Rozłóż wielomian na czynniki.- Zadanie 6: Matematyka z plusem 2. Zakres rozszerzony

Rozwiązanie

$a) x^{2} - 16 = x^{2} - 4^{2} = (x - 4) (x + 4)$

$b) 4 x^{2} - 5 = (2 x)^{2} - (5)^{2} = (2 x - 5) (2 x + 5)$

$c) 49 x^{4} - \frac{1}{81} = (7 x^{2})^{2} - (\frac{1}{9})^{2} = (7 x^{2} + \frac{1}{9}) (7 x^{2} - \frac{1}{9}) = (7 x^{2} + \frac{1}{9}) ((7 x)^{2} - (\frac{1}{3})^{2}) = (7 x^{2} + \frac{1}{9}) (7 x + \frac{1}{3}) (7 x - \frac{1}{3})$

$d) x^{7} - 100 x^{5} = x^{2} \cdot x^{5} - 100 \cdot x^{5} = (x^{2} - 100) \cdot x^{5} = x^{5} (x^{2} - 10^{2}) = x^{5} (x + 10) (x - 10)$

$e) x^{2} - 6 x + 9 = x^{2} - 2 \cdot x \cdot 3 + 3^{2} = (x - 3)^{2}$

$f) \frac{1}{9} x^{2} + \frac{1}{3} x + \frac{1}{4} = (\frac{1}{3} x)^{2} + 2 \cdot \frac{1}{3} x \cdot \frac{1}{2} + (\frac{1}{2})^{2} = (\frac{1}{3} x + \frac{1}{2})^{2}$

$g) x^{4} - 2 x^{2} + 1 = (x^{2})^{2} - 2 \cdot x^{2} \cdot 1 + 1^{2} = (x^{2} - 1)^{2} = (x^{2} - 1^{2})^{2} = ((x + 1) (x - 1))^{2} = (x + 1)^{2} (x - 1)^{2}$

$h) (x + 3)^{2} + 2 (x + 3) + 1 = ((x + 3) + 1)^{2} = (x + 3 + 1)^{2} = (x + 4)^{2}$

$i) x^{3} - 27 = (x - 3) (x^{2} + x \cdot 3 + 3^{2}) = (x - 3) (x^{2} + 3 x + 9)$

$j) x^{3} + 1 = (x + 1) (x^{2} - x \cdot 1 + 1^{2}) = (x + 1) (x^{2} - x + 1)$

$k) \frac{1}{27} x^{3} - 8 = (\frac{1}{3} x)^{3} - 2^{3} = (\frac{1}{3} x - 2) ((\frac{1}{3} x)^{2} + \frac{1}{3} x \cdot 2 + 2^{2}) = (\frac{1}{3} x - 2) (\frac{1}{9} x^{2} + \frac{2}{3} x + 4)$

$l) 64 x^{10} + x^{7} = 64 x^{3} \cdot x^{7} + x^{7} = x^{7} (64 x^{3} + 1) = x^{7} ((4 x)^{3} + 1^{3}) = x^{7} (4 x + 1) ((4 x)^{2} - 4 x \cdot 1 + 1^{2}) = x^{7} (4 x + 1) (16 x^{2} - 4 x + 1)$

Dagmara Kowalczuk

Nauczycielka matematyki

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