Last time, we saw how to divide polynomials using long division, which takes a lot of time and space. For the simplest cases, dividing by a monic linear polynomial \((x-c)\), there is a much more efficient method. We’ll again see both how to do it, and why it works; then we’ll see how it ties into the Remainder Theorem to efficiently evaluate a polynomial.
Deriving synthetic division from long division
We’ll start with this question from David in 1996:
Synthetic Division Please teach me about synthetic division! I am an over-the-hill college student and still don't understand it.
Doctor Rob answered, showing how we can condense long division:
Synthetic division is just a short-hand way of dividing a polynomial by a linear monic polynomial. In other words, one divides, using long division of polynomials, say:
c3x^2 + (c3r+c2)x + (c3r^2+c2r+c1)
---------------------------------------------------
x - r ) c3x^3 + c2x^2 + c1x + c0
c3x^3 - c3rx^2
-----------------
(c3r+c2)x^2 + c1x
(c3r+c2)x^2 - (c3r^2+c2r)x
--------------------------
(c3r^2+c2r+c1)x + c0
(c3r^2+c2r+c1)x - (c3r^3+c2r^2+c1r)
-----------------------------------
(c3r^3+c2r^2+c1r+c0)
This looks a little complicated; I’ll repeat each step he takes using a specific example.
$$\begin{array}{r l}
\phantom{x-2)}&\underline{\phantom{x^3+4\;\;}{\color{Blue}{x^2-2x+3}}} \\[-1pt]
x-{\color{Red}2}&){\color{DarkGreen}{x^3-4x^2+7x-4}} \\[-1pt]
& \phantom{)}\underline{x^3-2x^2} \\[-1pt]
& \phantom{)x^3}-2x^2+7x \\[-1pt]
& \phantom{x^3-}\underline{-2x^2+4x} \\[-1pt]
& \phantom{x^3-2x^2+}\;\;\;3x-4 \\[-1pt]
& \phantom{x^3-2x^2+}\;\;\;\underline{3x-6} \\[-1pt]
& \phantom{x^3-2x^2+3x-}\;\;\;{\color{Orchid}2}
\end{array}$$
The idea is to remove all the redundancy from this tableau.
1. By keeping columns of coefficients, we can eliminate all the powers of x, and the + signs between terms.
c3 (c3*r+c2) (c3*r^2+c2*r+c1)
------------------------------------------------
1 -r ) c3 c2 c1 c0
c3 - c3r
---------
c3r+c2 c1
c3r+c2 -(c3r^2+c2r)
-----------------------
c3r^2+c2r+c1 + c0
c3r^2+c2r+c1 - (c3r^3+c2r^2+c1r)
-----------------------------------
c3r^3+c2r^2+c1r+c0
Remove the x‘s, and make each line merely a list of terms:
$$\begin{array}{r l}
\phantom{1\;-2)}&\underline{\phantom{1\;\;\;\;\;}{\color{Blue}{1\;-2\;\;\;\;\;3\;\;}}} \\[-1pt]
1\;-{\color{Red}{2}}&){\color{DarkGreen}{1\;-4\;\;\;\;7\;-4}} \\[-1pt]
& \phantom{)}\underline{1\;-2\;} \\[-1pt]
& \phantom{)1\;}-2\;\;\;\;7 \\[-1pt]
& \phantom{1\;+}\underline{-2\;\;\;\;4\;} \\[-1pt]
& \phantom{1\;-2\;\;\;\;\;}3\;-4 \\[-1pt]
& \phantom{1\;-2\;\;\;\;\;}\underline{3\;-6} \\[-1pt]
& \phantom{1\;-2\;\;\;\;\;3\;-}\;\;{\color{Orchid}2}
\end{array}$$
2. The leading terms of the subtracted polynomials are always equal to the ones immediately above, so they are redundant and can be deleted.
This redundancy is because, as we saw last time, we are deliberately matching the entire first term in order to eliminate it. So we don’t need to repeat the number:
$$\begin{array}{r l}
\phantom{1\;-2)}&\underline{\phantom{1\;\;\;\;\;\;\;}{\color{Blue}{1\;-2\;\;\;\;\;3\;\;}}} \\[-1pt]
1\;-{\color{Red}{2}}&){\color{DarkGreen}{1\;-4\;\;\;\;\;7\;-4}} \\[-1pt]
& \phantom{)1\;}\underline{\;-2\;} \\[-1pt]
& \phantom{)1\;}-2\;\;\;\;\;7 \\[-1pt]
& \phantom{1\;+-2\;+}\underline{\;4\;} \\[-1pt]
& \phantom{)1\;-4\;}\;\;\;\;3\;-4 \\[-1pt]
& \phantom{1\;-2\;\;\;\;\;\;3\;\;\;}\underline{-6} \\[-1pt]
& \phantom{1\;-2\;\;\;\;\;\;3\;-}\;\;{\color{Orchid}2}
\end{array}$$
3. The terms "brought down" are redundant and can thus be eliminated:
c3 (c3r+c2) (c3r^2+c2r+c1)
-------------------------------------------------
1-r ) c3 c2 c1 c0
-c3r
---------
c3r+c2
-(c3r^2+c2r)
-----------------------
c3r^2+c2r+c1
-(c3r^3+c2r^2+c1r)
------------------------------------
c3r^3+c2r^2+c1r+c0
We can just subtract from the original copy of each coefficient.
$$\begin{array}{r l}
\phantom{1\;-2)}&\underline{\phantom{1\;\;\;\;\;\;\;}{\color{Blue}{1\;-2\;\;\;\;\;3\;\;}}} \\[-1pt]
1\;-{\color{Red}{2}}&){\color{DarkGreen}{1\;-4\;\;\;\;\;7\;-4}} \\[-1pt]
& \phantom{)1\;}\underline{\;-2\;} \\[-1pt]
& \phantom{)1\;}-2 \\[-1pt]
& \phantom{1\;+-2\;+}\underline{4} \\[-1pt]
& \phantom{)1\;-4\;}\;\;\;\;3 \\[-1pt]
& \phantom{1\;-2\;\;\;\;\;\;3\;\;\;}\underline{\;-6}\; \\[-1pt]
& \phantom{1\;-2\;\;\;\;\;\;3\;-}\;\;{\color{Orchid}2}
\end{array}$$
4. Instead of subtracting negatives, we can add positives. This means dropping all the - signs:
c3 (c3r+c2) (c3r^2+c2r+c1)
--------------------------------------------------
1-r ) c3 c2 c1 c0
c3r
----
c3r+c2
c3r^2+c2r
--------------
c3r^2+c2r+c1
c3r^3+c2r^2+c1r
---------------------
c3r^3+c2r^2+c1r+c0
Changing signs amounts to the common habit in long division of changing signs to avoid mistakes in the subtraction; we’re writing the opposite of the product.
$$\begin{array}{r l}
\phantom{1\;-2)}&\underline{\phantom{1\;\;\;\;\;\;\;}{\color{Blue}{1\;-2\;\;\;\;\;3\;\;}}} \\[-1pt]
1\;-{\color{Red}{2}}&){\color{DarkGreen}{1\;-4\;\;\;\;\;7\;-4}} \\[-1pt]
& \phantom{)1-}\underline{\;\;\;2} \\[-1pt]
& \phantom{)1\;}-2 \\[-1pt]
& \phantom{1\;+-2\;}\underline{-4} \\[-1pt]
& \phantom{)1\;-4\;}\;\;\;\;3 \\[-1pt]
& \phantom{1\;-2\;\;\;\;\;\;3\;\;\;\;\;}\underline{\;6\;} \\[-1pt]
& \phantom{1\;-2\;\;\;\;\;\;3\;-}\;\;{\color{Orchid}2}
\end{array}$$
5. The quotient above the line is the same as the quantities below the line (if we ignore the last one, which is the remainder), so it can be dropped. The horizontal line over the dividend can be removed, and all the other lines brought together to make a single horizontal line down below. The 1 is redundant, too, since it is always there, so it can be understood.
r ) c3 c2 c1 c0
--- c3r c3r^2+c2r c3r^3+c2r^2+c1r
-----------------------------------------------
c3 c3r+c2 c3r^2+c2r+c1 | c3r^3+c2r^2+c1r+c0
---------------------
This is the final form of synthetic division:
$$\begin{array}{r l}
\underline{{\color{Red}2}}|&{\color{DarkGreen}{1\;-4\;\;\;\;\;7\;-4}} \\[-1pt]
& \;\underline{\phantom{1\;-}2\;\;-4\;\;\;\;6\;\;} \\[-1pt]
& {\color{Blue}{1\;-2\;\;\;\;\;3}}\;\;\;|\underline{\;{\color{Orchid}2}\;}
\end{array}$$
Summary of the method, and another example
At this point, the mechanics of synthetic division can be readily seen: 1. Write r down, and put a line to the right and below it. This represents the divisor. 2. Further to the right, list the coefficients of the dividend polynomial. Don't forget to put in zeros if powers of the variable x are skipped. 3. Leave a blank line, then draw a horizontal line for the additions to follow. 4. Bring down the first coefficient below the line. 5. Multiply that last number by the number r, and put it above the line and in the next column to the right. 6. Add the two numbers in this column, and put the sum in the same column below the line. 7. If we are not out of columns, go back to step 5. 8. Draw a line to the left and below the last number in the last line. 9. The quotient is read off from the last line, coefficient by coefficient, left to right, ending at the vertical line. The remainder is the last number in the last line, to the right of the vertical line.
All the hard parts of long division (subtractions, and lots of writing) have been removed. You’ll be seeing a couple variations in the details (mainly where we draw lines).
Example: Divide (x - 3) into x^4 + x^3 - 7x + 11 3 | 1 1 0 -7 11 --- 3 12 36 87 ------------------- 1 4 12 29 |98 ---- So the quotient is x^3 + 4x^2 + 12x + 29, and the remainder is 98. If the divisor is x + 7, then put r = -7, for instance. If the divisor is of the form 3x - 14, then instead divide by x - 14/3, and then divide the quotient and remainder by 3, coefficient by coefficient. This obviously all works whatever the name of the variable (x in this message). Something more complicated must be used if the divisor has degree higher than 1.
To determine the degrees of the terms in the quotient, I count from the right starting at the vertical line: “0, 1, 2, 3”, so the 1 is the coefficient of \(x^3\).
What if the leading coefficient is not 1?
A question from 2005 expands on that comment near the end about dividing by \(3x-14\):
Synthetic Division when Coefficient of Linear Term is Not 1 I'm trying to use synthetic division to do 4x^3 - 7x^2 - 11x + 5 divided by 4x + 5. I can do this problem with regular long division, but can't figure out how to do it with synthetic division because the coefficient of the linear term in the divisor is not equal to 1. I only know how to use synthetic division with a linear coefficient equal to 1, but I'm told that there is a way to do this kind of division, too.
As we’ve seen, synthetic division applies directly only when the divisor has the form \((x-c)\); but it can be extended easily.
I answered:
Hi, Seth.
You have to rewrite the divisor in the right form, and then adjust:
4x^3-7x^2-11x+5 4x^3-7x^2-11x+5 4x^3-7x^2-11x+5
--------------- = --------------- = --------------- * 1/4
4x+5 4(x+5/4) x+5/4
So you'd divide by x + 5/4 synthetically, then divide all coefficients by 4.
So we divide the divisor by its leading coefficient, do the new division, and then divide the quotient by that leading coefficient to correct for the change. I didn’t show the work this way; it would look like this:
-5/4 | 4 -7 -11 5
| -5 15 -5
+----------------+----
4 -12 4 | 0
Divide by 4:
1 -3 1 r 0
This is the method Doctor Rob mentioned in his answer above. We can check by multiplying: $$\left(4x+5\right)\left(x^2-3x+1\right)=\left(4x^3-12x^2+4x\right)+\left(5x^2-15x+5\right)=4x^3-7x^2-11x+5$$
Note that although dividing by a fraction looks scary, fractions happened to cancel out throughout the work. That is typical when the quotient is designed to have integer coefficients, as in textbook problems! Can you see why? (Hint: consider that final division by 4; what does it imply when the resulting coefficients are integers?)
Equivalently, you can divide both the divisor and the dividend by 4 before using synthetic division:
4x^3-7x^2-11x+5 [4x^3-7x^2-11x+5]/4 x^3-7/4x^2-11/4x+5/4
--------------- = ------------------- = --------------------
4x+5 [4x+5]/4 x+5/4
Let's try it that way:
-5/4 | 1 -7/4 -11/4 5/4
| -5/4 15/4 -5/4
+------------------+----
1 -3 1 | 0
That worked pretty well! It won't always give whole numbers, of course, but you're likely to use it only where it will.
That involves fractions, which we avoided the first way; but they all have the same denominator, so they are easy. The benefit of this method is that you don’t have to remember to change your answer at the end.
Either way, the result is $$\frac{4x^3-7x^2-11x+5}{4x+5}=x^2-3x+1.$$
Dealing with a remainder
In that example, the remainder was zero; nothing I said deals directly with a non-zero remainder. Let’s think about that, by just adding a constant (the remainder) to our dividend. Suppose we want to divide \(4x^3-7x^2-11x+{\color{Red}8}\) by \(4x+5\) (so that the remainder should be 3).
By the first method (dividing divisor and quotient), we’d do this synthetic division:
$$\begin{array}{r l}
\underline{-\frac{5}{4}}|&4\;\;-7\;\;-11\;\;\;\;\;\;8 \\[-1pt]
& \underline{\phantom{4}\;\;-5\;\;\;\;\;\;15\;\;-5} \\[-1pt]
& 4\;-12\;\;\;\;\;\;\;4\;\;\;|\underline{\;\;3}
\end{array}$$
What we’ve done is:
$$\frac{4x^3-7x^2-11x+8}{4x+5}=\frac{4x^3-7x^2-11x+8}{x+\frac{5}{4}}\cdot\frac{1}{4}\\=\left(4x^2-12x+4+\frac{3}{x+\frac{5}{4}}\right)\cdot\frac{1}{4}=x^2-3x+1+\frac{3}{4x+5}.$$ Note that the remainder doesn’t have to be divided by 4, like the coefficients. (Doctor Rob was a little wrong there.)
By the second method (dividing both divisor and dividend), we’d do this:
$$\begin{array}{r l}
\underline{-\frac{5}{4}}|&1\;\;-\frac{7}{4}\;\;-\frac{11}{4}\;\;\;\;\;\frac{8}{4} \\[-1pt]
& \underline{\phantom{1}\;\;-\frac{5}{4}\;\;\;\;\;\;\frac{15}{4}\;-\frac{5}{4}} \\[-1pt]
& 1\;\;-3\;\;\;\;\;\;\;\;1\;\;\;|\underline{\;\;\frac{3}{4}}
\end{array}$$
(Note that I didn’t bother simplifying the constant term \(\frac{8}{4}=2\), expecting to be subtracting a fraction from it; keeping the same denominator saved work.)
What we’ve done is:
$$\frac{4x^3-7x^2-11x+8}{4x+5}=\frac{x^3-\frac{7}{4}x^2-\frac{11}{4}x+2}{x+\frac{5}{4}}\\=x^2-3x+1+\frac{\frac{3}{4}}{x+\frac{5}{4}}=x^2-3x+1+\frac{3}{4x+5}.$$ Note that this time I had to multiply the remainder by 4, though I didn’t have to change the coefficients.
I think I prefer the first method.
Synthetic division as an efficient way to evaluate a polynomial
Here is a 1997 question that starts from a different perspective:
Synthetic Division Dear Dr. Math, I need you to explain to me synthetic division. I am currently preparing for math during the summer. I ran into synthetic division, and I do not quite understand. If you could explain to me how it works and give an example it would be greatly appreciated! Thank you!
Doctor Jerry answered, focusing on a feature we haven’t pointed out yet:
Hi Barry, Synthetic division is an efficient arrangement of the arithmetic required to divide a polynomial by the monomial x-a. Although one can do this by long division, because the divisor is the simple polynomial x-a the work can be shortened. If f(x) = A*x^3+B*x^2+C*x+D, for example, synthetic division also provides an efficient way of calculating f(a). This is probably the more useful way of looking at synthetic division.
Sometimes the actual quotient is effectively a by-product of the evaluation – though often, as in using the rational-zero theorem, a very useful one. There we are initially just finding a way to get a value of zero, but then we use the quotient for the next step.
I'll give an illustration. Suppose f(x) = x^3 - 5x^2 + 2x - 10. If we want to calculate f(4), we may do this: f(4) = 4^3 - 5*4^2 + 2*4 - 10. Let's count the number of multiplications and additions required. 2 mults to get 4^2 1 more mult to get 4^3 1 mult to get 5*4^2 1 mult to get 2*4 3 adds to get 4^3 - 5*4^2 + 2*4 - 10 = -18
Note that he is assuming we use our heads and don’t just calculate each of \(4^2\) and \(4^3\) separately, but use the first to find the second.
We can write the same polynomial in an alternative form without explicitly raising x to any powers:
There's a better way. We write f(x) = x(x(x-5)+2)-10, which is called nested multiplication. Now count again. 1 add to get 4-5 1 mult to get 4(4-5) 1 add to get 4(4-5)+2 1 mult to get 4(4(4-5)+2) 1 final add to finish. The result, -18. That's 2 mults and 3 adds, compared to 5 mults and 3 adds above. For higher-degree polynomials, this can add up to big savings in computer or human time.
We’ve saved three multiplications, as well as the need to store the powers for later use.
But this method is exactly what synthetic division does:
So, the better way for evaluation is synthetic division. For doing it by hand the nested arithmetic can be done in a different format. You write
1 -5 2 -10 | 4
and then draw a line and bring down the first coefficient.
1 -5 2 -10 | 4
__________________
1
After that, you repeatedly multiply by 4 and add to the top line. I'll give the result.
1 -5 2 -10 | 4
4 -4 - 8
__________________
1 -1 -2 -18
If you look at x(x(x-5)+2)-10, with x = 4, you'll see that the arithmetic matches synthetic division.
The calculation looks more like the work of synthetic division if we write it as $$f(x)=((x-5)x+2)x-10,$$ or specifically as $$f(4)=((4-5)4+2)4-10.$$
Finally, one frequent use of synthetic division is to test numbers to see if they are roots of a polynomial. The number 4 is not a root since the last number generated (-18 in this case) is not zero. If you get 0, then the number tried is a root, assuming you didn't make any mistakes in arithmetic.
This is part of the work in applying the rational root theorem. Then, we usually go beyond merely testing for roots, and use the quotient to find further roots.
This method is sometimes called synthetic substitution, since the goal is not to divide but to evaluate after substituting a value for x.
Synthetic substitution and the Remainder Theorem
A 2016 question asks about the same idea:
Functional Similarities: Division by (x-c), and Evaluation at x = c Why does the same process work for synthetic division and synthetic substitution? What is the relationship between the two? Why is the process that is necessary to divide a polynomial by a term, such as x - 2, the same as if I'd like to find the value of the polynomial expression at x = 2? For example, Divide 4x^3 + 2x^2 + 5 by (x - 2) 2 | 4 2 0 5 |_____8__20__40_ 4 10 20 45 The answer is 4x^2 + 10x + 20 + 45/(x - 2). Also, if you want to know what 4x^3 + 2x^2 + 5 equals when x = 2, the answer is 45. I was surprised that these are the same process, and can't wrap my head around why that would be.
I answered;
Hi, Jen. There are two very different ways we can approach this. Multiplying both sides by that divisor, 4x^2 + 10x + 20 + 45/(x - 2) can also be expressed like so: (4x^3 + 2x^2 + 5) = (4x^2 + 10x + 20) * (x - 2) + 45 dividend = quotient *divisor + rem Now, what happens if we replace x with 2? The left side (the dividend) becomes the value of the polynomial, of course; on the right, (x - 2) becomes 0, and the whole thing reduces to 45. Thus, the value of the polynomial at x = 2 IS the remainder when you divide it by (x - 2). This is a theorem, called the remainder theorem, because it works for any polynomial and any constant.
The theorem says that, for any polynomial f and number c, the remainder from division by \((x-c)\) is the function value \(f(c)\).
So it's not so much that the process of synthetic division happens to produce the function value; rather, division (regardless of the method you use) produces that value as the remainder. Synthetic division happens to be a very efficient way to find the remainder, which is the function value.
So we can evaluate a polynomial by division, which would not be helpful if division were as hard as we expect it to be; but the method of synthetic division turns out to be easier than straight evaluation!
But if we re-visit the process of synthetic division, we find even more. Look at your work again: 2 | 4 2 0 5 |_____8__20__40_ 4 10 20 45 What you did was to start with the 4, then multiply that by 2 and add the 2; then multiply that by 2 and add the 0; and so on. We can put it all into a single expression -- which hides the quotient and just focuses on how we got the remainder -- like so: +------+------+------+--- coefficients of polynomial | | | | v v v v (((4)*2 + 2)*2 + 0)*2 + 5 | | | | v v v v 4 10 20 45 \_____________/ \__/ quotient rem
This amounts to what Doctor Jerry said above: synthetic division is nested multiplication.
More generally, for any value of x, we are doing this: (((4)*x + 2)*x + 0)*x + 5 This can be expanded (distributed) to give ((4*x + 2)*x + 0)*x + 5 (4*x^2 + 2*x + 0)*x + 5 4*x^3 + 2*x^2 + 0*x + 5 And that's the polynomial! So, not only is the remainder on division by x - c equal to the value of the polynomial at x = c, but the method itself is just an efficient way to arrange the polynomial itself. Both explanations show why synthetic division can be borrowed as a way to evaluate a polynomial (and called synthetic substitution, if you wish). And no work is even wasted in finding the quotient when you are just looking for one number; everything you write is part of the necessary work.
Amazing, isn’t it?
Evaluating by the remainder theorem
We’ll close with this 2005 question about the same type of problem under a different guise:
Applying the Remainder Theorem Given f(x) = 7x^4 + 9x^3 + 4x^2 - 4x + 16, use the Remainder Theorem to find f(2). I don't understand the theorem as a whole.
Doctor Wilko answered:
Hi Craig, Thanks for writing to Dr. Math! The remainder theorem says that when a polynomial f(x) is divided by (x-r), the remainder is f(r). In other words, the remainder of the division process has the same value you would get by plugging r into the function and evaluating it. So the remainder theorem, in your case, says that when the polynomial f(x) = 7x^4 + 9x^3 + 4x^2 - 4x + 16 is divided by (x - 2), the remainder is the same as what you get by evaluating f(2).
So applying the remainder theorem means evaluating a polynomial by division (synthetic or otherwise).
Let's see how to use this. Your question is something like,
"Use the remainder theorem to evaluate f(2)."
You're supposed to do the following:
Use division (synthetic or long) to divide the polynomial f(x) by (x-2) and note the remainder of 208. I recommend synthetic:
2 | 7 9 4 -4 16
14 46 100 192
--------------------
7 23 50 96 208
From this (because of the theorem), you can state that f(2) = 208 (without actually calculating f(2) directly).
With small numbers, I sometimes do this work entirely in my head: “7, times 2 is 14, plus 9 is 23; 23 times 2 is 46, plus 4 is 50; 50, times 2 is 100, minus 4 is 96; 96, times 2 is 192, plus 16 is … uh … 208.” Try that the other way! “2 to the 4th power is 16, times 7 is …”
Why is this useful? f(2) is somewhat complicated to calculate by hand with all the exponents, but I can get the answer to f(2) by doing a simpler problem (division of the polynomial f(x) by (x-2)) and taking note of the remainder. This remainder is the answer to f(2). The remainder theorem gives me an alternate way to calculate f(2).
Again, the amazing thing is that doing a division is easier than directly evaluating!
This also leads into another theorem, the factor theorem. Basically the factor theorem says that if you take a polynomial f(x) and calculate f(r), and f(r) = 0, then (x-r) is a factor of f(x). To relate this to your question, is f(2) = 0? No. We found that f(2) = 208. Therefore (x-2) is not a factor of f(x) = 7x^4 + 9x^3 + 4x^2 - 4x + 16. This means that (x-2) doesn't divide evenly into the polynomial f(x). To summarize this, the factor theorem gives you a quick way to see if some (x-r) is a factor of a polynomial f(x). Just calculate f(r). If f(r) = 0, then (x-r) is a factor of f(x).
This is what we do when we apply the rational zero theorem to factor a polynomial. We’ve come full circle.
Next time, we’ll take a side trip, looking at some interesting problems involving remainders that I found while searching for the questions we saw here.