Factorizacion De Polinomios De Cuarto Grado

Factorización de polinomios de grado 4

1x4+2x3+3x2+3x+3
x4+x3a+b+x24+ab+xb+3a+3
a+b=2 |
4+ab=5 |
B+3a=3 |
a=2-b |
b=3-3a |
b=3-6+3b |
-2b=-3 |
b=3/2 |
a=2-3/2 |
a=1/2 |4+1/2(3/2)=5 |
4+3/4=5 |
16+3/4=5 |
14/4=5 |



x4+2x3+5x2+3x+3
Si x=z-2/4=Remplazando
(Z-0.5)2(Z-0.5)2 |
(Z2-2Z(0.5)+0.52)(Z2-2(Z)(0.5)+0.52) |
Z2-1Z+0.25(Z2-Z+0.25) |(Z-Z3+0.25Z2-Z3+Z2-0.25=0.25Z-0.25Z+0.0675) |
(Z4-2Z3+1.5Z2-0.5Z+0.0625) |
(Z-0.5)4=

(Z-0.5)2=Z2-Z+0.2
Remplazando en el polinomio
x4+2x3+5x23x+3
z4-2z3+1.5z2-0.5z+0.0625+2z3-3z2+1.5z-0.255z2-5z+1.253+3z-1.5+3=0
z4+3.5z2-1z+2.5625=0
z4+3.5Z2-Z+2.5625=0
Z2+αZ+δZ+αZ+β=0
Z2+αZ+δZ2+αZ+βZ4-Z3α+Z2δZ3α-Z2α2+ZαδZ2β-Zαβ+δβZ4+Z2δ-α2-β+Zαδ-δβ+8β

δ-α2+β=3,5
αδ-αβ=-1
δβ=2,5626δ=3,5+α2-β
Remplazando
αδ-β=-1 |
α3,5+α2-β-β=-1 |
α3,5+δ2-2β=-1 |
3,5α+α3-2αβ=-1 |
β=X3+3,5α |
(3,5+α+α2-β)β=2,5625 |3,5+α3-α3+3,5α+12αα3+3,5α+12α=4
7α+α3-α3+3,5α-12αα3+3,5α+12α=2,56
α3+3,5α-12αα3+3,5α+12α=2,5625
α6+3,5α4+α3+3,5α4+12,25α2+3,5α-α3-3,5α-1=4α2(K,IOKI) |
α6+7α4+12,25α2-1=10,25α2 |
α6+7αα4+2α2-1=0 |
SiendoA=α2Remplazamos |

A3+7A2+2A-1 |
Encontramos A=Z |
P(0)=-1 |
P(1)=9 |
P(0,25)=-00,046875 |
P(0,30)=0,257 |

0,25+0,302=P0,275=0,100 |
0,275+2,52=P0,25625=-0,025 |0,2625+0,252=P-0,259375=7,12*10-3 |
0,254375+0,256252=P0,2578125=-1,967*10-3 |
0,259375+0,25781252=P0,25859375=-1,96 |
0,25854375+0,2593752=P0,2589843754.84*10-3 |
|
α=0,508905 |
δ=3,5+α2-β |δ=3,5+0,258984-2,86194 |
δ=0,8970 |

A=0,258484375 |
B=α3+3,5α+12α |
B=0,13174846+1,7811677111,017810149 |
B=2,86194 |
(Z3-αZ+δ)(Z2+αZ+β) |(Z2-6,5089Z+0,8970)(Z2+0,508905Z+2,86194) |

Siendo x=z-0,5 X*Z=X+0,5
X27X+0,25-0,508905X-0,2544+0,8470(X2+X+0,25+0,508405X0,2544+2,8919)
X2+0,49095X+0,8426(X2+1,508905X+3,36634)
Comprobando el polinomio...