Question

Fie polinoamele P(X)=-X4-8X3+X2-3X si Q(X)=X3+5X2.
1)Efectuati:
P(X)+Q(X)=S(X); P(X)-Q(X)=D(X); Q(X)-P(X)=R(X).

Answer 1

Răspuns:

$S(X)= -X^4-7X^3+6X^2-3X$

$D(X)=X^4-9X^3 -4X^2-3X$

$R(X)=X^4+ 9X^3+4X^2+3X$

Explicație pas cu pas:

Fie polinoamele :

P(X) = -X⁴-8X³+X²-3X

Q(X) = X³+5X²

1) Efectuati:

• P(X)+Q(X)=S(X);
• P(X)-Q(X)=D(X);
• Q(X)-P(X)=R(X)

$P(X)=-X^{4}-8X^{3}+X^{2}-3X \\ Q(X)=X^{3}+5X^{2} \\ \\ P(X)+Q(X)=S(X)\\\\(-\ X^{4}}-\underline{8X^{3}}+\bold{X^{2}}-3X) + (\underline{X^{3}}+\bold{5X^{2}}) \\ \\S(X)= -X^4-7X^3+6X^2-3X$

----------------------------

$P(X)-Q(X)=D(X) \\\\ (X^{4}-8X^{3}+X^{2}-3X) - (X^{3}+5X^{2}) = \\ \\X^{4}-\bold{8X^{3}}+\underline{X^{2}}-3X - \bold{X^3}-\underline{5X^2} = \\ \\ D(X)=X^4-9X^3 -4X^2-3X$

----------------------------

$Q(X)-P(X)=R(X) \\\\ (X^3+5X^2) - (-X^4-8X^3+X^2-3X) \\ \\ \bold{X^3}+\underline{5X^2} + X^4+\bold{8X^3}-\underline{X^2}+3X = \\ \\ R(X)=X^4+ 9X^3+4X^2+3X$