Decompõe em fatores os polinómios
Monómios e polinómios: Matematicamente Falando 8 - Pág. 139 Ex. 2
Enunciado
Decompõe em fatores os polinómios seguintes.
| a) | \({x^2} – 6x + 9\) |
| b) | \(4{x^2} + 4x + 1\) |
| c) | \({a^2} + 2ab + {b^2}\) |
| d) | \({y^2} – 25\) |
| e) | \(4{a^2} – 1\) |
| f) | \(8{x^3}y – 2x{y^3}\) |
| g) | \(2{x^2} + 12x + 18\) |
| h) | \(3{a^2}x + 6ax + 3x\) |
| i) | \({x^3} – x\) |
| j) | \(2a – 16{a^2}\) |
| k) | \(x\left( {x – 1} \right) + 2\left( {x – 1} \right)\) |
| l) | \(4\left( {2b + 1} \right) – {b^2}\left( {2b + 1} \right)\) |
| m) | \({a^2}\left( {a – 2} \right) – 2a\left( {a – 2} \right) + \left( {a – 2} \right)\) |
Resolução
Os polinómios seguintes estão decompostos em fatores.
| a) | \({x^2} – 6x + 9 = {\left( {x – 3} \right)^2} = \left( {x – 3} \right)\left( {x – 3} \right)\) |
| b) | \(4{x^2} + 4x + 1 = {\left( {2x + 1} \right)^2} = \left( {2x + 1} \right)\left( {2x + 1} \right)\) |
| c) | \({a^2} + 2ab + {b^2} = {\left( {a + b} \right)^2} = \left( {a + b} \right)\left( {a + b} \right)\) |
| d) | \({y^2} – 25 = \left( {y + 5} \right)\left( {y – 5} \right)\) |
| e) | \(4{a^2} – 1 = \left( {2a + 1} \right)\left( {2a – 1} \right)\) |
| f) | \(8{x^3}y – 2x{y^3} = 2xy\left( {4{x^2} – {y^2}} \right) = 2xy\left( {2x + y} \right)\left( {2x – y} \right)\) |
| g) | \(2{x^2} + 12x + 18 = 2\left( {{x^2} + 6x + 9} \right) = 2{\left( {x + 3} \right)^2} = 2\left( {x + 3} \right)\left( {x + 3} \right)\) |
| h) | \(3{a^2}x + 6ax + 3x = 3x\left( {{a^2} + 2a + 1} \right) = 3x{\left( {a + 1} \right)^2} = 3x\left( {a + 1} \right)\left( {a + 1} \right)\) |
| i) | \({x^3} – x = x\left( {{x^2} – 1} \right) = x\left( {x + 1} \right)\left( {x – 1} \right)\) |
| j) | \(2a – 16{a^2} = 2a\left( {1 – 8a} \right)\) |
| k) | \(x\left( {x – 1} \right) + 2\left( {x – 1} \right) = \left( {x – 1} \right)\left( {x + 2} \right)\) |
| l) | \(4\left( {2b + 1} \right) – {b^2}\left( {2b + 1} \right) = \left( {2b + 1} \right)\left( {4 – {b^2}} \right) = \left( {2b + 1} \right)\left( {2 + b} \right)\left( {2 – b} \right)\) |
| m) | \({a^2}\left( {a – 2} \right) – 2a\left( {a – 2} \right) + \left( {a – 2} \right) = \left( {a – 2} \right)\left( {{a^2} – 2a + 1} \right) = \left( {a – 2} \right){\left( {a – 1} \right)^2} = \left( {a – 2} \right)\left( {a – 1} \right)\left( {a – 1} \right)\) |
