Maths Expertes : Équations Polynômiales - Binôme de Newton & Factorisation

u∈Cv∈Cn∈N
(u+v)n=
n
X
k=0 n
kun−kvk
n
n= 0 u v (u+v)0= 1
P0
k=0 0
ku0−kvk=0 0u0v0= 1
n∈Nu v n
(u+v)n+1 = (u+v)(u+v)n
= (u+v)
n
X
k=0 n
kun−kvk
=u
n
X
k=0 n
kun−kvk+v
n
X
k=0 n
kun−kvk
=
n
X
k=0 n
kun+1−kvk+
n
X
k=0 n
kun−kvk+1
=
n
X
k=0 n
kun+1−kvk+
n+1
X
k=1 n
k−1un+1−kvk
=
n
X
k=1 n
kun+1−kvk+
n
X
k=1 n
k−1un+1−kvk+n 0un+1v0+n nu0vn+1
=
n
X
k=1 n
k+n k−1un+1−kvk+n+ 1 0un+1v0+n+ 1 n+ 1u0vn+1
=
n
X
k=1 n+ 1
kun+1−kvk+n+ 1 0un+1v0+n+ 1 n+ 1u0vn+1
=
n+1
X
k=0 n
kun−kvk
n
(2 + i)3=3
023i0+3 122i1+3 221i2+3 320i3= 8 + 12i−6−i= 2 + 11i
(1 −i)5=5
015i0−5 114i1+5 213i2−5 312i3+5 411i4−5 510i5= 1 −5i−10 + 10i+ 5 −i=−4+4i

a b c a ̸= 0 az2+bz +c= 0 ∆ = b2−4ac
∆>0
z1=−b−√∆
2az2=−b+√∆ 2a
∆=0
z0=−b
2a
∆<0
z1=−b−ip|∆|
2az2=z1
a b c a ̸= 0 z1z2az2+bz +c= 0
z1=z2z∈C
az2+bz +c=a(z−z1)(z−z2)
z2−2z+5 = 0 ∆ = −16 z1= 1−2i z2=
1+2i z2−2z+5 = (z−(1−2i))(z−(1+2i))
n a0. . . anan̸= 0
n
P(z) =
n
X
k=0
akxk=a0+a1x+a2x2+. . . +anxn
P(z) = 0 n
z∈Cα∈Cn
zn−αn= (z−a)
n−1
X
k=0
zn−1−kαk= (z−α)(zn−1+zn−2α+. . . +zαn−2+αn−1)
zn−αn

zn−1 = (z−1)(zn−1+zn−2+. . . + 1)
α∈C∗n∈N∗
(z−α)
n−1
X
k=0
zn−1−kαk=
n−1
X
k=0
zn−kαk
| {z }
(1)
−
n−1
X
k=0
zn−1−kαk+1
| {z }
(2)
=zn+ zn−1α+ . . . + zαn−1
| {z }
(1)
− zn−1α− . . . − zαn−1−αn
| {z }
(2)
=zn−αn
n∈N∗P(z)=0 n
α∈CP Q n −1
z
P(z) = (z−α)Q(z)
P z −α
z−α
P(z) = Pn
k=0 akzkn+ 1 a0a1. . . an
an̸= 0 α∈CP(α) = 0
z∈C
P(z) = P(z)−P(α)
=
n
X
k=0
akzk−
n
X
k=0
akαk
=
n
X
k=0
ak(zk−αk)
= (z−α)
n
X
k=0
ak k−1
X
l=0
zk−1−lαl!
| {z }
Q(z)
= (z−α)Q(z)
an̸= 0 Q n −1
P(z) = z3−1z∈C
z3−1=(z−1)(z2+z+ 1)
P(z) = z2−4z−2z∈C
P(z)=(z−2)(z+ 2)

n∈NP n n
n∈NPnn n
nPnn∈N∗
P0P0
n∈NPnP n + 1
P0< n + 1
P α Q n
z∈CP(z) = (z−α)Q(z)Q n
P n + 1
Pn+1
nPn
P(z) = z3−2z2+z−2P(2) = 0
P P z −2z∈CP(z)=(z−2)(z2+ 1)
2i−i
P(z) = z4−4z2−z+ 2
P(2) = 0
P(2) = 0 P z −2
Q Q(z) = az3+bz2+cz +d z P (z)=(z−2)Q(z)
P(z) = (z−2)(az3+bz2+cz +d)
P(z) = az4+bz3+cz2+dz −2az −2bz2−2cz −2d
=az4+ (b−2a)z3+ (c−2b)z2+ (d−2c)z−2d
a= 1 d=−1
b−2=0
c−2b=−4
−1−2c=−1
b= 2 c= 0
P(z)=(z−2)(z3+ 2z2−1)
P(z) = (z−2)(z3+ 2z2−1)