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### Resultant of Two Polynomials

```
Date: 09/13/2001 at 08:45:50
From: Lyubomir Ahtarov
Subject: Abstract algebra

Can you give me an example of how to find the resultant of two
polynomials?
```

```
Date: 09/13/2001 at 14:10:45
From: Doctor Rob
Subject: Re: Abstract algebra

Thanks for writing to Ask Dr. Math, Lyubomir.

There are several ways to compute the resultant of two polynomials

m                     m
f(x) = SUM a(i)*x^i = a(m)*PRODUCT (x-c(i)),
i=0                   i=1

n                     n
g(x) = SUM b(j)*x^j = b(n)*PRODUCT (x-d(j)).
j=0                   j=1

1. Use the definition:

n       m
R(f,g) = a(m)^n*b(n)^m*PRODUCT PRODUCT (c(i)-d(j)),
j=1     i=1

m
= a(m)^n*PRODUCT g(c(i)),
i=1

n
= (-1)^(m*n)*b(n)^m*PRODUCT f(d(j)).
j=1

You don't need to know the roots c(i) or d(j) explicitly. Just
substitute them symbolically into one of the above products, expand
the result, and simplify using the facts that f(c(i)) = 0 and
g(d(j)) = 0.

2. Use the (m+n)-by-(m+n) Sylvester determinant:

|a(n) a(n-1)  ...   ...   a(0)   0    0   ...   0  |
| 0    a(n)  a(n-1) ...   ...   a(0)  0   ...   0  |
|...   ...    ...   ...   ...   ...  ...  ...  ... |
R(f,g) = | 0    ...     0    a(n) a(n-1) ...  ...  a(1) a(0)| (mth
|b(m) b(m-1)  ...   ...   b(0)   0    0   ...   0  |  row)
| 0    b(m)  b(m-1) ...   ...   b(0)  0   ...   0  |
|...   ...    ...   ...   ...   ...  ...  ...  ... |
| 0    ...     0    b(m) b(m-1) ...  ...  b(1) b(0)|

3. Use the following recursions to reduce the degrees of the
arguments of the resultant until it can be directly evaluated.
Here k is any constant.

a. R(f,g) = (-1)^(m*n)*R(g,f),
b. R(f1*f2,g) = R(f1,g)*R(f2,g),
c. R(f,g1*g2) = R(f,g1)*R(f,g2),
d. R(f,k*g) = k^m*R(f,g),
e. If k*g(x) = q(x)*f(x) + r(x), where p = degree(r) < m,
then R(f,k*g) = a(m)^(n-p)*R(f,r).
f. R(f,k) = k^m.

Examples of all three methods for the two polynomials

f(x) = 2*x^2 - 5*x + 1,           m = 2,
g(x) = 3*x^3 - 2*x^2 + 2*x + 1,   n = 3,

follow.

1. Use the second product formula for the resultant:

R(f,g) = a(2)^3*g(c(1))*g(c(2)),
= 2^3*(3*c(1)^3-2*c(1)^2+2*c(1)+1)*
(3*c(2)^3-2*c(2)^2+2*c(2)+1),
= 72*c(1)^3*c(2)^3 - 48*c(1)^3*c(2)^2 + 48*c(1)^3*c(2) +
24*c(1)^3 - 48*c(1)^2*c(2)^3 + 32*c(1)^2*c(2)^2 -
32*c(1)^2*c(2) - 16*c(1)^2 + 48*c(1)*c(2)^3 -
32*c(1)*c(2)^2 + 32*c(1)*c(2) + 16*c(1) + 24*c(2)^3 -
16*c(2)^2 + 16*c(2) + 8,
= 72*[c(1)*c(2)]^3 - 48*[c(1)+c(2)]*[c(1)*c(2)]^2 -
64*[c(1)*c(2)]^2 + 48*[c(1)+c(2)]^2*[c(1)*c(2)] -
104*[c(1)+c(2)]*[c(1)*c(2)] + 64*[c(1)*c(2)] +
24*[c(1)+c(2)]^3 - 16*[c(1)+c(2)]^2 + 16*[c(1)+c(2)] + 8

Now use c(1)*c(2) = -a(1)/a(2) = 1/2, c(1) + c(2) = a(0)/a(2) = 5/2:

R(f,g) = 72*(1/2)^3 - 48*(5/2)*(1/2)^2 - 64*(1/2)^2 +
48*(5/2)^2*(1/2) - 104*(5/2)*(1/2) + 64*(1/2) +
24*(5/2)^3 - 16*(5/2)^2 + 16*(5/2) + 8,
= 9 - 30 - 16 + 150 - 130 + 32 + 375 - 100 + 40 + 8,
= 338

|2 -5  1  0  0|   |2  -5    1    0   0|
|0  2 -5  1  0|   |0   2   -5    1   0|
2. R(f,g) = |0  0  2 -5  1| = |0   0    2   -5   1|,
|3 -2  2  1  0|   |0 11/2  1/2   1   0|
|0  3 -2  2  1|   |0   0  11/2  1/2  1|

|2 -5   1    0    0 |
|0  2  -5    1    0 |     |  2   -5    1 |
= |0  0   2   -5    1 | = 4*|57/4 -7/4   0 |,
|0  0 57/4 -7/4   0 |     |  0  57/4 -7/4|
|0  0   0  57/4 -7/4|

|  0    0  1352/3249|
= 4*|57/4 -7/4     0    |  = 4*(57/4)^2*(1352/3249),
|  0  57/4   -7/4   |

= 338.

3. g(x) = f(x)*(6*x+11)/4 + (57*x-7)/4, so r(x) = (57*x-7)/4, p = 1.

R(f,g) = 2^2*R(f,r) = 4*R(f,r),
= 4*[(-1)^(2*1)*R(r,f)],
= 4*R(r,f).

f(x) = (8*x/57-1084/3249)*r(x) + 1352/3249, so r'(x) = 1352/3249,
p' = 0.

R(f,g) = 4*R(r,f),
= 4*[(57/4)^2*R(r,r')],
= (57^2/4)*R(r,r'),
= (57^2/4)*1352/3249 = 338.

I hope this is what you were seeking.  If not, write again.

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Modern Algebra

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