liczby -1, 1 oraz 3 są pierwiastkami wielomianu w(x) stopnia trzeciego i w (5)=-96. Zatem wspolczynn.... Question from @szejkii

November 2018 1 14 Report

liczby -1, 1 oraz 3 są pierwiastkami wielomianu w(x) stopnia trzeciego i w (5)=-96. Zatem wspolczynnik przy x^3 ma wartosc:

a. 2 b. -2 c. 0,5 d. -0.5

Roma

$W(x) = ax^3 + bx^2 + cx + d$

W(-1) = 0

$W(-1) = a \cdot (-1)^3 + b \cdot (-1)^2 + c \cdot (-1) + d = -a+ b - c + d \\\ -a+ b - c + d = 0$

W(1) = 0

$W(1) = a \cdot 1^3 + b \cdot 1^2 + c \cdot 1 + d = a+ b + c + d \\\ a+ b + c + d = 0$

W(3) = 0

$W(1) = a \cdot 3^3 + b \cdot 3^2 + c \cdot 3 + d = 27a+ 9b + 3c + d \\\ 27a+ 9b + 3c + d = 0$

W(5) = - 96

$W(5) = a \cdot 5^3 + b \cdot 5^2 + c \cdot 5 + d = 125a+ 25b + 5c + d \\\ 125a+ 25b + 5c + d = -96$

$\begin{cases} -a+ b - c + d = 0\\a+ b + c + d = 0\\27a+ 9b + 3c + d = 0\\125a+ 25b + 5c + d = -96 \end{cases}$

Dodając stronami 1 i 2 równanie otrzymujemy:

$\begin{cases} 2b + 2d = 0 \\a+ b + c + d = 0\\27a+ 9b + 3c + d = 0\\125a+ 25b + 5c + d = -96 \end{cases} \\\\ \begin{cases} 2d = -2b /:2 \\a+ b + c + d = 0\\27a+ 9b + 3c + d = 0\\125a+ 25b + 5c + d = -96 \end{cases} \\\\ \begin{cases} d = -b \\a+ b + c -b = 0\\27a+ 9b + 3c -b = 0\\125a+ 25b + 5c -b = -96 \end{cases} \\\\ \begin{cases} d = -b \\a+c = 0\\27a+ 8b + 3c = 0\\125a+ 24b + 5c = -96 \end{cases} \\\\ \begin{cases} d = -b \\ c = -a\\27a+ 8b - 3a = 0\\125a+ 24b - 5a = -96 \end{cases}$

$\begin{cases} d = -b \\ c = -a\\24a+ 8b = 0 \ /: (-8) \\120a+ 24b = -96 \ /:24 \end{cases} \\\\ \begin{cases} d = -b \\ c = -a\\-3a- b = 0 \\ 5a+ b = -4 \end{cases} \\\\ \begin{cases} d = -b \\ c = -a\\b = -3a \\ 5a -3a = -4 \end{cases} \\\\ \begin{cases} d = -b \\ c = -a\\b = -3a \\ 2a = -4 \ /:2 \end{cases} \\\\ \begin{cases} d = -b \\ c = -a\\b = -3a \\ a = - 2 \end{cases} \\\\ \begin{cases} a = -2 \\ b = 6 \\ c = 2 \\ d = -6 \\ \end{cases}$