Rozłóż wielomian na czynniki...- Zadanie 4.13: Prosto do matury 2. Zakres podstawowy

Rozwiązanie

$a) Q (x) = x^{3} - x^{2} + x - 1 = x^{2} (x - 1) + (x - 1) = (x - 1) (x^{2} + 1)$

$b) Q (x) = x^{3} - 5 x^{2} - 4 x + 20 = x^{2} (x - 5) - 4 (x - 5) = (x - 5) (x^{2} - 4) = (x - 5) (x - 2) (x + 2)$

$c) Q (x) = 8 x^{3} - 20 x^{2} - 18 x + 45 = 4 x^{2} (2 x - 5) - 9 (2 x - 5) = (2 x - 5) (4 x^{2} - 9) = (2 x - 5) (2 x - 3) (2 x + 3)$

$d) Q (x) = 11 \frac{1}{4} x^{3} - 6 \frac{3}{4} x^{2} - 20 x + 12 = \frac{45}{4} x^{3} - \frac{27}{4} x^{2} - 20 x + 12 = \frac{1}{4} (45 x^{3} - 27 x^{2} - 80 x + 48) = \frac{1}{4} (9 x^{2} (5 x - 3) - 16 (5 x - 3)) = \frac{1}{4} (5 x - 3) (9 x^{2} - 16) = \frac{1}{4} (5 x - 3) (3 x - 4) (3 x + 4)$

$e) Q (x) = 14 x^{3} + 2 x^{2} - 21 x - 3 = 2 x^{2} (7 x + 1) - 3 (7 x + 1) = (7 x + 1) (2 x^{2} - 3) = 2 (7 x + 1) (x^{2} - \frac{3}{2}) = 2 (7 x + 1) (x - \frac{3}{2}) (x + \frac{3}{2}) = 2 (7 x + 1) (x - \frac{6}{2}) (x + \frac{6}{2}) = \frac{1}{2} (7 x + 1) \cdot 2 (x - \frac{6}{2}) \cdot 2 (x + \frac{6}{2}) = \frac{1}{2} (7 x + 1) (2 x - 6) (2 x + 6)$

$f) Q (x) = 6 x^{3} + 8 x^{2} - 18 x - 62 = x^{2} (6 x + 22) - 6 (3 x + 2) = 2 x^{2} (3 x + 2) - 6 (3 x + 2) = (3 x + 2) (2 x^{2} - 6) = (3 x + 2) \cdot 2 (x^{2} - 3) = (3 x + 2) \cdot 2 (x - 3) (x + 3)$

$g) Q (x) = 3 x^{3} - x^{2} + 48 x - 4 = x^{2} (3 - 1) + 43 x - 4 = x^{2} (3 - 1) + 4 (3 - 1) = (3 - 1) (x^{2} + 4)$

$h) Q (x) = \frac{1}{4} x^{4} - \frac{1}{8} x^{3} - 3 x^{2} + 1 \frac{1}{2} x = \frac{1}{4} x^{3} (x - \frac{1}{2}) - 3 x (x - \frac{1}{2}) = (x - \frac{1}{2}) (\frac{1}{4} x^{3} - 3 x) = (x - \frac{1}{2}) \cdot x \cdot (\frac{1}{4} x^{2} - 3) = (x - \frac{1}{2}) \cdot x \cdot (\frac{1}{2} x - 3) (\frac{1}{2} x + 3)$

$i) Q (x) = x^{5} + 2 x^{4} + x^{3} - x^{2} - 2 x - 1 = x^{3} (x^{2} + 2 x + 1) - (x^{2} + 2 x + 1) = (x^{2} + 2 x + 1) (x^{3} - 1) = (x + 1)^{2} (x^{3} - 1) = (x + 1)^{2} (x - 1) (x^{2} + x + 1)$

$j) Q (x) = 8 x^{5} - 32 x^{4} + 32 x^{3} + x^{2} - 4 x + 4 = 8 x^{3} (x^{2} - 4 x + 4) + (x^{2} - 4 x + 4) = (x^{2} - 4 x + 4) (8 x^{3} + 1) = (x - 2)^{2} (8 x^{3} + 1) = (x - 2)^{2} ((2 x)^{3} + 1^{3}) = (x - 2)^{2} (2 x + 1) (4 x^{2} - 2 x + 1)$

$k) Q (x) = 9 x^{5} + 12 x^{4} + 4 x^{3} - 9 x^{2} - 12 x - 4 = x^{3} (9 x^{2} + 12 x + 4) - (9 x^{2} + 12 x + 4) = x^{3} (3 x + 2)^{2} - (3 x + 2)^{2} = (3 x + 2)^{2} + (x^{3} - 1)$

$l) Q (x) = x^{6} - 4 x^{5} + 4 x^{4} + 4 x^{2} - 16 x + 16 = x^{4} (x^{2} - 4 x + 4) + 4 (x^{2} - 4 x + 4) = x^{4} (x - 2)^{2} + 4 (x - 2)^{2} = (x - 2)^{2} \cdot (x^{4} + 4)$

Aleksandra Filipowska

Nauczycielka matematyki

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