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Further Maths
Core pure
Polynomials
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Created by
Jack H
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Cards (8)
a
2
+
a^2 +
a
2
+
b
2
=
b^2 =
b
2
=
(
a
+
i
b
)
(
a
−
i
b
)
(a + ib)(a - ib)
(
a
+
ib
)
(
a
−
ib
)
If one solution to a polynomial with
real
coefficients is
complex,
then another solution will be its
conjugate
(x - z)(x - z*) =
x
2
−
2
x
R
e
(
z
)
+
x^2 -2xRe(z) +
x
2
−
2
x
R
e
(
z
)
+
∣
z
∣
2
|z|^2
∣
z
∣
2
For a quadratic with roots p and q:
∑
p
=
\sum p =
∑
p
=
−
b
/
a
-b/a
−
b
/
a
∑
p
q
=
\sum pq =
∑
pq
=
c
/
a
c/a
c
/
a
For a cubic with roots p, q and r:
∑
p
=
\sum p =
∑
p
=
−
b
/
a
-b/a
−
b
/
a
∑
p
q
=
\sum pq =
∑
pq
=
c
/
a
c/a
c
/
a
∑
p
q
r
=
\sum pqr =
∑
pq
r
=
−
d
/
a
-d/a
−
d
/
a
For a quartic with roots p,q,r and s:
∑
p
=
\sum p =
∑
p
=
−
b
/
a
-b/a
−
b
/
a
∑
p
q
=
\sum pq =
∑
pq
=
c
/
a
c/a
c
/
a
∑
p
q
r
=
\sum pqr =
∑
pq
r
=
−
d
/
a
-d/a
−
d
/
a
∑
p
q
r
s
=
\sum pqrs =
∑
pq
rs
=
e
/
a
e/a
e
/
a
If an equation in x has a root x = p, and if we make a
substitution
u = f(x), then the resulting equation in u has a
root
u
=
f(p)
To carry out substitution:
Create an
equation
showing the relationship between
u
and
x
, and make
x
the subject
Replace
x
for
u
in the
polynomial
using the
relationship
Manipulate to get a
polynomial
in
u
with the required
order
Rearrange to get a
polynomial
that is equal to
0