Rozłóż wielomian w na czynniki i wyznacz...- Zadanie 1: MATeMAtyka 2. Zakres podstawowy i rozszerzony

Rozwiązanie

$a) w (x) = 0, 01 x^{5} - x = 0, 01 x (x^{4} - 100) = 0, 01 x (x^{2} - 10) (x^{2} + 10) = 0, 01 x (x - 10) (x + 10) (x^{2} + 10)$

Obliczamy pierwiastki wielomianu:

$w (x) = 0$

$0, 01 x (x - 10) (x + 10) (x^{2} + 10) = 0$

$0, 01 x = 0 ∣ : 0, 01 lub x - 10 = 0 lub x + 10 = 0 lub x^{2} + 10 = 0$

$x = 0 lub x = 10 lub x = - 10 lub r. sprzeczne x^{2} = - 10 < 0$

$x = 0 lub x = 10 lub x = - 10$

$b) w (x) = x^{3} - 6 x^{2} + 9 x = x (x^{2} - 6 x + 9) = x (x - 3)^{2}$

Obliczamy pierwiastki wielomianu:

$w (x) = 0$

$x (x - 3)^{2} = 0$

$x = 0 lub (x - 3)^{2} = 0$

$x = 0 lub x - 3 = 0$

$x = 0 lub x = 3$

$c) w (x) = x^{4} + 4 x^{3} - 5 x^{2} = x^{2} (x^{2} + 4 x - 5) = x^{2} (x^{2} - x + 5 x - 5) = x^{2} [x (x - 1) + 5 (x - 1)] = x^{2} (x - 1) (x + 5)$

Obliczamy pierwiastki wielomianu:

$w (x) = 0$

$x^{2} (x - 1) (x + 5) = 0$

$x^{2} = 0 lub x - 1 = 0 lub x + 5 = 0$

$x = 0 lub x = 1 lub x = - 5$

$d) w (x) = (x^{2} + 5 x - 3)^{2} - 9 = (x^{2} + 5 x - 3)^{2} - 3^{2} = (x^{2} + 5 x - 3 + 3) (x^{2} + 5 x - 3 - 3) = (x^{2} + 5 x) (x^{2} + 5 x - 6) = x (x + 5) (x^{2} - x + 6 x - 6) = x (x + 5) [x (x - 1) - 6 (x - 1)] = x (x + 5) (x - 1) (x + 6)$

Obliczamy pierwiastki wielomianu:

$w (x) = 0$

$x (x + 5) (x - 1) (x + 6) = 0$

$x = 0 lub x + 5 = 0 lub x - 1 = 0 lub x + 6 = 0$

$x = 0 lub x = - 5 lub x = 1 lub x = - 6$

$e) w (x) = (2 x^{2} - 1)^{2} - (1 - 6 x^{2})^{2} = [(2 x^{2} - 1) - (1 - 6 x^{2})] [(2 x^{2} - 1) + (1 - 6 x^{2})] = (2 x^{2} - 1 - 1 + 6 x^{2}) (2 x^{2} - 1 + 1 - 6 x^{2}) = (8 x^{2} - 2) \cdot (- 4 x^{2}) = 8 (x^{2} - \frac{1}{4}) \cdot (- 4 x^{2}) = - 32 x^{2} (x^{2} - \frac{1}{4}) = - 32 x^{2} (x - \frac{1}{2}) (x + \frac{1}{2})$

Obliczamy pierwiastki wielomianu:

$w (x) = 0$

$- 32 x^{2} (x - \frac{1}{2}) (x + \frac{1}{2}) = 0$

$- 32 x^{2} = 0 ∣ : (- 32) lub x - \frac{1}{2} = 0 lub x + \frac{1}{2} = 0$

$x^{2} = 0 lub x = \frac{1}{2} lub x = - \frac{1}{2}$

$x = 0 lub x = \frac{1}{2} lub x = - \frac{1}{2}$

$f) w (x) = 25 x^{6} - 10 x^{4} + x^{2} = x^{2} (25 x^{4} - 10 x^{2} + 1) = x^{2} [(5 x^{2})^{2} - 2 \cdot 5 x^{2} \cdot 1 + 1^{2}] = x^{2} (5 x^{2} - 1)^{2} = x [(5 x - 1) (5 x + 1)]^{2} = x (5 x - 1)^{2} (5 x + 1)^{2}$