Wyznacz W(x)+P(x), W(x)-P(x) - Zadanie 1: Matematyka 2. Zakres podstawowy. Po gimnazjum

Rozwiązanie

$a)$

$W (x) + P (x) = 3 x^{5} - 4 x^{3} + 5 + (- 7) =$

$= 3 x^{5} - 4 x^{3} - 2$

$W (x) - P (x) = 3 x^{5} - 4 x^{3} + 5 - (- 7) =$

$= 3 x^{5} - 4 x^{3} + 5 + 7 =$

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

$W (x) \cdot P (x) = (3 x^{5} - 4 x^{3} + 5) \cdot (- 7) =$

$= - 21 x^{5} + 28 x^{3} - 35$

$b)$

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

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

$= - x^{4} - x^{3} - 4 x$

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

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

$= - x^{4} - 3 x^{3} + 4 x$

$W (x) \cdot P (x) = (- x^{4} - 2 x^{3}) \cdot (x^{3} - 4 x) =$

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

$= - x^{12} + 4 x^{5} - 2 x^{6} + 8 x^{4} =$

$= - x^{12} - 2 x^{6} + 4 x^{5} + 8 x^{4}$

$c)$

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

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

$= x^{3} - 3 x$

$W (x) - P (x) = (2 x^{2} - 3 x) - (x^{3} - 2 x^{2}) =$

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

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

$W (x) \cdot P (x) = (2 x^{2} - 3 x) \cdot (x^{3} - 2 x^{2}) =$

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

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

$= 2 x^{5} - 7 x^{4} + 6 x^{3}$

$d)$

$W (x) + P (x) = (- \frac{1}{2} x^{3} + 4 x^{2} + \frac{1}{4}) + (8 x^{2} - 6) =$

$= - \frac{1}{2} x^{3} + 4 x^{2} + \frac{1}{4} + 8 x^{2} - 6 =$

$= - \frac{1}{2} x^{3} + 12 x^{2} - 5 \frac{3}{4}$

$W (x) - P (x) = (- \frac{1}{2} x^{3} + 4 x^{2} + \frac{1}{4}) - (8 x^{2} - 6) =$

$= - \frac{1}{2} x^{3} + 4 x^{2} + \frac{1}{4} - 8 x^{2} + 6 =$

$= - \frac{1}{2} x^{3} - 4 x^{2} + 6 \frac{1}{4}$

$W (x) \cdot P (x) = (- \frac{1}{2} x^{3} + 4 x^{2} + \frac{1}{4}) \cdot (8 x^{2} - 6) =$

$= (- \frac{1}{2} x^{3} + 4 x^{2} + \frac{1}{4}) \cdot 8 x^{2} + (- \frac{1}{2} x^{3} + 4 x^{2} + \frac{1}{4}) \cdot (- 6) =$

$= - 4 x^{5} + 32 x^{4} + 2 x^{2} + 3 x^{3} - 24 x^{2} - \frac{6}{4} =$

$= - 4 x^{5} + 32 x^{4} + 3 x^{3} - 22 x^{2} - 1, 5$

Agnieszka Nowak

Nauczycielka matematyki

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