Em matemática , a resultante de dois polinômios mônicos
P
{\displaystyle P}
Q
{\displaystyle Q}
corpo
k
{\displaystyle k}
r
e
s
(
P
,
Q
)
=
∏
P
(
x
)
=
0
∏
Q
(
y
)
=
0
(
y
−
x
)
,
{\displaystyle \mathrm {res} (P,Q)=\prod _{P(x)=0}\prod _{Q(y)=0}(y-x),\,}
Das diferenças de suas raízes, de onde
x
{\displaystyle x}
y
{\displaystyle y}
fecho algébrico de
k
{\displaystyle k}
p
{\displaystyle p}
q
{\displaystyle q}
p
deg
Q
q
deg
P
.
{\displaystyle p^{\deg Q}q^{\deg P}.\,}
r
e
s
(
P
,
Q
)
=
∏
Q
(
y
)
=
0
P
(
y
)
{\displaystyle \mathrm {res} (P,Q)=\prod _{Q(y)=0}P(y)\,}
E esta expressão permanece invariante sem
P
{\displaystyle P}
Q
{\displaystyle Q}
r
e
s
(
P
,
Q
)
=
(
−
1
)
deg
P
⋅
deg
Q
⋅
r
e
s
(
Q
,
P
)
{\displaystyle \mathrm {res} (P,Q)=(-1)^{\deg P\cdot \deg Q}\cdot \mathrm {res} (Q,P)}
r
e
s
(
P
⋅
R
,
Q
)
=
r
e
s
(
P
,
Q
)
⋅
r
e
s
(
R
,
Q
)
{\displaystyle \mathrm {res} (P\cdot R,Q)=\mathrm {res} (P,Q)\cdot \mathrm {res} (R,Q)}
Se
P
′
=
P
+
R
∗
Q
{\displaystyle P'=P+R*Q}
deg
P
′
=
deg
P
{\displaystyle \deg P'=\deg P}
r
e
s
(
P
,
Q
)
=
r
e
s
(
P
′
,
Q
)
{\displaystyle \mathrm {res} (P,Q)=\mathrm {res} (P',Q)}
Se
X
,
Y
,
P
,
Q
{\displaystyle X,Y,P,Q}
X
=
a
00
⋅
P
+
a
01
⋅
Q
,
Y
=
a
10
⋅
P
+
a
11
⋅
Q
{\displaystyle X=a_{00}\cdot P+a_{01}\cdot Q,Y=a_{10}\cdot P+a_{11}\cdot Q}
então
r
e
s
(
X
,
Y
)
=
det
(
a
00
a
01
a
10
a
11
)
deg
P
⋅
r
e
s
(
P
,
Q
)
{\displaystyle \mathrm {res} (X,Y)=\det {\begin{pmatrix}a_{00}&a_{01}\\a_{10}&a_{11}\end{pmatrix}}^{\deg P}\cdot \mathrm {res} (P,Q)}
r
e
s
(
P
−
,
Q
)
=
r
e
s
(
Q
−
,
P
)
{\displaystyle \mathrm {res} (P_{-},Q)=\mathrm {res} (Q_{-},P)}
P
−
(
z
)
=
P
(
−
z
)
{\displaystyle P_{-}(z)=P(-z)}
