Apresenta os polinómios na forma reduzida
Monómios e polinómios: Matematicamente Falando 8 - Pág. 147 Ex. 11
Enunciado
Apresenta cada um dos polinómios seguintes na forma reduzida.
| a) | \(15x – {\left( {x + 7} \right)^2}\) |
| b) | \(x\left( {x – 1} \right) – {\left( {x – 2} \right)^2}\) |
| c) | \(\left( {x + 2} \right)\left( {x – 3} \right) + {\left( {x + 1} \right)^2}\) |
| d) | \({\left( {x + \frac{1}{2}} \right)^2} – {\left( {x – \frac{1}{2}} \right)^2} – \frac{3}{4}\left( {x – 1} \right)\left( {x + 1} \right)\) |
| e) | \(2x\left( {{x^2} + 3x – \frac{1}{2}} \right)\) |
| f) | \(\left( {n – 2} \right)\left( {n + 3} \right)\) |
| g) | \(\left( {1 – y – {y^2}} \right)\left( {y + 2} \right)\) |
| h) | \(\left( {{t^2} – t + 1} \right)\left( {{t^3} – 2t + 5} \right)\) |
Resolução
Apresenta-se abaixo cada um dos polinómios na forma reduzida.
| a) | \[\begin{array}{*{20}{l}}{15x – {{\left( {x + 7} \right)}^2}}& = &{15x – \left( {{x^2} + 14x + 49} \right)}\\{}& = &{ – {x^2} + x – 49}\end{array}\] |
| b) | \[\begin{array}{*{20}{l}}{x\left( {x – 1} \right) – {{\left( {x – 2} \right)}^2}}& = &{{x^2} – x – \left( {{x^2} – 4x + 4} \right)}\\{}& = &{3x – 4}\end{array}\] |
| c) | \[\begin{array}{*{20}{l}}{\left( {x + 2} \right)\left( {x – 3} \right) + {{\left( {x + 1} \right)}^2}}& = &{{x^2} – 3x + 2x – 6 + \left( {{x^2} + 2x + 1} \right)}\\{}& = &{2{x^2} + x – 5}\end{array}\] |
| d) | \[\begin{array}{*{20}{l}}{{{\left( {x + \frac{1}{2}} \right)}^2} – {{\left( {x – \frac{1}{2}} \right)}^2} – \frac{3}{4}\left( {x – 1} \right)\left( {x + 1} \right)}& = &{\left( {x + \frac{1}{2} + x – \frac{1}{2}} \right)\left( {x + \frac{1}{2} – x + \frac{1}{2}} \right) – \frac{3}{4}\left( {{x^2} – 1} \right)}\\{}& = &{2x \times 1 – \frac{3}{4}{x^2} + \frac{3}{4}}\\{}& = &{ – \frac{3}{4}{x^2} + 2x + \frac{3}{4}}\end{array}\] |
| e) | \[\begin{array}{*{20}{l}}{2x\left( {{x^2} + 3x – \frac{1}{2}} \right)}& = &{2{x^3} + 6{x^2} – x}\end{array}\] |
| f) | \[\begin{array}{*{20}{l}}{\left( {n – 2} \right)\left( {n + 3} \right)}& = &{{n^2} + 3n – 2n – 6}\\{}& = &{{n^2} + n – 6}\end{array}\] |
| g) | \[\begin{array}{*{20}{l}}{\left( {1 – y – {y^2}} \right)\left( {y + 2} \right)}& = &{y + 2 – {y^2} – 2y – {y^3} – 2{y^2}}\\{}& = &{ – {y^3} – 3{y^2} – y + 2}\end{array}\] |
| h) | \[\begin{array}{*{20}{l}}{\left( {{t^2} – t + 1} \right)\left( {{t^3} – 2t + 5} \right)}& = &{{t^5} – 2{t^3} + 5{t^2} – {t^4} + 2{t^2} – 5t + {t^3} – 2t + 5}\\{}& = &{{t^5} – {t^4} – {t^3} + 7{t^2} – 7t + 5}\end{array}\] |
![Um quadrado [ABCD]](https://www.acasinhadamatematica.pt/wp-content/uploads/2018/04/9V2Pag92-1a-720x340.png)
