Question

2. Resolva as adições dos polinômios abaixo:
a) (a² + 6a) + ( - a² - a) =
b) (3x² + 2x) + (2x + 1) =

Answer 1

Retiramos os parênteses, organizamos os monômios e resolvemos:

Item A)

$(a^{2} \ + \ 6a) \ + \ ( \ - \ a^{2} \ - \ a) \\\\ a^{2} \ + \ 6a \ - \ a^{2} \ - \ a \\\\ a^{2} \ - \ a^{2} \ + \ 6a \ - \ a \\\\ 6a \ - \ a \\\\ \boxed{ \ 5a \ }$  

Item B)

$(3x^{2} \ + \ 2x \ ) \ + \ (2x \ + \ 1) \\\\ 3x^{2} \ + \ 2x \ + \ 2x \ + \ 1 \\\\ 3x^{2} \ + \ 4x \ + \ 1 \ = \ 0 \\\\ \\\\ \Delta \ \ = \ \ b^{2} \ - \ 4ac \\\\ \Delta \ \ = \ \ 4^{2} \ - \ 4 \ . \ 3 \ . \ 1 \\\\ \Delta \ \ = \ \ 16 \ - \ 12 \\\\ \Delta \ \ = \ \ 4$

$x \ \ = \ \ \frac{-b \ \ \± \ \ \sqrt{\Delta}}{2a} \\\\ x \ \ = \ \ \frac{-4 \ \ \± \ \ \sqrt{4}}{2 \ . \ 3} \\\\ x \ \ = \ \ \frac{-4 \ \ \± \ \ 2}{6}$

•   $x' \ \ = \ \ \frac{-4 \ \ + \ \ 2}{6} \ \ \ \ \Rightarrow \ \ \ \ \ x' \ \ = \ \ \frac{2^{:2}}{6^{:2}} \ \ \ \ \Rightarrow \ \ \ \ \boxed{ \ x \ = \ \frac{1}{3}}$
•  $x'' \ \ = \ \ \frac{-4 \ - \ 2}{6} \ \ \ \ \Rightarrow \ \ \ \ \ x'' \ \ = \ \ -\frac{6}{6} \ \ \ \ \Rightarrow \ \ \ \ \boxed{ \ x'' \ = \ -1 \ }$