Question

2(p+1)^2-3(1-p)^2= Alguèm poderia m ajudar nessa questão??

Answer 1

$\large\begin{array}{l} \mathtt{2(p+1)^2-3(1-p)^2}\\\\ =\mathtt{2\cdot (p+1)(p+1)-3\cdot (1-p)(1-p)}\\\\\\ \texttt{Aplicando a distributiva nas multiplica\c{c}\~oes,}\\\\ =\mathtt{2\cdot [(p+1)\cdot p+(p+1)\cdot 1]-3\cdot [(1-p)\cdot 1-(1-p)\cdot p]}\\\\ =\mathtt{2\cdot [p^2+p+p+1]-3\cdot [1-p-p+p^2]}\\\\ =\mathtt{2\cdot [p^2+2p+1]-3\cdot [1-2p+p^2]} \end{array}$

$\large\begin{array}{l} =\mathtt{2p^2+4p+2-3+6p-3p^2}\\\\ =\mathtt{2p^2-3p^2+4p+6p+2-3}\\\\ =\boxed{\begin{array}{c}\mathtt{-p^2+10p-1} \end{array}}\quad\quad\checkmark \end{array}$


$\large\begin{array}{l} \texttt{D\'uvidas? Comente.}\\\\ \texttt{Bons estudos!} \end{array}$