Rozłóż...- Zadanie 34: Matematyka z plusem 2. Zakres podstawowy i rozszerzony

Rozwiązanie

$a) 10 x^{6} + 30 x^{5} + 5 x^{4} + 15 x^{3} = 10 x^{5} (x + 3) + 5 x^{3} (x + 3) =$

$= (x + 3) (10 x^{5} + 5 x^{3}) = (x + 3) 5 x^{3} (2 x^{2} + 1) = 5 x^{3} (x + 3) (2 x^{2} + 1)$

$b) 40 x^{7} - 100 x^{6} + 8 x^{5} - 20 x^{4} = 20 x^{6} (2 x - 5) + 4 x^{4} (2 x - 5) =$

$= (20 x^{6} + 4 x^{4}) (2 x - 5) = 4 x^{4} (5 x^{2} + 1) (2 x - 5)$

$c) 15 x^{8} + 5 x^{7} + 105 x^{6} + 35 x^{5} = 5 x^{7} (3 x + 1) + 35 x^{5} (3 x + 1) =$

$= (3 x + 1) (5 x^{7} + 35 x^{5}) = (3 x + 1) 5 x^{5} (x^{2} + 7) = 5 x^{5} (3 x + 1) (x^{2} + 7)$

$d) 18 x^{5} - 8 x^{3} = 2 x^{3} (9 x^{2} - 4) =$

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

$e) 8 x^{6} - 18 x^{4} = 2 x^{4} (4 x^{2} - 9) =$

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

$f) \frac{5 x ^{7}}{4} - 5 x^{5} = 5 x^{5} (\frac{1}{4} x^{2} - 1) =$

$= 5 x^{5} ((\frac{1}{2} x)^{2} - 1^{2}) = 5 x^{5} (\frac{1}{2} x - 1) (\frac{1}{2} x + 1)$

$g) 3 x^{5} - 15 x^{4} + 6 x^{3} - 30 x^{2} = 3 x^{4} (x - 5) + 6 x^{2} (x - 5) =$

$= (x - 5) (3 x^{4} + 6 x^{2}) = (x - 5) 3 x^{2} (x^{2} + 2) = 3 x^{2} (x - 5) (x^{2} + 2)$

$h) 8 x^{7} - 72 x^{5} - 4 x^{6} + 36 x^{4} = 8 x^{5} (x^{2} - 9) - 4 x^{4} (x^{2} - 9) =$

$= (x^{2} - 9) (8 x^{5} - 4 x^{4}) = (x^{2} - 3^{2}) 4 x^{4} (2 x - 1) = 4 x^{4} (x + 3) (x - 3) (2 x - 1)$

$i) 3 x^{8} - 12 x^{6} - 6 x^{7} + 24 x^{5} = 3 x^{6} (x^{2} - 4) - 6 x^{5} (x^{2} - 4) =$

$= (x^{2} - 4) (3 x^{6} - 6 x^{5}) = (x^{2} - 2^{2}) 3 x^{5} (x - 2) = 3 x^{5} (x - 2) (x + 2) (x - 2) =$

$= 3 x^{5} (x - 2)^{2} (x + 2)$

$j) - 6 x^{5} + 15 x^{4} + 24 x^{3} - 60 x^{2} = - 3 x^{4} (2 x - 5) + 12 x^{2} (2 x - 5) =$

$= (2 x - 5) (- 3 x^{4} + 12 x^{2}) = (2 x - 5) 3 x^{2} (- x^{2} + 4) =$

$= 3 x^{2} (2 x - 5) (2^{2} - x^{2}) = 3 x^{2} (2 x - 5) (2 - x) (2 + x)$

$k) 10 x^{5} + 15 x^{4} - 2 x^{3} - 3 x^{2} = 5 x^{4} (2 x + 3) - x^{2} (2 x + 3) =$

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

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

$l) 12 x^{10} + 24 x^{9} - 3 x^{8} - 6 x^{7} = 12 x^{9} (x + 2) - 3 x^{7} (x + 2) =$

$= (x + 2) (12 x^{9} - 3 x^{7}) = (x + 2) 3 x^{7} (4 x^{2} - 1) =$

$= 3 x^{7} (x + 2) ((2 x)^{2} - 1^{2}) = 3 x^{7} (x + 2) (2 x - 1) (2 x + 1)$

Paulina Adamska

Nauczycielka matematyki

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