you know Descartes' rule of signs? That'll give you a clue for what kind of roots to expect.

Answer 6

no

Answer 7

Ok, first, we have the polynomial in standard form (always want to have that first), so count the sign changes between terms. e.g. 2x^5 is positive, -9x^4 is negative, so there is a sign change between the first two terms.

Answer 8

yea i know that

Answer 9

Ok, and you can also substitute (-x) in for x and count the number of sign changes in f(-x).
So how many sign changes are in f(x) and how many are in f(-x)?

Answer 10

4

Answer 11

There are actually 5 for f(x). How many for f(-x)?

Answer 12

A shortcut for seeing f(-x) is that the terms with odd exponents will change signs, but the ones with even exponents won't.

Answer 13

3

Answer 14

If you plug in (-x) for x, f(-x)= -2x^5 -9x^4 -12x^3 -10x -3. All the terms are negative so there are zero sign changes in f(-x).
So what this tells us is that there are no negative real roots, and there may be 1, 3, or 5 positive real roots. There are going to be a total of 5 roots since it's a 5th degree polynomial, but there could be pairs of imaginary/complex roots as well.

Answer 15

The reason this helps is that when we go hunting for roots, we won't bother checking the negative side of the x-axis because there are not going to be any roots there. We will only test positive numbers.
Next step is Rational Root Theorem (RRT); know how to do that?

Answer 16

no

Answer 17

www.myalgebra.com

Answer 18

Take factors of the constant at the end and divide them by the factors of the leading coefficient at the beginning.
i.e. factors of -3 divided by factors of 2. Those are both prime so there won't be many.

Answer 19

RRT says that if there are any rational roots, then the only ones possible are {-3, -2, -3/2, -1, -1/2, 1/2, 1, 3/2, 2, 3}, but DRS told us that there can't be any negative real roots, so we can throw out half of that list. :-)

Answer 20

Oops, sorry, -2 and 2 shouldn't be on that list.

Answer 21

The list of possible real rational roots is {1/2, 1, 3/2, 3}, so when we use synthetic division to test possible roots, those are the ones we will test.

Answer 22

(This all goes a lot quicker once you get used to it, I promise!)

Answer 23

ok......

Answer 24

Ok, time to rock the synthetic division. Are you familiar with that process?

Answer 25

YES!!!!

Answer 26

Sweet! It is a very useful tool. What you want to do is synthetically divide the polynomial by each of those possible rational roots in turn, until you find one. You'll know it's a root if the 'remainder' is zero.

Answer 27

Tip: for regular-looking polynomials like this one, roots usually hang out near zero, so test the small numbers first.

Answer 28

thank you so much

Answer 29

Your welcome; my pleasure. I love this stuff. :">
Let me know when you find a root, and I'll tell you the next step (unless you know it already).

Answer 30

your a life saver :)

Answer 31

I'd hate to think that your life was actually in danger from this problem. :-p
Is this for an algebra-2 course? pre-calc?

Answer 32

pre-calc

Answer 33

Cool. This is covered a bit in algebra-2 as well, but now you'll learn even more powerful techniques for making equations bend to your will.
The first value I tested was x=1 and it worked. Did you get that as well?

Answer 34

yes

Answer 35

Ok, now for some super-secret tech. Once you find a root, you no longer need to test values with the original polynomial.

Answer 36

really

Answer 37

After dividing by (x-1), the quotient function is q(x)=2x^4 -x^3 +5x^2 -7x +3 (a reduced polynomial), so you can synthetically divide q(x) by the next value you want to test and it'll work the same way.
There is a drawback, though . . .

Answer 38

wht?

Answer 39

First, the good news: synthetic division is also called synthetic evaluation because of the factor-root theorem and the remainder theorem, the 'remainders' that you get with synthetic division are the y-values that go with the x-values that you are testing, so if the number you test doesn't give a remainder of zero, it does give you the coordinates of a point on the graph of the function.

Answer 40

As you go along testing possible roots, you can use those same numbers and plot them on a graph, and the shape of the graph will point in the direction of the next possible root.

Answer 41

wow

Answer 42

Told you it was useful stuff!
Now the bad news: that synthetic evaluation trick only works when you synthetically divide the original polynomial, the reduced polynomials that you get as quotients are equivalent systems (meaning they have the same roots), but they are not the same function you started with and will give you different y-values, so if you want to plot points, you have to synthetically divide the original polynomial.

Answer 43

this is too much

Answer 44

But, if you're not too interested in what the graph looks like and you don't feel you need it to help you find roots, then ignore that and continue to use the reduced polynomials to find roots (because it's easier)

Answer 45

Yeah, it's a lot, but just take it one step at a time and don't worry about all the extra details.

Answer 46

Thank you again for this!!!

Answer 47

I've found 3 roots so far. How many did you get?

Answer 48

3

Answer 49

Ok, last bit of good news: Since it's a 5th degree polynomial, there are going to be 5 roots. Since there are only 2 roots left to find - stop doing synthetic division!

Answer 50

After finding the following roots {1, 3, 1/2}, the reduced polynomial at the bottom of my synthetic division chart is 2x^2+2. Is that what you have too?

Answer 51

yes

Answer 52

Cool.
Now you'll probably remember from more basic algebra, that you can't factor this quadratic, and if you set it equal to zero and try to solve it, you'll end up having to take the square-root of a negative number. These last two roots are imaginary numbers and you can't use synthetic division to find them.

Once you've reduced the polynomial to a quadratic, find the last two with the quadratic formula. It'll find the roots no matter what.

Answer 53

I'm sure you'll see easily that for 2x^2+2=0, the solutions are x=-i and x=i since you end up taking the square-root of -1. This makes the complete set of zeroes for f(x):
{1/2, 1, 3, -i, i}
You now have complete ownage and domination over that polynomial. Feel free to tease it and call it nasty names, for you have defeated it soundly and it can no longer do you harm.

Answer 54

:)

Answer 55

Think you got the hang of it now?

Answer 56

yep your awesome