Question

Alguém sabe fazer isso por favor

Answer 1

Olá Ingrid,

para a multiplicação, aplique a distributiva, para a subtração, troque o sinal da última parcela:

$P(x)+Q(x)=( x^{3}+3x^2-x)+( x^{2} -x+2)\\ P(x)+Q(x)=x^3+3x^2+x^2-x-x+2\\\\ \boxed{P(x)+Q(x)=x^3+4x^2-2x+2}$

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$P(x)+R(x)-Q(x)=(x^3+3x^2-x)+(x-5)-(x^2-x+2)\\ P(x)+R(x)-Q(x)=x^3+3x^2-x+x-5-(x^2-x+2)\\ P(x)+R(x)-Q(x)=x^3+3x^2-5-x^2+x-2\\ P(x)+R(x)-Q(x)=x^3+3x^2-x^2+x-5-2\\\\ \boxed{P(x)+R(x)-Q(x)=x^3+2x^2+x-7}$

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$P(x)*Q(x)=(x^3+3x^2-x)*(x^2-x+2)\\ P(x)*Q(x)=x^5-x^4+2x^3+3x^4-3x^3+6x^2-x^3+x^2-2x\\ P(x)*Q(x)=x^5-x^4+3x^4+2x^3-3x^3-x^3+6x^2+x^2-2x\\\\ \boxed{P(x)*Q(x)=x^5+2x^4-2x^3+7x^2-2x}$

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$P(x)*R(x)=(x^3+3x^2-x)*(x-5)\\ P(x)*R(x)=x^4-5x^3+3x^3-15x^2-x^2+5x\\\\ \boxed{P(x)*R(x)=x^4-2x^3-16x^2+5x}$

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$Q(x)*R(x)=(x^2-x+2)*(x-5)\\ Q(x)*R(x)=x^3-5x^2-x^2+5x+2x-10\\\\ \boxed{Q(x)*R(x)=x^3-6x^2+7x-10}$

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$R(x)+Q(x)-P(x)=(x-5)+(x^2-x+2)-(x^3+3x^2-x)\\ R(x)+Q(x)-P(x)=x^2-3-(x^3+3x^2-x)\\ R(x)+Q(x)-P(x)=x^2-3x^2+x-3\\\\ \boxed{R(x)+Q(x)-P(x)=-2x^2+x-3}$

Bons estudos =))