Question

find the solution to the equation using fundamental theorem of algebra, rational root, descartes rule, and factor therom. equation is x^3-31x^2+300x-900

substitute 0 for the function and graph.

Answer 1

Fundamental theorem of algebra says that if if we have a polynomial \(P(x)\) of degree \(n\) we will have exactly \(n\) roots (though some may have the same value). This coupled with the definition of a root implies that we can write our polynomial in factored form
\[P(x) = (x-r_1)(x-r_2)...(x-r_n)\].

Rational root theorem says that if we have our polynomial written in expanded form, with the highest order term first and the constant term last, our roots are a subset of the factors of the constant term divided by the factors of the highest order term. Our life is made a bit easier here, as the highest order term has a coefficient of 1, so we only need to consider the factors of the constant term (-900).

Descartes' Rule of Signs tells us how many complex and real roots we can expect to find.
Write out the polynomial in the usual fashion with highest order terms first and count the number of sign changes. This number, possibly reduced by a multiple of 2, will give the number of positive roots.

Now replace the polynomial P(x) with P(-x) and repeat the process of counting sign changes. This will give us the number of negative roots, which may also be reduced by a multiple of 2.

Finally, the difference between the total number of roots (given by the order of the highest order term) and the sum of the positive and negative roots will be the number of possible complex roots.

Answer 2

With that in mind, can you tell me the number of roots this equation will have?

Answer 3

would it hvae 3?

Answer 4

Yes, there are a total of 3 roots. How many of them could be complex?

Answer 5

im completley confused

Answer 6

Well, one thing to know is that complex roots always come in conjugate pairs if the coefficients of your polynomial are restricted to real numbers (as they are in this case — we don't have any i terms floating about). That means that if we have 3 roots, we can either have 3 real roots, or we can have 1 real root and a pair of complex roots.

The complex roots come in conjugate pairs: \((a+bi),(a-bi)\) and when multiplied together the \(i\) terms disappear:
\[(a-bi)(a+bi) = a^2+abi - abi -b^2i^2 = a^2-b^2i^2\]but \(i^2=-1\)\[a^2-b^2i^2=a^2-b^2(-1) = a^2+b^2\]

Answer 7

so would i have a pair of complex ?

Answer 8

Well, if you have any complex roots, they will come in pairs. We don't know yet whether we do have any. So, we can either have 3 real roots, or we can have 1 real root + 1 conjugate pair of complex roots.

Answer 9

ok...

Answer 10

how would i solve the equation using the fundamental therom of algebra?

Answer 11

Well, you can't solve it using just the FToA. That's just one of the tools we'll use to find all the roots. It tells us there must be 3 of them in this case. Now its work is done, and it can kick back while the other theorems do the heavy lifting — think of it as management :-)

Answer 12

oh so we have to use all them together to do the problem?

Answer 13

Let's use Descartes' Rule of Signs now to see if we can classify the roots a bit more.

Our polynomial is \[P(x) = x^3-31x^2+300x-900\] (written with the exponents in descending order)
Scanning from left to right, how many times does the sign change?

Answer 14

3

Answer 15

Okay, so that means we can have a maximum of 3 positive roots. We can also have 1 positive root + 2 complex roots. Those are the only combinations possible. 3 sign changes in the original polynomial, less a multiple of 2 complex roots. The only multiples of 2 that fit are 0*2 and 1*2, which gives us 3 real, positive roots / 0 complex roots, or 1 real, positive root / 2 complex roots. Agreed?

Answer 16

If we'd had 5 sign changes, then the possible combinations would be 5 real, positive roots, 3 real, positive roots + 2 complex roots, or 1 real, positive root + 4 complex roots.

Answer 17

oh ok

Answer 18

Now, let's rewrite \(P(x)\) as \(P(-x)\) by substituting \(-x\) wherever we find \(x\):
\[P(-x) = (-x)^3-31(-x)^2+300(-x)-900 = \]
What do you get when you simplify that?

Answer 19

p(-x)=-x^3-31x+300x-900??

Answer 20

Close: shouldn't that be -300x?

Answer 21

oh i see ..the -x would change it from positive to negative?

Answer 22

(careful attention to detail here is really important as mistakes can leave you hopelessly confused later!)


\[P(-x) = (-x)^3-31(-x)^2+300(-x)-900 =\]\[ (-1)^3x^3-31(-1)^2x^2+300(-1)^1x^1-900=\]\[-1x^3-31(1)x^2+300(-1)x-900 =\]\[-x^3-31x^2-300x-900\]

Now, how many sign changes do we encounter scanning that from left to right?

Answer 23

1?

Answer 24

No, it starts out as - and stays that way, so there are 0 sign changes.

Answer 25

\(x^3-31x^2-300x-900\) would have 1 sign change

Answer 26

0 sign changes in P(-x) means that we have 0 possible negative, real roots. Had the number been larger, we would have to allow for the possibility of pairs of complex roots, just like we did for the earlier scan.

So, bottom line: we either have 3 positive, real roots, or 1 positive, real root and 2 complex roots (in a conjugate pair). Progress :-)

Answer 27

Now comes the tedious part, I'm afraid.

Answer 28

:/ im so bad at math

Answer 29

Nah, I prefer to think of it differently: you just haven't encountered the right explanation yet. Maybe today will be your lucky day :-)

Answer 30

well so far, youve helped more then my online teacher all year lol

Answer 31

Let's multiply some polynomials and see if we can identify some patterns.
\[(x-a)(x-b) = x^2-ax-bx + ab = x^2-(a+b)x + ab\]\[(x+a)(x+b) = x^2 + ax + bx + ab = x^2+(a+b)x + ab\]\[(x-a)(x+b)(x-c) = x^3 +(b-a-c)x^2+(ac-ab-bc)x + abc\]

Notice that the highest order \(x\) term always has a coefficient of 1? And the constant term is always the product of the constant terms in the binomials we multiplied?

Answer 32

ok i see

Answer 33

brb need to bring the horses from the field to the barn and ill be back! less then 10 minutes

Answer 34

back

Answer 35

Well, that's what the Rational Root Theorem is all about. If we know the constant term in our polynomial is 1, for example, and the order of our polynomial is 2, that means our roots must be \(\pm1\) (I'm assuming here that the highest order term has a coefficient of 1 to keep things a bit simpler). Let's consider some possibilities:
\[x^2-1 = (x-1)(x+1)\]\[x^2-2x+1 = (x-1)(x-1)\]\[x^2+2x+1 = (x+1)(x+1)\]

With the first one, we have 1 sign change in \(P(x)\) and 1 sign change in \(P(-x) = x^2-1\) so we have 1 positive real root and 1 negative real root and no complex roots, which is easily seen looking at the factors.

With the second one, we have 2 sign changes in \(P(x)\) and 0 sign changes in \(P(-x) = x^2+2x+1\) so we either have 2 real, positive roots, or 0 real, positive roots and 2 complex roots. As you can see, the former is the correct choice.

With the third one, we have no sign changes in \(P(x)\) and 2 sign changes in \(P(-x) = x^2-2x+1\) so there are 0 real, positive roots and 2 real, negative roots.

\[x^2+1=(x+i)(x-i) \]
With the fourth one, we have 0 sign changes in \(P(x)\) and 0 sign changes in \(P(-x) = x^2+1\) but we know by the Fundamental Theorem that we must have 2 roots, so if they aren't real roots, they must be complex, and indeed we have a pair of complex roots \(x = \pm i\).

Answer 36

That's a bit of a mouthful to chew on, I agree :-)

Answer 37

so i have a pair of complex roots? is that all we had to do for this question?

Answer 38

No, this is just warming up for the fireworks :-)

Answer 39

Let's say our polynomial is \(P(x) = x^2 + 4x + 4\)

FToA says 2 roots
DRoS says 2 negative real roots
RRT says that the roots we should test are \(\pm 1, \pm2, \pm4\).

RRT says what it says because as we previously observed, the constant term is produced by the multiplication of the various constant terms, and those constant terms are just the roots of the equation (an assist by the Factor Theorem there). Unfortunately, we often have more potential roots candidates than actual roots, because the constant term may not have only 1 possible factoring. Our 4, for example, can be made by 1*4, 2*2, 4*1, -1*-4, -2*-2, -4*-1. Those can't all be roots because we only get 2 roots and that's 6 possibilities.

Answer 40

so we hahave to narrown them down to 2?

Answer 41

In this case, using the DRoS and RRT together, we can rule out the positive root candidates without testing, because DRoS said 0 positive roots. But that still leaves us with two possible factorings:
\[(x+2)(x+2)\]\[(x+1)(x+4)\]
Now, I'm confident that you can look at those and see even without multiplying that the first one is the proper choice.

Answer 42

But let's say the numbers were not so friendly, DRoS didn't narrow down the list so tightly, and we had a polynomial with terms up to \(x^{19}\) instead of the trivial one I provided. How would we proceed?

Answer 43

If we plug a root into P(x), P(x) = 0 by definition. So, we take our list of candidates, sharpen our pencil, and pick one of the candidates to plug into P(x) to see if the result is 0. If it is, we got lucky, and we've found a root! If it isn't, then we mutter some words that maybe shouldn't be used if small children are present, and pick another candidate.

Once we do have a root \(r_n\) in hand, we can simplify the polynomial by dividing it by \((x-r_n)\) and starting the process over again. This is where the Factor Theorem helps us out.

Answer 44

As we divide off each root, the size of that constant should shrink, and as it shrinks the number of possible roots/factors to try does also.

Answer 45

im with you

Answer 46

Eventually, if we didn't make any mistakes, we'll be left with \((x-r_0)\) as our polynomial and we can go crack open a cold beverage as a reward :-)

Answer 47

So here's the rub with the problem you have to solve: 900 has quite a few factors! 900 = 2*2*3*3*5*5, which means any of 2,3,5,6,10,15,30,45,60, 75, 90, etc. are potential roots (and the negative values as well, unless DRoS has ruled them out). My advice is to try bigger numbers first, in the hopes of narrowing that search space down quickly.

Answer 48

You don't necessarily want to go really big, however; I would try the products of two of the prime factors first, so 2*3 = 6, 2*5 = 10, 3*5 =15, along with the individual prime factors 2, 3, and 5. Let's do a few trials:

\[P(2) = (2)^3-31(2)^2+300(2)-900 = 8-31*4+600-900 \ne 0\](you can often get a sense that a given candidate isn't going to be a root without doing the arithmetic exactly)
\[P(3) = (3)^3-31(3)^2+300(3)-900 = 27-31*9 + 900 - 900 \ne 0\]
\[P(5) = (5)^3-31(5)^2+300(5)-900 = 125-31(25)+1500-900 \ne 0\]
So we've eliminated 2, 3, and 5 as possible roots. We don't have to try them in the future, either.

How 2*3 = 6?
\[P(6) = (6)^3-31(6)^2+300(6)-900 = 216-31*36 + 1800-900 \]\[= 216-1116+1800-900\]All of our trailing digits look like they might combine to 0, so it's worth figuring this one out exactly, and what do you know, it turns out to be 0! We've found our first root!

Now we can divide \[\frac{P(x)}{(x-6)}\] to get a simplified polynomial to use for the rest of the process. Polynomial long division or synthetic division would be the tool of choice here. Are you familiar with them?

Answer 49

ive done it a few times...

Answer 50

Let's see what you've got :-)

What do you get when you divide \[\frac{x^3-31x^2+300x-900}{x-6}\]

Answer 51

x^2-25x+150?

Answer 52

Yes!

Answer 53

Okay, let's run the process again. \[P(x) = x^2-25x+150\]
By FToA, how many total roots?
By DRoS, how many positive, negative, complex roots?

Answer 54

2 roots?

Answer 55

Yes, go on...

Answer 56

2 complexx roots

Answer 57

Well, that's one possibility. What about 2 positive roots?

Answer 58

We have two sign changes in P(x), so that means 2 positive, real roots, or 0 positive, real roots and 2 complex roots, right? No sign changes in P(-x), so no negative roots.

Answer 59

right

Answer 60

Our constant term is 150, which has prime factors of what?

Answer 61

2x3x5^2

Answer 62

Exactly. And we already tried 2, 3, and 5 before, so we don't need to try them again, as they weren't roots. A number that *is* a root should be tried again, because we may have roots with a multiplicity > 1. We can try 6 again, or we could try 2*5=10 or 3*5=15 or 5*5=25, your call.

Answer 63

(no need to try any negative numbers because of DRoS, but often we don't get that lucky break)

Answer 64

i guess you could also try 2*3*5 = 30 or 3*5*5 = 75 or 2*3*5*5= 150...

Answer 65

so should we try 2*5=10?

Answer 66

Sometimes you'll have an approximate idea of the places where the curve crosses the x-axis, which can help in further limiting the number of trials. For example, if we knew that all of the crossings happened between -20 and 50, there wouldn't be any point in trying 150 as a candidate.

Answer 67

Sure, go for it! Plug x=10 into our new, sleeker polynomial.

Answer 68

p(x)=(10)^2-25(10)+150

Answer 69

which simplifies to...

Answer 70

p(x)=100-250+150
p(x)=300

Answer 71

100-250+150 = 300? Is that your final answer? :-)

Answer 72

yea

Answer 73

Isn't that exactly the same as 100 + 150 - 250?

Answer 74

but 100+150-250 would equal 0

Answer 75

Yes, that's right! And if P(x) = 0, what does that mean about x?

Answer 76

that it ewuals 0?

Answer 77

No. It means that x is a root...

Answer 78

If \(P(x) = x^2-25x+150\) and \(P(10) = 10^2-25(10)+150 = 0\), then \(x=10\) is a root of \(P(x)\).

Answer 79

And that means we can divide (factor) out \((x-10)\) from \(P(x)\) to get an even simpler polynomial to solve for our final root.
\[\frac{x^2-25x+150}{x-10}=\]

Answer 80

x-15?

Answer 81

Right. And what do the FToA, RRT, and DRoS say about that?

Answer 82

its a root

Answer 83

Yes, there's 1 sign change, so 1 positive root, 0 negative roots, 0 complex roots. And the value of that root is?

Answer 84

15?

Answer 85

Yes!

Answer 86

what does the fundamental therom of algebra tell us about the equatio?

Answer 87

Here's a graph of the three different polynomials. Note how all three cross the x-axis at x=15, 2 cross the x-axis at x=10, and 1 crosses at x=6.

Answer 88

Fundamental Theorem of Algebra says that \(P(x) = x-15\) has 1 root, because the highest power of \(x\) is 1.

Answer 89

so the possible rational solutions are 15,2,10?

Answer 90

Where'd that 2 come from?

Answer 91

No, the roots are just the x values where the polynomial crosses the x-axis (y=0).

Answer 92

its only suppose to be 15 and 10..pressed a extra button

Answer 93

We have 3 roots, though — don't forget the first one we found!

Answer 94

having the y-axis drawn in the diagram where it was is perhaps a bit confusing, here's an unlabeled version that doesn't have that problem:

Answer 95

so 10,15,6 are the possible solutions

Answer 96

Not possible solutions, they are the solutions.

Answer 97

and the possiblepositive negative and complex solutions were