Question

Dany jest wielomian w(x)=axdo5+bxdo2+c

Znajdz wartości współczynników a,b oraz c jesli wiadomo ze W(pierwiastek z 2)=4 W(-pierwiastek z 2)=-12 i W(0)=-10

Answer 1

4=a*(√2)⁵+b*(√2)²+c

-12=a*(-√2)⁵+b*(-√2)²+c

-10=a*0⁵+b*0²+c

 

4=4√2a+2b+c

-12=-4√2a+2b+c

-10=c

 

4=4√2a+2b-10

-12=-4√a+2b-10

 

4√2a+2b=14

-4√2a+2b=-2

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4b=12

b=3

 

4√2a+2*3=14

4√2a=8

a=2/√2

a=2√2/2

a=√2

 

Answer 2

W(x) = a x^5 + b x^2 + c

W( p(2)) = 4  i  W( -p(2)) = -12  i W( 0) = - 10

Mamy

W( p(2)) = a* ( p(2))^5 + b*( p(2)) ^2 + c = 4a p(2) + 2b + c

W( -p(2)) = a*( -p(2))^5 + b*( -p(2))^2 + c = - 4a p(2) + 2b + c

W( 0) = a*0 + b*0 + c = c

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c = - 10

4ap(2) + 2b - 10 = 4

-4 a p(2) + 2b - 10 = -12

-------------------------------- II   i  III dodajemy stronami

4b - 20 =  - 8

4b = 20 - 8 = 12

b = 3

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4 a p(2) = 4 + 10 - 2b = 14 - 2*3 = 14 - 6 = 8  / : 4 p(2)

a = p(2)

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Odp. a = p(2), b = 3, c = - 10

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