Question

Se da #polinomul p(x)=x¹⁰⁰+x⁵⁰+3x¹⁷+3x³+2x²-1. Determinati restul impartirii acestui polinom prin x²(x²-1) .

Answer 1

$\displaystyle Sa~notam~catul~cu~Q~si~restul~cu~R.\\ \\ P(x)=x^2(x^2-1)Q(x)+R(x) \\ \\ \deg R < \deg \big(x^2(x^2-1) \big)=4,~deci~\deg R \le 3.\\ \\ Rezulta~R(x)=ax^3+bx^2+cx+d,~unde~a,b,c,d \in \mathbb{R}. \\ \\ (Atentie:~a,b,c,d~pot~sa~fie~nule;~gradul~lui~R~nu~e~neaparat~3.) \\ \\ Avem~P(-1)=R(-1),~P(0)=R(0)~si~P(1)=R(1),~deci~trei \\ \\ conditii.~Noua~ne~trebuie~patru,~pentru~ca~avem~patru~coeficienti \\ \\ de~aflat.$

$\displaystyle Cea~de-a~patra~conditie~nu~apare~direct~din~cauza~ca~x^2(x^2-1) \\ \\ are~o~radacina~dubla.~No~problem!~Observam~ca~\\ \\ P(x)-R(x)=x^2(x^2-1)Q(x)~;~adica~P(x)-R(x)~are~radacina \\ \\ cel~putin~dubla~0.~Rezulta~(P-R)'(0)=0.~Adica~P'(0)-R'(0)=0. \\ \\ De~aici~e~simplu.$

$\displaystyle P'(x)=100x^{99}+50x^{49}+51x^{16}+9x^2+4x \\ \\ R'(x)=3ax^2+2bx+c \\ \\ \begin{cases} R(-1)=P(-1) \\ R(0)=P(0) \\ R(1)=P(1) \\ R'(0)=P'(0) \end{cases} \Leftrightarrow ~~~ \begin{cases} -a+b-c+d=-3 \\ d=-1 \\ a+b+c+d=9 \\ c=0 \end{cases}. \\ \\ \\ Deci \left \{ {{-a+b=-2} \atop {a+b=10}} \right. .~Rezulta~a=6~si~b=4. \\ \\ R(x)=6x^3+4x^2-1.$