In searching for questions about polynomial division, I ran across several about problems where you are given the remainders when an unknown polynomial is divided by two or three different small polynomials, and have to find the remainder when it is divided by a different, but related, polynomial (typically the product of the others). We’ll see four different approaches (from three different Math Doctors – progressively getting easier), and then solve our initial problem all four ways.
A problem, and two hints
We’ll start with a question from 1995, during the first year of Ask Dr. Math:
Polynomial Problem The problem is: A remainder of 9 results when the polynomial p(x) is divided by x - 2. A remainder of -3 results when p(x) is divided by x + 2. A remainder of 3 results when p(x) is divided by x - 1. Find the remainder when p(x) is divided by (x - 2)(x + 2)(x - 1). Could you just give us a hint first?
We aren’t asked to find the polynomial \(p(x)\); in fact, there will be infinitely many that would work. All we want is this remainder (which, by the way, we can expect not to be a number, but a quadratic polynomial, since we are dividing by a cubic). But how can we possibly figure that out? In fact, how can we be sure that every such polynomial will leave the same remainder?
Doctor Ethan answered, doing just what was asked:
It's Ethan, and after a little discussion with my friend Ken, and a little help from a professor here, we got it. The idea is really neat. Here is the hint. I will start with a pretty vague hint, and if you need more then write back. All right, we have this function, and we want to know the remainder when we divide by that product. I think that it will be easier if we just think of it as trying to find the smallest polynomial that we can that has the given properties. That will be the remainder. Okay, so here is the process. It may not make sense to you because I don't want to give away too much. Let us call f(x) our function. Well then, if f(x)/(x-2) has remainder 9, then f(2) must be 9. Similarly f(-2) = -3, and f(1) = 3. I will let you think about why that is true since you said that you just wanted a hint. I will also let you think about how we can use this fact to find the answer. Write back if this isn't enough of a hint.
Evidently Ruth never wrote back. Is it enough of a hint for you?
The first hint is not essential, but is interesting: Whatever remainder \(f(x)\) we come up with, we could put it in place of \(p(x)\), and everything in the problem would still be true. That’s because two polynomials have the same remainder for a given divisor, if and only if they differ by a multiple of that divisor.
That is, if $$p(x)=q(x)(x-2)(x+2)(x-1)+f(x),$$ then our remainder \(f(x)\) itself will satisfy all the conditions for \(p(x)\), because it differs from \(p(x)\) by \(q(x)(x-2)(x+2)(x-1)\), which is a multiple of each of the original three divisors, so that \(f(x)\) and \(p(x)\) leave the same remainder for each division! (Next time, we’ll see some problems where that is not true.)
The second hint is the Remainder Theorem, which we saw last time; it says that, for any polynomial f and number c, the remainder from division of \(f(x)\) by \((x−c)\) is the function value \(f(c)\).
After seeing four explicit methods for solving similar problems, we’ll come back and solve this problem using each of them. Along the way, we’ll see both hints in action.
A simpler problem, solved laboriously
A 1996 question has only two given remainders, making it easier:
Division of Unknown Polynomials I am an eighth grade student at the Willmar Junior High School in Willmar, Minnesota. I have a math question that I need help on. When a polynomial P(x) is divided by x-1, the remainder is 3. When P(x)is divided by x-2, the remainder is 5. Find the remainder when P(x) is divided by x^2-3x+2. Thank you, Jacob
The first thing you should observe is that the new divisor \(x^2-3x+2\) is \((x-1)(x-2)\), the product of the initial divisors. That will be true of all of the problems we look at here.
Method 1: Remainder arithmetic
Doctor Steven answered, giving a slightly complicated solution, using the definition of a remainder, and facts about remainders:
This is a tough problem for an eighth grade class! First let's look at what we know: 1. We know P(x) when divided by (x-1) gives a remainder of 3. So P(x) = f(x)*(x-1) + 3. (We don't care what f(x) is) 2. We know P(x) when divided by (x-2) gives a remainder of 5. So P(x) = g(x)*(x-2) + 5, also.(We don't care what g(x) is either!) So P(x) = f(x)*(x-1) + 3 = g(x)*(x-2) + 5. Subtract 5 from every part of this three part equation to get: P(x) - 5 = f(x)*(x-1) - 2 = g(x)*(x-2) + 0.
Not quite clear? We’re calling the two quotients \(f(x)\) and \(g(x)\), and the two given remainders imply that $$f(x)(x-1)-2=g(x)(x-2).$$
Well, we also know that (x-1) = 1*(x-2) + 1. So the remainder of x-1 when divided by x-2 is +1.
This just says that because \((x-1)\) is 1 more than \((x-2)\), the remainder is 1. We’ll use that soon.
Now check this out for some higher mathematics (really though it's
pretty easy):
___ ___
Look at 4| 5 ; its remainder is 1. Now look at 4| 3 ; its remainder is 3.
____
Now look at 4| 15 ; its remainder is 3, or 3*1, the product of the remainders of 5/4 and 3/4. This works for anything that is multiplied together.
This shows, by example, that remainders (using the same divisor) multiply: The remainder of a product is the product of the remainders. Technically, that’s not quite true, because, for examples, the remainders of 7 and 15 after division by 4 are both 3, but the remainder of \(7\times15=105\) is not \(3\times3=9\), but 1 (because \(105=4\times26+1\)). Really, the rule is that the remainders will differ by a multiple of the divisor. That’s all we’ll really need.
These rules for remainders, also called modular arithmetic, are discussed further in Casting Out Nines: Why It Works. The associated notation is useful; for example, we write \(27\equiv3\pmod{4}\) to say that the remainder when we divide 27 by 4 is 3 (because \(27=4\cdot6+3\), or equivalently \(27-3=4\cdot6\)). We read this as “27 is congruent to 3, modulo 4”, meaning that 27 and 3 leave the same remainder, and since \(3<4\), it is the remainder.
Stated in those terms, if \(a\equiv b\pmod{m}\) and \(c\equiv d\pmod{m}\), then \(ac\equiv bd\pmod{m}\).
In our problem, we are told that $$P(x)\equiv3\pmod{x-1}\\P(x)\equiv5\pmod{x-2}.$$ We’ve also seen that $$x-1\equiv1\pmod{x-2}.$$ Here we are working with polynomials rather than numbers, but the ideas still work. Our conclusion so far is that \(f(x)(x-1)-2=g(x)(x-2)\); in modular terms, $$f(x)(x-1)\equiv2\pmod{x-2},$$ since the difference is a multiple of \((x-2)\).
So f(x)*(x-1) - 2 better not have a remainder when divided by (x-2) (since it equals the far righthand side, which definitely doesn't have a remainder when divided by (x-2) ).
This tells us that f(x) better have a remainder of 2 when divided by (x-2) (since rem(2)*rem(1) - rem(2) = rem(0) ).
So f(x) = m(x)*(x-2) + 2. (We don't care what m(x) is.)
So we conclude that $$f(x)\equiv f(x)(1)\equiv f(x)(x-1)\equiv2\pmod{x-2},$$ because \(x-1\equiv1\pmod{x-2}\).
Alternatively, without using mod, we could say that $${\color{Red}{f(x)(x-1)+3}}={\color{SeaGreen}{g(x)(x-2)+5}}\\f(2)(2-1)+3=g(2)(2-2)+5\\f(2)+3=5\\f(2)=2,$$
So (using the Remainder Theorem) \(f(x)\equiv2\pmod{x-2}\); that is, \(f(x)\) has a remainder of 2.
Now we have: P(x) - 5 = [m(x)*(x-2) + 2](x-1) - 2 = g(x)(x-2) + 0. But all we need to look at now is: P(x) - 5 = [m(x)*(x-2) + 2](x-1) - 2. Add 5 to both sides to get: P(x) = [m(x)*(x-2) + 2](x-1) + 3. Then simplify on the right to get: P(x) = m(x)*(x-2)*(x-1) + (2x - 2) + 3. Simplify some more to get: P(x) = m(x)*(x^2 - 3x + 2) + 2x + 1. So the remainder of P(x) when divided by x^2 - 3x + 2 is (2x + 1). Whew! That was a toughy. ;)
Presumably they were not already familiar with this kind of problem (which would be true of me as well!) and had to work it out from scratch; that makes their work a little harder than it needs to be.
Let’s put this all together, using modular notation and skipping a couple side trails:
We are given that $$p(x)\equiv3\pmod{x-1}\\p(x)\equiv5\pmod{x-2}$$ Equivalently, $$P(x)={\color{Red}{f(x)(x-1)+3}}\\P(x)={\color{SeaGreen}{g(x)(x-2)+5}}$$ where \(f(x)\) and \(g(x)\) are the quotients.
Combining these, $${\color{Red}{f(x)(x-1)+3}}={\color{SeaGreen}{g(x)(x-2)+5}}\\f(x)(x-1)=g(x)(x-2)+2\\f(x)(x-1)\equiv2\pmod{x-2}$$
But \(x-1\equiv1\pmod{x-2}\), so this implies that \(f(x)\equiv2\pmod{x-2}\).
Therefore $$f(x)={\color{Blue}{m(x)(x-2)+2}}$$ where \(m(x)\) is another quotient.
Therefore $$P(x)={\color{Red}{f(x)(x-1)+3}}=\left[{\color{Blue}{m(x)(x-2)+2}}\right](x-1)+3\\=m(x)(x-2)(x-1)+2(x-1)+3\\=m(x)(x-2)(x-1)+2x+1$$
So the remainder is \(2x+1\).
We can check the answer by observing that this itself is a polynomial whose remainder by the product of the divisors is \(2x+1\); and its remainders by the two individual divisors are, as required, 3 and 5 respectively: $$2x+1={\color{Red}{2(x-1)+3}}\\2x+1={\color{SeaGreen}{2(x-2)+5}}$$ (You could determine this by synthetic division.)
The next method will improve on this one.
A larger problem, and two methods
A 1998 question, from George, is like our first question, with three given remainders:
Polynomial and Remainder An unknown polynomial f(x) of degree 37 yields a remainder of 1 when divided by x-1, a remainder of 3 when divided by x-3, a remainder of 21 when divided by x-5. Find the remainder when f(x) is divided by (x-1)(x-3)(x-5).
Method 2: Remainder Theorem, Iterated
Doctor Rob answered with two methods; the first is similar to the previous method, but without modular arithmetic, and extended to the larger problem:
Okay! There are a couple of ways of solving this.
The first method uses the Remainder Theorem:
f(x) = (x-a)g(x) + f(a).
The first condition says that
f(x) = (x-1)g(x) + 1.
The second condition says that
f(x) = (x-1)g(x) + 1 = (x-3)h(x) + 3.
Substituting in x = 1 tells you that
1 = -2h(1) + 3,
so h(1) = 1, so
h(x) = (x-1)k(x) + 1.
Then f(x) = (x-3)[(x-1)k(x) + 1] + 3
= (x-1)(x-3)k(x) + x.
Here we substituted the value corresponding to the first divisor in order to learn something about the second quotient, giving us the remainder when we divide by the product of those first two divisors.
We could equally well have used the value, 3, for the second divisor to learn about the first quotient: $$f(x)={\color{Red}{(x-1)g(x)+1}}={\color{SeaGreen}{(x-3)h(x)+3}}\\(3-1)g(3)+1=(3-3)h(3)+3\\2g(3)+1=3\\g(3)=1\\g(x)={\color{Orange}{(x-3)n(x)+1}}\\f(x)=(x-1)[{\color{Orange}{(x-3)n(x)+1}}]+1=(x-1)(x-3)n(x)+(x-1)+1=(x-1)(x-3)n(x)+x$$ again.
Now we bring in the third divisor:
The third condition says that f(x) = (x-5)m(x) + 21. Substituting in x = 5 tells you that 21 = (5-1)(5-3)k(5) + 5, so k(5) = 2, so k(x) = (x-5)n(x) + 2. Then f(x) = (x-1)(x-3)[(x-5)n(x) + 2] + x, = (x-1)(x-3)(x-5)n(x) + 2(x-1)(x-3) + x, = (x-1)(x-3)(x-5)n(x) + 2x^2 - 7x + 6. Thus the remainder sought is 2x^2 - 7x + 6.
At each step, we write the equation for a division, substituting a value to obtain a new remainder, write a new equation, and substitute that in the first.
Again, we can confirm the answer, by showing that this remainder satisfies all three conditions:
$$2x^2-7x+6={\color{Red}{(2x-5)(x-1)+1}}\\2x^2-7x+6={\color{SeaGreen}{(2x-1)(x-3)+3}}\\2x^2-7x+6={\color{Blue}{(2x+3)(x-5)+21}}$$
Method 3: Chinese Remainder Theorem
The second method uses the Chinese Remainder Theorem.
f(x) = 1 (mod x-1),
f(x) = 3 (mod x-3),
f(x) = 21 (mod x-5).
First we must find any polynomials a(x), b(x) and c(x) such that
a(x)*(x-3)(x-5) + b(x)*(x-1)(x-5) + c(x)*(x-1)(x-3) = 1
a(x) = 1/8, b(x) = -1/4, and c(x) = 1/8 will do. Then the answer is given by
f(x) = 1*a(x)*(x-3)(x-5) +
3*b(x)*(x-1)(x-5) +
21*c(x)*(x-1)(x-3) (mod (x-1)(x-3)(x-5)),
= 1*(1/8)(x-3)(x-5) +
3*(-1/4)(x-1)(x-5) +
21*(1/8)(x-1)(x-3) (mod (x-1)(x-3)(x-5)),
= x^2(1/8-3/4+21/8) +
x(-1+9/2-21/2) +
(15/8-15/4+63/8) (mod (x-1)(x-3)(x-5)),
= 2x^2 - 7x + 6 (mod (x-1)(x-3)(x-5)),
and the remainder is 2x^2 - 7x + 6.
We discuss the Chinese Remainder Theorem in Chinese Remainders With and Without the Theorem; here it is being applied to polynomials rather than to numbers. As I say there, “The theorem proper, e.g. as presented in Wikipedia, is just an existence theorem, not explicitly saying how to solve the problem,” but this is a common method for solving such problems, as explained here. I won’t try to fully explain its application here, but I can point out that finding functions a, b, and c (which turn out to be constants) is identical to what you would do to write $$\frac{1}{(x-1)(x-3)(x-5)}$$ as a sum of partial fractions $$\frac{A}{x-1}+\frac{B}{x-3}+\frac{C}{x-5};$$ and a key idea is that, for example, the remainder of \(f(x)\) on division by \((x-5)\) will be 21 because $$f(x)=1A(x-3)(x-5)+3B(x-1)(x-5)+21C(x-1)(x-3),$$ and $$f(5)=1A(5-3)(5-5)+3B(5-1)(5-5)+21C(5-1)(5-3)\\=0+0+21C(4)(2)=21(8C)=21,$$ which is just what we need.
It is comforting that the answers agree! Note that the degree is immaterial, as long as it is at least 2.
In particular, we can obtain a degree-37 polynomial that satisfies all the conditions by adding the product of any degree-34 polynomial by \((x-1)(x-3)(x-5)\) to the remainder \(2x^2-7x+6\).
A simple problem and a quick method
Here’s another 1998 question with two problems, the second fits our topic, while the first prepares us for it:
Remainder and Factor Theorem Dear Dr. Maths, How do I answer the following questions? 1) 2x^4 - 3x^3 - 18x^2 - 7 divided by x^2 - 1 2) When the polynomial p(x) is divided by (x-1), the remainder is 5, and when p(x) is divided by (x-2), the remainder is 7. Find the remainder, when p(x) is divided by (x-1)(x-2). Thank you Dr Maths!
Doctor Sonya first stated a theorem:
Dear Yanzhen,
To solve this problem, there is an important theorem you need to know, called the Remainder and Factor theorem. As I see from the subject line, you've already heard of this, but I'll restate it anyway.
The theorem says that when you divide a polynomial f(x) by another one g(x), there is a quotient q(x) and a remainder r(x). These are related by the equation
f(x) = g(x)q(x) + r(x)
The remainder r(x) either is zero, or has degree lower than the divisor g(x).
For example, if g(x) is a linear polynomial, then r(x) should be a constant, and if g(x) is quadratic, then r(x) should be a linear polynomial of the form Ax+B.
I'm not going to prove this theorem, but if you want more information on it, take a look in the Dr. Math archives, or write us back.
This is actually what is commonly called the Division Algorithm (though it is really a theorem, not an algorithm), and it mostly defines what we mean by division (while stating that it can always be done), and therefore defines “remainder”.
Method 4: Undetermined coefficients
The first problem could be solved just by long division; but the theorem provides a useful foundation for the second problem:
All right, now that this theorem's been stated, here's how we'll use it in your problems.
(1) For your first problem:
Write f(x) = 2x^4 - 3x^3 - 18x^2 - 7. If the remainder is Ax + B when this is divided by x^2 - 1 = (x-1)(x+1), we have
f(x) = (x-1)(x+1)q(x) + Ax + B.
This may not look like we've simplified anything, but wait, what happens if you plug x = 1 into both sides of the equation? You get:
f(1) = (1-1)(1+1)q(1) + 1A + B
Now evaluate both sides and see what you get. Try the same thing with x = -1. Doing this will give you two equations for A and B, which you can then solve. Once you have A and B, plug their values back into the remainder Ax + B, and you'll be done.
The key here is that the remainder on division by a degree-2 polynomial will be a degree-1 (or less) polynomial), so we can call it \(Ax+B\) for now, and then try to determine what the coefficients are.
When we let \(x=1\), the equation simplifies to $$f(1)=(0)(2)q(1)+(1)A+B=A+B$$ And since \(f(1)=2-3-18-7=-26\), we have $$A+B=-26$$
When we let \(x=-1\), it simplifies to $$f(1)=(-2)(0)q(1)+(-1)A+B=-A+B$$ And since \(f(-1)=2+3-18-7=-20\), we have $$-A+B=-20$$
We can easily solve this system of equations to find that \(A=-3\) and \(B=-23\).
I imagine Doctor Sonya read the first problem as asking only for the remainder (as the second does), so the answer is just \(r(x)=-3x-23\). That is probably what was intended.
But we can check the answer by actually doing the division:
$$\begin{array}{r l}
\phantom{x^2+0x-1}&\underline{\phantom{2x^4-3x^3-1}2x^2-3x-16\;}\\[-1pt]
x^2+0x-1&)2x^4-3x^3-18x^2+0x-7\\[-1pt]
& \phantom{)}\underline{2x^4\phantom{-3x^3-}-2x^2\;}\\[-1pt]
& \phantom{)2x^4}-3x^3-16x^2+0x\\[-1pt]
& \phantom{2x^4}\;\;\underline{-3x^3\phantom{-16x^2-}+3x\;}\\[-1pt]
& \phantom{2x^4-3x^3}\;\;-16x^2-3x-7\\[-1pt]
& \phantom{2x^4-3x^3}\;\;\;\underline{-16x^2\phantom{-3x\;}+16\;\;}\\[-1pt]
& \phantom{2x^4-3x^3-16x^2}\;-3x-23
\end{array}$$ The remainder is just what we said.
Applying it to our kind of problem
The important thing is that this same approach applies to the problems we’re interested in:
(2) For your second problem, I'm going to write out what the problem says in terms of the remainder and factor theorem. p(x) = (x-1)q(x) + 5 p(x) = (x-2)Q(x) + 7 (Notice here that q(x) and Q(x) are different quotients.) Now, try plugging 1 into the first equation and 2 into the second equation. You'll get that: p(1) = 5 and that p(2) = 7 .
This is what the Remainder Theorem tells us: The remainder on division by \((x-c)\) is \(f(c)\),
Now, suppose the remainder is Ax + B when p(x) is divided by (x-1)(x-2). We write
p(x) = (x-1)(x-2)q(x) + Ax + B.
Here, q(x) is the quotient. Putting x = 1 and x = 2 respectively will give you two equations for A and B, which you can then solve simultaneously to find out what Ax + B is. When you plug in, remember that you already know what p(1) and p(2) are.
Let’s do that:
When we let \(x=1\), the equation simplifies to $$p(1)=(0)(-1)q(1)+(1)A+B=A+B$$ And since \(p(1)=5\), we have $$A+B=5$$
When we let \(x=2\), it simplifies to $$p(2)=(1)(0)q(1)+(2)A+B=2A+B$$ And since \(p(2)=7\), we have $$2A+B=7$$
We can easily solve this system of equations to find that \(A=2\) and \(B=3\). So the remainder is \(2x+3\).
Solving our first problem by each method
Now let’s solve the first problem, using what we’ve learned.
We were given $$p(x)={\color{Red}{f(x)(x-2)+9}}\\p(x)={\color{SeaGreen}{g(x)(x+2)-3}}\\p(x)={\color{Blue}{h(x)(x-1)+3}},$$ or equivalently, $$p(2)=9; p(-2)=-3; p(1)=3$$
We want to find the remainder on division by the product, \((x-2)(x+2)(x-1)\); that is, \(r(x)\) such that $$p(x)=q(x)(x-2)(x+2)(x-1)+r(x).$$
Method 1 (remainder arithmetic)
We can extend the method from above, and also replace some applications of the Remainder Theorem with modular arithmetic:
We are given that $$p(x)\equiv 9\pmod{x-2}\\p(x)\equiv-3\pmod{x+2}\\p(x)\equiv3\pmod{x-1}$$
Combining the first two, $${\color{Red}{f(x)(x-2)+9}}\equiv-3\pmod{x+2}\\f(x)(x-2)\equiv-12\pmod{x+2}$$
But since \(x-2\equiv-4\pmod{x+2}\), this implies that $$-4f(x)\equiv-12\pmod{x+2}\\f(x)\equiv3\pmod{x+2}.$$
Therefore $$f(x)={\color{Orange}{m(x)(x+2)+3}},$$ and $$p(x)={\color{Red}{f(x)(x-2)+9}}=\left[{\color{Orange}{m(x)(x+2)+3}}\right](x-2)+9\\=m(x)(x+2)(x-2)+3(x-2)+9\\=m(x)(x+2)(x-2)+3x+3$$
Now we use the third remainder to introduce the third factor:
Since $$p(x)\equiv3\pmod{x-1}\\x+2\equiv3\pmod{x-1}\\x-2\equiv-1\pmod{x-1},$$ we see that the last line implies that $$p(x)\equiv m(x)(3)(-1)+3x+3\pmod{x-1}\\3\equiv -3m(x)+3x+3\pmod{x-1}\\m(x)\equiv x\pmod{x-1}$$ But \(x\equiv1\pmod{x-1}\), so we find that $$m(x)\equiv 1\pmod{x-1}$$
Therefore $$m(x)={\color{Orchid}{n(x)(x-1)+1}}$$ and
$$p(x)=m(x)(x+2)(x-2)+3x+3\\=[{\color{Orchid}{n(x)(x-1)+1}}](x+2)(x-2)+3x+3\\=n(x)(x-1)(x+2)(x-2)+(x+2)(x-2)+3x+3\\=n(x)(x-1)(x+2)(x-2)+x^2+3x-1$$
So the remainder is \(x^2+3x-1\).
Method 2 (iterated Remainder Theorem)
Combining the first two facts, $${\color{Red}{f(x)(x-2)+9}}={\color{SeaGreen}{g(x)(x+2)-3}}\\f(2)(2-2)+9=g(2)(2+2)-3\\g(2)=3$$
So $$g(x)={\color{Orange}{m(x)(x-2)+3}}$$ and $$p(x)={\color{SeaGreen}{g(x)(x+2)-3}}\\=[{\color{Orange}{m(x)(x-2)+3}}](x+2)-3\\=m(x)(x-2)(x+2)+3x+3$$
Now using the third remainder, $$p(1)=m(1)(1-2)(1+2)+3(1)+3=-3m(1)+6;$$ since \(p(1)=3\), \(m(1)=1\), and $$m(x)={\color{Orchid}{n(x)(x-1)+1}}.$$ So $$p(x)=[{\color{Orchid}{n(x)(x-1)+1}}](x-2)(x+2)+3x+3\\=n(x)(x-2)(x+2)(x-1)+(x-2)(x+2)+3x+3\\=n(x)(x-2)(x+2)(x-1)+x^2+3x-1$$ and the remainder is, again, \(x^2+3x-1\).
Method 3 (Chinese Remainder Theorem)
We want to find parameters so that $$A(x+2)(x-1)+B(x-2)(x-1)+C(x-2)(x+2)=1$$ Setting x to each of the values 2, -2, 1, we find that \(A=\frac{1}{4}\), \(B=\frac{1}{12}\), and \(C=-\frac{1}{3}\).
Then, inserting the remainders, $$p(x)={\color{Red}{9}}A(x+2)(x-1)+{\color{SeaGreen}{(-3)}}B(x-2)(x-1)+{\color{Blue}{3}}C(x-2)(x+2)\\=9\cdot\frac{1}{4}(x+2)(x-1)-3\frac{1}{12}(x-2)(x-1)-3\frac{1}{3}C(x-2)(x+2)\\=\frac{9}{4}(x+2)(x-1)-\frac{1}{4}(x-2)(x-1)-C(x-2)(x+2)\\=\frac{1}{4}\left[(9x^2+9x-18)-(x^2-3x+2)-(4x^2-16)\right]\\=\frac{1}{4}\left[4x^2+12x-4\right]=x^2+3x-1$$ and the remainder is, yet again, \(x^2+3x-1\).
Method 4 (undetermined coefficients)
We can very easily extend this method to three given remainders (and a cubic final divisor) by recognizing that the remainder can have degree up to 2.
Suppose that $$p(x)=q(x)(x-2)(x+2)(x-1)+Ax^2+Bx+C.$$
Then $$p(2)=4A+2B+C=9\\p(-2)=4A-2B+C=-3\\p(1)=A+B+C=3$$
We find that \(A=1\), \(B=3\), and \(C=-1\), so the remainder is (no surprise!) \(x^2+3x-1\).
By the way, I suspect that Doctor Ethan, who answered the first question, probably had this method in mind: Just look for the simplest polynomial with \(p(2)=9\), \(p(-2)=-3\), and \(p(1)=3\), which will be a quadratic. It’s conceptually that simple (but we learned a lot more taking the long way).
Confirmation
As before, we can confirm the answer, by showing that our remainder satisfies all three conditions, making it, as promised, the smallest of all possible \(p(x)\):
$$x^2+3x-1={\color{Red}{(x+5)(x-2)+9}}\\x^2+3x-1={\color{SeaGreen}{(x+1)(x+2)-3}}\\x^2+3x-1={\color{Blue}{(x+4)(x-1)+3}}$$
(I got the quotients, and confirmed the remainders, by synthetic division.)
And, of course, $$x^2+3x-1=0(x-2)(x+2)(x-1)x^2+3x-1,$$ so we get the claimed remainder, with quotient 0. But what we’ve shown is that we’d get the same remainder for any such \(p(x)\).
Next time, we’ll look at a couple more complicated problems of this sort.