Division algorithm for polynomials
Given two polynomials n and d, where the degree of n exceeds the degree of d, the
following is true:
n x   d  x  
n x 
R x 
 Q x  
d x 
d x 
The division yields a quotient polynomial Q and a remainder polynomial R, which is expressed
as a fraction with the original divisor d.
Note these facts about the degrees of these polynomials:


Q has degree equal to the difference between the degrees of n and d
the degree of R is strictly less than that of d, but non-negative
Example 1:
x 2  5x  10
x 3
x 2
Polynomial long division:
x  3 x  5 x  10
 x 2 3 x 
2
Thus
x 2  5x  10
4
 x 2
x 3
x 3
2 x  10
 2 x 6 
4
The quotient has degree 1, and the remainder has degree 0.
Division of the form
n x 
, can be performed synthetically: “same content, less symbolic”
x c
Here is how that looks for the example above,
Leading coefficient
drops down
x 2  5x  10 x 2  5x  10

x 3
x   3
From left:
Multiply –3 by previous entry on row below
Write here, add to entry above
x+
+r
Write that here, and move right and repeat.
The last term in this “subquotient” is the remainder, while the first numbers are the coefficients
of what is called the depressed polynomial, which begins one degree lower than n and ends
with its own constant term. You can see that in blue, and the remainder labeled with green.
An equivalent form of the division algorithm comes when we “clear” the denominator d:
nx   Qx   dx   Rx 
This form is most helpful if we focus on the case where d is a linear binomial of the form x – c
The equations would then be:
n x 
R x 
 Q x  
x c
x c
nx   Qx   x  c   Rx 
We can discern several notable facts, but let’s start with this:
Since R must have degree strictly less than the degree of x – c, this tells us that R has degree 0,
and is a constant, like 4 was in Example 1.
So we can rewrite our equation, if we wish, as
nx   Qx   x  c   K
Is anything notable about the constant K? Yes, indeed – we can identify it very explicitly.
The equation is true for all x – it is an identity – and, in particular, is true for x = c itself.
nc   Qc c  c   K
nc   0  K
nc   K
That is, the constant remainder K is the output of the dividend function at x = c
You can see that above, in Example 1:
nx   x 2  5x  10 and n 3   32  5 3  10  4
Now we can rewrite one last time, for this special case
nx   Qx   x  c   K
n x 
:
x c
nx   Qx   x  c   nc 
We’re now in a position to summarize the good outcomes of this division algorithm discussion.
After that, we will look at how to go about finding roots, with examples incorporating synthetic
division and depressed polynomials.
Polynomial Roots
n x 
equals nc 
x c
1.
The remainder of
2.
c is a root of n if and only if x – c is a factor of n
3.
A polynomial of degree n can have no more than n roots.
4.
Every polynomial has at least one root.
5. Every complex polynomial f with real coefficients can be factored into the product of
exactly n linear factors, allowing for repeats and the occurrence of complex conjugate*
factors.
f  z   an z n  an1z n1  an2 z n2    a2 z 2  a1z  a0  K z  r1 p1 z  r2 p2 z  rm pm
The powers p1, p2 ,, pm are called the multiplicities of their respective roots.
Ideas about proof
1. If c is a root of n, by definition nc   0 and nx   Qx   x  c   0  Qx   x  c  ; hence x – c
is a factor of n. Conversely, if x – c is a factor of n, then the remainder nc   0 , proving c is a
root of n.
2. The division algorithm will only “work” a maximum of n times with a linear binomial divisor.
3, 4, 5. What we’re saying here is pretty much the Fundamental Theorem of Algebra. Its
proof is beyond our scope.
Complex conjugate factors
Claim: Let c be a complex root of a function f with real coefficients.
Then the complex conjugate of c, symbolized by c , must also be a root of f.
Proof: We examine how a polynomial looks and behaves at x = c:
f  x   an x n  an1x n1  an2 x n2    a2 x 2  a1x  a0
f c   0  anc n  an1c n1  an2c n2    a2c 2  a1c  a0
anc n  an1c n1  an2c n2    a2c 2  a1c  a0  0
Now, conjugate both sides.
anc n  an1c n1  an2c n2    a2c 2  a1c  a0  anc n  an1c n1  an2c n2    a2c 2  a1c  a0  0
anc n  an1c n1  an2c n2    a2c 2  a1c  a0  f c   0
The reason “conjugation” passes so easily through the symbols is that is merely “sign-switching,”
which respects the order of operations at that “+/–” level.
With all this talk about roots, it’s about time we discussed how to find them…
Possible Rational Roots
If f is a polynomial with integer coefficients, then any possible rational number roots of f will
have the form: a0 a , where f  x   an x n  an1x n1    a1x  a0
n
Thus, for example, the polynomial gx   2x 4  6 x 3  7 x 2  5x  20 has possible rational roots
among the set of numbers with forms
 1,  2,  4 ,  5,  10 ,  20
1
or
 1,  2,  4 ,  5,  10 ,  20
2
which, canceling “repeats,” yields
 0.5,  1,  2,  2.5,  4,  5,  10,  20
Now synthetic division can be used to check if any of these are roots, and then a depressed
polynomial could be attacked by the same principles.
Don’t bother with this one, though, because there are none to be found.
Examples:
2.
x 3  2 x 2  5x  6
Possible Rational Roots (PRR):
x=1
Bingo!
1, 2 , 3 , 6
1
1 1 2 5 6
1 1  6
1 1  6 0
 1,  2,  3,  6
The depressed polynomial x 2  x  6 factors as x  3x  2 , revealing the final roots: 3 and –2
You may wish to check that continuing the synthetic division will reveal it as well!
Three roots: x = 1, –2 , 3
3.
PRR:
x 5  2 x 3  2 x 2  3x  2
1, 2
1
 1,  2
1 1 0 2 2 3 2
1 1 1 1  2
1 1 1 1  2 0
x=1
Depressed polynomial:
x 4  x3  x2  x  2
The depressed polynomial, x 4  x 3  x 2  x  2 , has the same set of PRR.
1 1 1 1 1  2
1 2 1 2
1 2 1 2 0
x=1
Depressed polynomial:
x 3  2x 2  x  2
The depressed polynomial, x 3  2 x 2  x  2 , has the same set of PRR.
2
x = –2
1
2 1 2
2 0 2
1 0 1 0
Depressed polynomial:
x2 1
This depressed polynomial has complex conjugate roots ± i
Five roots:
4.
PRR:
x = 1 (twice), –2, i, –i
2x 4  3x 3  12 x 2  7 x  6
1,  2 , 3 ,  6
1,  2
  12 ,  1,  3 2 ,  2,  3,  6
2
x=2
PRR:
1, 3
1,  2
2 3  12  7 6
4 14
4 6
2 7 2 3
0
  12 ,  1,  3 2 ,  3
Depressed polynomial:
2x 3  7 x 2  2x  3
3
x = –3
PRR:
1
1, 2
7
2 3
6 3 3
2 1 1 0
Depressed polynomial:
2x 2  x  1
  12 ,  1
1
2
x=½
2
2 1 1
1 1
2 2 0
The remaining depressed polynomial, 2x + 2, has root x = –1
Four roots:
x = 2, –3 , ½, –1
Yes, it’s tedious and may be very time-consuming. But it’s worth it!
Note that for Examples 2 to 4, the polynomial given is now expressible with only linear factors:
x 3  2x 2  5x  6  x  1x  2x  3
x 5  2x 3  2x 2  3x  2  x  12 x  2x  i x  i 
2x 4  3x 3  12 x 2  7 x  6  2x  1x  12x  3x  2
Also, notice how one can now “construct” a polynomial based solely on roots, although care
must be taken to account for constant factors. (such as the leading 2 in Example 4)
Example 5:
Find an equation for a 4th degree polynomial P with roots of multiplicity one at x = 3 and x = –2,
and a root of multiplicity two at x = 1, with y-intercept 0,24 
Solution:
We know from the roots and multiplicities that Px   K x  3x  2x  12 , where K is a
constant.
The y-intercept tells us that P0  24  K  321  6K , hence K = – 4
Therefore Px   4x  3x  2x  12 