Question

Sendo P(x) = x³ + 2x² e Q(x) = X^4 – x³ + 2x – 1, calcule: a) [P(x)]² b) P(x) . Q(x)

 

 

Ajudem por favor.

Answer 1

$<var>\text{a) }[P(x)]^2 =(x^3+2x^2)^2=(x^3)^2+2.x^3.2x^2+(2x^2)^2= </var>$

$<var>=x^{3.2}+4x^{3+2}+4x^{2.2}=x^6+4x^5+4x^4=x^4(x^2+4x+4)=</var>$

=$<var>=x^4(x+2)^2</var>$

 

$<var>\text{b) }P(x).Q(x) = (x^3+2x^2)(x^4-x^3+2x-1)=</var>$

$<var>=x^7-x^6+2x^4-x^3+2x^6-2x^5+4x^3-2x^2=</var>$

$<var>=x^7+x^6-2x^5+2x^4+3x^3-2x^2</var>$

Answer 2

$\text{P}(\text{x})=\text{x}^3+2\text{x}^2$

 

 

a)

 

$[\text{P}(\text{x})]^2=(\text{x}^3+2\text{x}^2)^2=(\text{x}^3)^2+2\cdot\text{x}^3\cdot2\text{x}^2+(2\text{x}^2)^2=\text{x}^6+4\text{x}^5+4\text{x}^4$

 

$[\text{P}(\text{x})]^2=\text{x}^4\cdot(\text{x}^2+4\text{x}+4)=\text{x}^4\cdot(\text{x}+2)^2$

 

 

 

$\text{Q}(\text{x})=\text{x}^4-\text{x}^3+2\text{x}-1$

 

 

b)

 

$\text{P}(\text{x})\cdot\text{Q}(\text{x})=(\text{x}^3+2\text{x}^2)\cdot(\text{x}^4-\text{x}^3+2\text{x}-1)$

 

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

 

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

 

$\text{P}(\text{x})\cdot\text{Q}(\text{x})=\text{x}^2\cdot(\text{x}^5+\text{x}^4-2\text{x}^3+2\text{x}^5+3\text{x}-2)$