Rozłóż sumę algebraiczną na czynniki.- Zadanie 7: Matematyka poznać, zrozumieć 3. Zakres podstawowy. Po gimnazjum

Rozwiązanie

$a) 16 x^{2} - 8 x y + y^{2} = (4 x - y)^{2}$

$b) - x^{2} - 23 x - 3 = - (x^{2} + 23 x + 3) = - (x + 3)^{2}$

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

$d) x^{4} y^{4} - 81 = (x^{2} y^{2} - 9) (x^{2} y^{2} + 9) = (x y - 3) (x y + 3) (x^{2} y^{2} + 9)$

$e) (x + y)^{2} - 16 y^{2} = (x + y - 4 y) (x + y + 4 y) = (x - 3 y) (x + 5 y)$

$f) 1 - (a + b)^{2} = (1 - (a + b)) (1 + (a + b)) = (1 - a - b) (1 + a + b)$

$g) 9 (a + b)^{2} - (a - b)^{2} = (3 (a + b) - (a - b)) (3 (a + b) + (a - b)) =$

$= (3 a + 3 b - a + b) (3 a + 3 b + a - b) = (2 a + 4 b) (4 a + 2 b)$

$h) 16 - m^{2} + 2 m n - n^{2} = 16 - (m^{2} - 2 m n + n^{2}) = 4^{2} - (m - n)^{2} =$

$= (4 - (m - n)) (4 + (m - n)) = (4 - m + n) (4 + m - n)$

$i) 2 x (a - b) + 3 y (b - a) = (a - b) \cdot (2 x - 3 y)$

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

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

$= (x + 2 y) \cdot (5 - 4 x + 2 x) = (x + 2 y) (5 - 2 x)$

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

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

$m) 2 b - 4 c - 5 a b + 10 a c = 2 (b - 2 c) - 5 a (b - 2 c) = (b - 2 c) (2 - 5 a)$

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

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

$o) 7 x (x - 1) + 4 x^{2} (x - 1) = (x - 1) \cdot (7 x + 4 x^{2}) = (x - 1) \cdot x (7 + 4 x)$

$p) (4 x^{2} + 4 x + 1) - (25 x^{2} - 10 x + 1) = (2 x + 1)^{2} - (5 x - 1)^{2} =$

$= (2 x + 1 - (5 x - 1)) (2 x + 1 + (5 x - 1)) = (2 x + 1 - 5 x + 1) (2 x + 1 + 5 x - 1) = (- 3 x + 2) \cdot 7 x$

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

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

Magdalena Matusik

Nauczycielka matematyki

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