There are several ways that we can deduce the actual form of a polynomial. If we are given enough values at specific points, we can use Lagrange interpolation or method of differences to figure out the polynomial. However, if we were instead given various equations which hold, then the more natural way is to apply the remainder-factor theorem.
Introductory application of remainder-factor theorem:
Consider a generalized -degree polynomial:
In order to determine all the coefficients, , we need to know at least points of . And it is most common that we determine these coefficients by solving a system of linear equations simultaneously. However, as becomes larger, it becomes more and more impractical to solve them by hand. So is there a better alternative? Yes, there is, and it involves direct application of the remainder-factor theorem. To illustrate this point, let us consider a polynomial in the following:
Consider a quadratic polynomial that satisfies
and we are interested to find the unknowns in the quadratic polynomial, . Recall that by the remainder theorem, we have remainder of upon division by to be , respectively. This tells us we need to solve the following equations simultaneously:
However, solving them simultaneously can get cumbersome and/or tedious. So is there an alternative approach? Yes, there is! We could, in fact, avoid solving these equations simultaneously, and the way to do so is to apply the second part of the remainder-factor theorem: factor theorem.
Let be a polynomial such that for some constant . Then is a factor of . Conversely, if is a factor of , then .
But how is it related to our polynomial? Well, If we consider a dummy polynomial such that it has roots of 1, 2, and 3. Then, by the factor theorem, we have for some constant . Similarly, we know that for . Thus, like , the equation has roots of 1, 2, and 3 as well. Hence, we can find the relationship between and to get . With that we know,
But since we know is a quadratic polynomial, the coefficient of in must be 0, so is forced to be . Then we have . So, we have successfully found the unknown values of and from the polynomial of , and we did not have to solve any equations below simultaneously, which would have otherwise wasted plenty of our time.
Let's try out another similar example that applies this very concept:
Let denote a -degree polynomial such that . Is there sufficient information to find the numerical value Why or why not?
It might be tempting to say that because for , it naturally follows that , but this is not necessarily true because we cannot extrapolate these data to assume that the trend continues as well. In fact, we can show that by recalling the fundamental theorem of algebra, but that is a discussion for another day. Now, back to the question: how do we find To find , we need to find the exact formula for first. Since it is given that is a -degree polynomial, we have for unknown constants and . And we need to find all these unknowns first! Quite a handful of steps, isn't it? Fear not! Let's apply what we learned earlier.
Consider a dummy polynomial such that it has roots , then
Similarly, knowing that has roots as well, . Hence, we can find the relationship between and to get , or equivalently,
So . But wait! Can't we push further? In fact, we can't push any further; this is because we only know 4 points for a degree-4 polynomial, when in fact we need to know at least points for an -degree polynomial.
We are given additional information regarding this polynomial:
As shown in the last example above, we know that to determine all the coefficients of an -degree polynomial, we need to know at least points. But is it mandatory to know this extra point to determine all its coefficients? Not exactly, because we could have known other facts as well. For example, if we know that the additional information that is monic, then from the equation , the leading coefficient has to be equal to by the definition of a monic polynomial, and thus we can determine the value of to be . Below is some common additional information regarding a polynomial that it might not be that obvious what we should be doing in the first place.
(i) We are given the value of a leading coefficient (the leading coefficient is 1 if the polynomial is monic).
(ii) We are given the constant term of the polynomial.
A degree 5 polynomial
- leaves a remainder of 1 when divided by
- leaves a remainder of 2 when divided by
- leaves a remainder of 3 when divided by
- leaves a remainder of 4 when divided by
- leaves a remainder of 5 when divided by .
What is if is monic? What is if its constant term is 240 instead?
Observe that is divisible by . Since these linear factors are distinct, we have
We are given that is a degree-5 monic polynomial, which tells us that . Then
Now, if the constant term of is 240 instead, then we know that , which implies
Therefore, in this case
Are you confident to test out these newly acquired skills? Give it a try with the following problems:
- Applying to a subset of the values, where it's clear what the polynomial form is.
- Test against the other given values.
- Cases where the "odd one out" is in the sequence, like when .
Suppose we are given conditions for a function to have a pattern for some numbers that is not a polynomial. Then it isn't as easy; for example, if a -degree monic polynomial for then we can't have
because then we won't have it as a polynomial. So what do we do then? Let's see the following examples:
Let a monic -degree polynomial satisfy
First, if is a monic cubic polynomial, then is a monic quartic polynomial. We see that
where the factor of was added to make it a quartic. How do we find Note that must divide the RHS for to be a polynomial, so by the remainder-factor theorem
Therefore, we finally have
Let's see another example of this kind.
is a monic quadratic polynomial such that
Since is a monic quartic polynomial, we have
We substitute this back in to get
We don't need to expand the whole thing as we just need the coefficient of to be 0 for to divide
So we have
A monic -degree polynomial satisfies
We remember that . So it is a polynomial. We can easily find to be
where is a monic -degree polynomial. Find
We recall 2 things: and is a polynomial and is zero if So We can write
We find that Thus, since the degree is we can use RFT to get
We know how to solve systems of equations systematically by substitution and elimination. When the equations exhibit a clear pattern, however, we can write them in terms of polynomials, thus converting the system into a polynomial problem.
Here is a conversion problem:
Notice the pattern here is that the coefficient of is the square of the coefficient of , while the coefficient of stays constant. We can summarize this observation with a quadratic as follows:
When is this conversion useful? If the problem asks to find the value of expressions like or , then it suffices to find and , which can be quickly done using the method of difference. Imagine having to solve for first and then plugging them in to find the answer! This method becomes more appealing when the number of variables is large. A long list of equations turns into points
Here is another conversion problem:
Notice that the coefficients here are consecutive squares:
By now you'll get the idea. When in doubt, try to model one of the equations using polynomials and see if the other equations follow suit. In fact, constructing polynomials is not limited to just linear equations. Be creative!