Form some third degree equation that has the roots -3, -1 and 2.

1. Jul 14, 2014

TheSodesa

1. The problem statement, all variables and given/known data

What the title says. There's a b part to the problem, but of course I can't move on to it until I understand what is going on here.

2. Relevant equations

A third degree polynomial is of the form f(x) = ax3 + bx2 + cx + d

This information was not given in the question, but I'm assuming that it is necessary for solving this question.

3. The attempt at a solution

I tried substituting the roots into the function ( f(-3), f(-1) and f(2)), but soon realized, that I have 4 unknowns to deal with, not 3, so that wasn't gong to cut it.

f(-3) = -27a + 9b -3c + d
f(-1) = -a + b - c + d
f(2) = 8a + 4b + 2c + d

That annoying constant d is what I can't deal with, with just these 3 equations.

The answer at the back of the book states that any equation of the form f(x) = a(x+3)(x+1)(x-2) would suffice, but apparently I don't understand the theory of polynomials well enough to get how they came to this conclusion. I see the coefficient a , but where are the coefficients b and c, and the constant d? How can they just ignore them, or did they?

What am I missing here?

2. Jul 14, 2014

Mentallic

Try graphing

$$y=x^2-2$$

$$y=2x^2-4 = 2(x^2-2)$$

$$y=-x^2+2 = -(x^2-2)$$

or generally,

$$y=a(x^2-2)$$

What you'll notice is that all of these quadratics have the same roots which are $x=\pm\sqrt{2}$. The a coefficient just changes how steep the quadratic is, and its direction (if a<0 then it grows downwards etc.).

So a cubic polynomial of the form

$$y=(x-p)(x-q)(x-r)$$

will have roots p,q,r but so will

$$y=a(x-p)(x-q)(x-r)$$

for whatever value a is (but avoid making it 0).

3. Jul 14, 2014

Staff: Mentor

As Mentallic explained, there are infinitely many 3rd degree polynomials with these roots. So technically your approach is perfectly valid, you can assume any value for d - make it 1, or 2, or whatever, then calculate all other unknowns.

Not that it makes much sense, when there is much faster route (shown by Mentallic).

Actually expressing polynomials as a(x-p)(x-q)... is often quite convenient and worth of remembering.

4. Jul 14, 2014

TheSodesa

So I should start with the assumption, that the cubic polynomial is of the form f(x) = (x-p)(x-q)(x-r)? Why?

The annoying thing about this book is that it's a revision book, an they don't provide much in therms of theory, but instead simply state formulas and theorems without proving them. I don't know if this is a stupid question, but why did you (and the book) put minus signs inside the parentheses instead of plus signs, e.g. (x+p)(x+q)(x+r)?

(x+q)(x+p)(x+r)
= (x2 + x(q+p) + qp)(x+r)
= (x3 + x2(q+p) + xqp + rx2 + rx(q+p) + rqp)
= (x3 + x2(q+p+r) + x(r(q+p) + qp) + rqp)

(x-p)(x-q)(x-r)
= (x2 -x(q+p) + qp)(x-r)
= (x3 -x2(q+p) + xqp - rx2 + rx(q+p) - rqp)
= (x3 - x2(q+p-r) + x(r(q+p) + qp) + rqp)

You can expand both of these products, and they both are 3rd degree polynomials (possibly with a few mistakes. Lots of signs to worry about :grumpy:). Why is the minus sign preferred over the plus sign?

Last edited: Jul 14, 2014
5. Jul 14, 2014

Staff: Mentor

Expressing the polynomial as (x-p)(x-q)... guarantees it will have roots p and q - as (x-p) equals zero, any product containing (x-p) will equal zero for x=p (same logic for q).

That's why this way of expressing polynomials is so convenient.

6. Jul 14, 2014

TheSodesa

Hmm, this is starting to make sense.

Now if I could just have an answer to a slightly unrelated question, namely: "Is there a standard method for changing the polynomial from the sum form ax3 + bx2 + cx + d into it's product form, assuming you don't know the roots or the coefficients or can it not be done in that case?"

I absolutely understand that the function g(x) = (x-q)(x-p)(x-r) = 0 if any of the 3 constants (they don't look like coefficients when the polynomial is written as a product) is equal to x. My problem with this question seems to be, that I'm used to seeing polynomials in the sum form, and haven't though of the product of, say, two binomials as a polynomial, as silly as that might sound. In my mid, if it isn't written as the sum of n number of terms, it's not a polynomial, or at the very least didn't use to be.

I guess I should have started out by saying, that since I'm trying to solve a question that involves a 3rd degree polynomial, I'm going to write it as the product of 3 terms, that each contain the first power of x and the subtraction of a constant term from that x.

Last edited: Jul 14, 2014
7. Jul 14, 2014

Staff: Mentor

There are formulas for polynomials up to fourth degree. If you want more information look for Abel–Ruffini theorem and Galois theory - but it is probably more than you can swallow.

Form doesn't matter. Besides, every of these products can be converted into the sum form. Going the other way is much more difficult.

8. Jul 14, 2014

TheSodesa

Funnily enough, I've heard of the Abel-Ruffini theorem, just not by name. At the high school level, the Finnish school system doesn't really put much emphasis on memorizing names, at least not in math. That's not to say that I actually understood where the formula (I only ever saw one) for solving these higher level polynomial came from or that I still remember it.

It just bugs me that only the sum form seems intuitive to me. I would not have thought to even try to form a factorized polynomial based on the information that I had. I didn't know that finding the roots of a polynomial is equivalent to polynomial factorization into factors of degree 1.

Last edited: Jul 14, 2014
9. Jul 14, 2014

Mentallic

That's what learning is all about.

10. Jul 14, 2014

TheSodesa

Now if I only could understand it properly.

As an example, there's a formula in my book that states that any second degree polynomial can be factored as follows:

ax2 + bx + c = a(x-x1)(x-x2)

Looks very similar to what you guys have posted, and what the answer at the back of my book is saying, but so far I've been unable to find how this is possible. I feel like once I get the idea behind this, as in the middle steps between the two sides of the equation, I can finally solve the problem. I can see, that it's saying a is a factor in all of the terms, but beyond that...

11. Jul 14, 2014

Mentallic

To be honest, I'm a little unsure about what your confusion actually is, so I'll just explain everything that I feel might need explaining so you can hopefully then point out what you need more clarification on.

$$ax^2+bx+c$$

We can solve for x by using the quadratic formula, which would then give us

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

and what I've noticed is that many students don't realize it's equivalent to

$$x=\frac{-b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}$$

which means that we have two roots,

$$x_1=\frac{-b}{2a}+\frac{\sqrt{b^2-4ac}}{2a}$$

and

$$x_2=\frac{-b}{2a}-\frac{\sqrt{b^2-4ac}}{2a}$$

And you can check for yourself that these work. Just plug x1 for x into the quadratic

$$ax^2+bx+c$$

and confirm that it equals 0, and then do the same for x2.

Now, we also have the factored form for quadratics which is

$$a(x-x_1)(x-x_2)$$

and in this form, it's very easy to confirm that x1 and x2 are indeed roots because if you plug x1 in, you get

$$a(x_1-x_1)(x_1-x_2) = a\times 0 \times (x_1-x_2) = 0$$

and similarly for x2, the expression is 0. For no other x values would the expression be 0, because at least one of the factors must be zero for that to happen, which happens only when x=x1 or x=x2 - and the constant a is assumed to be non-zero else it wouldn't be a quadratic.

If you expand

$$a(x-x_1)(x-x_2)=a\left(x-\frac{-b+\sqrt{b^2-4ac}}{2a}\right)\left(x-\frac{-b-\sqrt{b^2-4ac}}{2a}\right)$$

which would take quite a bit of effort but is definitely doable, then you'll find it simplifies back into $ax^2+bx+c$ as it should. Essentially what I'm trying to say is that both forms are equivalent.

$$ax^2+bx+c=a(x-x_1)(x-x_2)$$

where x1,x2 are the factors of the quadratic.

Finally, everything here can be extended to work for any nth degree polynomial.

12. Jul 14, 2014

TheSodesa

Sorry for being so difficult.

The exact issue I'm having now is the fact that ax2 + bx + c can't simply be rearranged as a(x-x1)(x+x2) for obvious reasons. The coefficient b and constant c just disappear and x1 and x2 just appear out of nowhere.

Same with ax3 + bx2 + cx + d = a(x-x1)(x-x2)(x-x3). The expressions can't simply be rearranged to form the factorized version.

I guess I'm asking if this is just one of those basic truths that has been confirmed through experimentation, and it just can't be shown that there are polynomials that couldn't be factorized like this.

Last edited: Jul 14, 2014
13. Jul 14, 2014

TheSodesa

I can see from Mentallic's latest post that it is easy enough to move from the already factorized form to the sum form, but going the other way around is the issue for me. What I've been meaning to ask is, is this sort of like it is with derivatives and integrals, in that the only reason we have integration formulas at all is because of the fundamental theorem of calculus?

As in, in order to complete this problem, should I have just known, that when factorizing a polynomial, it is going to have as many factors as it's highest term. A 4th degree polynomial would have 4 factors, (x-q)(x-w)(x-e)(x-r), a 3rd degree polynomial is always going to have 3 factors (x-a)(x-b)(x-c), and so on.

And we know all of this because people have spent time multiplying these in order to get them to the sum form?

Just realized this is going way off topic. :/

Last edited: Jul 14, 2014
14. Jul 14, 2014

D H

Staff Emeritus
They don't "just appear out of nowhere". Mentallic's post showed exactly how those $x_1$ and $x_2$ can be computed given a quadratic $ax^2+bx+c$.

Actually, they can. Finding the roots of a cubic equation is a bit more complicated than is finding the roots of a quadratic equation, but it is still doable. Similarly, there are formulae for finding the roots of a quartic equation. What about higher order polynomials? That's your next question.

It turns out that there are no general purpose methods for factoring fifth order polynomials or higher. This discovery marked a very important turning point in mathematics.

That a general mechanism for solving fifth degree and higher polynomials in a closed form doesn't exist does not however mean that such equations are unsolvable. It just means that the roots cannot be expressed in a nice, simple closed form. They most certainly are "solvable". The basis for this claim is the Fundamental Theorem of Algebra. One consequence of this theorem is that an nth degree polynomial with real or complex coefficients has n roots in the complex numbers, counting multiplicity. What "multiplicity" means: Consider $x^2-2x+1$. This can be factored as $(x-1)(x-1)$. The root $x=1$ is a double root. Note that you might have to consider complex numbers. For example, $x^2+1$ has no roots in the reals. It has two roots in the complex numbers, i and -i.

15. Jul 14, 2014

TheSodesa

Alright.

Thanks everyone for your assistance. I think I've gathered what's important out of this thread.

I'm going to attempt a solution later today, based on what I've learned here.

16. Jul 14, 2014

Ray Vickson

You are looking at it all wrong: the coefficients $a,b,c$ do not "disappear" in the factored form $a(x-x_1)(x-x_2)$. The roots $x_1$ and $x_2$ are given by formulas involving $a,b,c$, as others have already pointed out to you.

There are theorems that ALL polynomials of degree $n$ have $n$ roots (but some roots are possibly repeated---so there may be fewer than $n$ different ones). However, some polynomials have roots that are "complex numbers", so cannot be factored using only ordinary, real numbers. For example, the polynomial $p(x) = x^2 + 1$ has no real factors, but can be factored over the complex numbers as $p(x) = (x+i)(x-i)$, where $i = \sqrt{-1}$ is the unit imaginary number.

None of these things require "experimentation"; these basic facts have been known for about 200 years. However, in practical terms, if you really want to factorize a polynomial of high degree (say, $n \geq 5$) you may need to resort to numerical methods, because it has been proven that for general polynomials of degree 5 or more there is no general, finite algebraic formula for the roots in terms of the coefficients. In other words, it would not matter how many thousands of pages you used to try to write down a formula for the roots, and it would not matter how intelligent you were: it just cannot be done at all. Believe it or not, those are facts that can be proven! Nevertheless, the polynomial DOES HAVE $n$ roots (possibly complex); it is just that we cannot write down a formula to find them---even though we know they exist. (There are finite formulas for the roots in the special cases of degrees $n = 1,2,3,4$ but not for larger $n$.)

Last edited: Jul 14, 2014