− 9 | 10 | |||
są odpowiednio równe | i | . | ||
16 | 81 |
1 | 9 | |||
W( | ) = − | |||
2 | 16 |
1 | 10 | |||
W( | ) = | |||
3 | 81 |
193 | ||
p = | ||
62 |
37 | ||
q = | ||
62 |
193 | 37 | |||
w(x) = x4 − | x2 −4x + | |||
62 | 62 |
3 | 655598 | |||
jak obliczyłam w( | ) to wynik jest | |||
2 | 3906 |
1 | ||
p = | ||
2 |
3 | ||
q = | ||
2 |
1 | 4p | 32 | 16q | 9 | ||||||
W(1/2) = | − | − | + | = − | ||||||
16 | 16 | 16 | 16 | 16 |
1 | 9p | 108 | 81q | 10 | ||||||
W(1/3) = | − | − | + | = | ||||||
81 | 81 | 81 | 81 | 81 |
⎧ | −2p + 8q − 11 = 0 | |
⎩ | −p + 9q − 13 = 0 |
⎧ | −2p + 8q − 11 = 0 | |
⎩ | 2p − 18q + 26 = 0 |
3 | 27 − 26 | 1 | ||||
−10q + 15 = 0 −−−> q = | −−−> p = 9q − 13 = | = | ||||
2 | 2 | 2 |
1 | 3 | |||
W(x) = x4 − | x2 − 4x + | |||
2 | 2 |
3 | 9 | |||
W( | ) = − | |||
2 | 16 |
1 | 3 | 7 | 52 | |||||
W(x) : (−2x + 3) = ( − | x3 − | − | x − | ) + R | ||||
2 | 4 | 8 | 16 |
9 | ||
R = − | ||
16 |
3 | 9 | |||
Wystarczy tyle W( | ) = − | = R (na mocy tw. o reszcie) nie musisz wykonywać tego | ||
2 | 16 |