Save
...
Further Maths
Core pure
Roots of polynomials
Save
Share
Learn
Content
Leaderboard
Learn
Created by
Sophie Gaved
Visit profile
Cards (9)
If p and q are roots of a polynomial
a
x
2
+
ax^2 +
a
x
2
+
b
x
+
bx +
b
x
+
c
=
c =
c
=
0
0
0
, then p + q =
-b
/
a
and p x q =
c
/
a.
If p, q and r are roots of a polynomial
a
x
3
+
ax^3 +
a
x
3
+
b
x
2
+
bx^2 +
b
x
2
+
c
x
+
cx +
c
x
+
d
=
d =
d
=
0
0
0
, then
∑
p
\sum p
∑
p
=
-b
/
a
,
∑
p
q
\sum pq
∑
pq
=
c
/
a
and
∑
p
q
r
\sum pqr
∑
pq
r
=
-d
/
a.
If p, q, r and s are roots of a polynomial
a
x
4
+
ax^4 +
a
x
4
+
b
x
3
+
bx^3 +
b
x
3
+
c
x
2
+
cx^2 +
c
x
2
+
d
x
+
dx +
d
x
+
e
=
e =
e
=
0
0
0
, then
∑
p
\sum p
∑
p
=
-b
/
a,
∑
p
q
\sum pq
∑
pq
=
c
/
a
,
∑
p
q
r
\sum pqr
∑
pq
r
=
-d
/
a
and
∑
p
q
r
s
\sum pqrs
∑
pq
rs
=
e
/
a
.
1/p + 1/q +...+ 1/n =
∑
p
\sum p
∑
p
/
∑
p
q
\sum pq
∑
pq
for any number of roots.
p
n
p^n
p
n
x
q
n
q^n
q
n
x...x
N
n
N^n
N
n
=
∑
p
q
.
.
.
N
\sum pq...N
∑
pq
...
N
for any number of roots.
p
2
+
p^2 +
p
2
+
q
2
+
q^2 +
q
2
+
.
.
.
+
...+
...
+
n
2
n^2
n
2
=
(
∑
p
)
2
(\sum p)^2
(
∑
p
)
2
-
2
∑
p
q
\sum pq
∑
pq
for any number of roots.
p
3
+
p^3 +
p
3
+
q
3
q^3
q
3
=
(
p
+
q
)
2
(p + q)^2
(
p
+
q
)
2
-
3
(p
x
q)(p
+
q) only for quadratics.
p
3
+
p^3 +
p
3
+
q
3
+
q^3 +
q
3
+
r
3
r^3
r
3
=
(
p
+
q
+
r
)
3
(p + q + r)^3
(
p
+
q
+
r
)
3
-
3
(p
+
q
+
r)(
∑
p
q
\sum pq
∑
pq
)
+
3
(p
x
q
x
r) only
for
cubics
If a polynomial has roots p, q, r and s, the
polynomial
with roots
a p
+ b,
a
q + b, a r +
b and a
s + b is given
by
f((w
-
b
)/
a
).