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.

- anonymous

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- whpalmer4

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.

- whpalmer4

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

- anonymous

would it hvae 3?

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## More answers

- whpalmer4

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

- anonymous

im completley confused

- whpalmer4

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 \(i^2=-1\)\[a^2-b^2i^2=a^2-b^2(-1) = a^2+b^2\]

- anonymous

so would i have a pair of complex ?

- whpalmer4

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.

- anonymous

ok...

- anonymous

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

- whpalmer4

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 :-)

- anonymous

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

- whpalmer4

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?

- anonymous

3

- whpalmer4

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?

- whpalmer4

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.

- anonymous

oh ok

- whpalmer4

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?

- anonymous

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

- whpalmer4

Close: shouldn't that be -300x?

- anonymous

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

- whpalmer4

(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?

- anonymous

1?

- whpalmer4

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

- whpalmer4

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

- whpalmer4

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 :-)

- whpalmer4

Now comes the tedious part, I'm afraid.

- anonymous

:/ im so bad at math

- whpalmer4

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

- anonymous

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

- whpalmer4

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?

- anonymous

ok i see

- anonymous

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

- anonymous

back

- whpalmer4

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 \(x = \pm i\).

- whpalmer4

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

- anonymous

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

- whpalmer4

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

- whpalmer4

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.

- anonymous

so we hahave to narrown them down to 2?

- whpalmer4

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.

- whpalmer4

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?

- whpalmer4

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.

- whpalmer4

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.

- anonymous

im with you

- whpalmer4

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 :-)

- whpalmer4

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.

- whpalmer4

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?

- anonymous

ive done it a few times...

- whpalmer4

Let's see what you've got :-)
What do you get when you divide \[\frac{x^3-31x^2+300x-900}{x-6}\]

- anonymous

x^2-25x+150?

- whpalmer4

Yes!

- whpalmer4

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?

- anonymous

2 roots?

- whpalmer4

Yes, go on...

- anonymous

2 complexx roots

- whpalmer4

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

- whpalmer4

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.

- anonymous

right

- whpalmer4

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

- anonymous

2x3x5^2

- whpalmer4

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.

- whpalmer4

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

- whpalmer4

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

- anonymous

so should we try 2*5=10?

- whpalmer4

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.

- whpalmer4

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

- anonymous

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

- whpalmer4

which simplifies to...

- anonymous

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

- whpalmer4

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

- anonymous

yea

- whpalmer4

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

- anonymous

but 100+150-250 would equal 0

- whpalmer4

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

- anonymous

that it ewuals 0?

- whpalmer4

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

- whpalmer4

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

- whpalmer4

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.
\[\frac{x^2-25x+150}{x-10}=\]

- anonymous

x-15?

- whpalmer4

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

- anonymous

its a root

- whpalmer4

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

- anonymous

15?

- whpalmer4

Yes!

- anonymous

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

- whpalmer4

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.

##### 1 Attachment

- whpalmer4

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

- anonymous

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

- whpalmer4

Where'd that 2 come from?

- whpalmer4

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

- anonymous

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

- whpalmer4

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

- whpalmer4

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:

##### 1 Attachment

- anonymous

so 10,15,6 are the possible solutions

- whpalmer4

Not possible solutions, they are the solutions.

- anonymous

and the possiblepositive negative and complex solutions were

- anonymous

1 positi e

- anonymous

positive

- whpalmer4

As it turned out, there were no negative roots, nor were there any complex roots. By DRoS on the original polynomial (and its offspring), we never had the possibility of any negative roots.

- anonymous

thanks!

- whpalmer4

You're welcome. I understand this stuff a little bit better each time I explain it to someone else :-)

- anonymous

now time to take my final!

- whpalmer4

At the risk of confusing you, there's a trick you can do when evaluating the polynomials to make the arithmetic a bit less tedious. Say we had our original polynomial \[P(x) = x^3-31x^2+300x-900\]To evaluate that, we're going to take our test value, cube it, subtract 31 * test value squared, add 300 * test value, and finally subtract 900. It's possible to write that polynomial in a different way that gives the same answer but with less arithmetic.
\[x^3-31x^2+300x-900 = x(x^2-31x+300)-900 \]\[= x(x(x-31)+300)-900 = ((x-31)x+300)x-900\]
Note that we can construct that polynomial just by reading across the coefficients of the polynomial. The only catch is that if there is a "missing" term, we need to put in a 0 there, so doing this with \[P(x) = x^3-31x^2-900 = x^3-31x^2+0x-900\]gives us\[((x-31)x+0)x-900\]
This is easier to evaluate on a calculator or by hand because you have fewer multiplications and additions, and the intermediate numbers tend to be smaller.
Let's do P(6) this way with the original polynomial:
\[((6-31)6+300)6-900 \rightarrow ((-25)6+300)6-900\rightarrow(-150+300)6-900\]\[\rightarrow150*6-900=0\]Now doing it the usual way:\[6*6*6-31*6*6+300*6-900\]The clever way has 2 multiplications and 3 addition/subtractions. The usual way has 5 multiplications and 3 addition/subtractions. As you have more terms, the usual way gets even worse. The clever way is known as Horner's Method, after an 18th century British mathematician named (wait for it...drum roll please....) Horner.

- whpalmer4

Good luck on the final!

- anonymous

thanks!

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