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MAXImum [283]
3 years ago
5

Please help......

Mathematics
2 answers:
Leviafan [203]3 years ago
6 0

Answer:

2 0 4 14 30

Step-by-step explanation:

liubo4ka [24]3 years ago
5 0

(1)

f(x) = ax^3 + bx^2 + cx + d  

In table attached, x and f(x) values are shown

a = 3rd difference/3! =  12/6 = 2

f(0) = d = -4

f(2) =  2(2)^3 + b(2)^2 + c(2) - 4 = -2

12 + 4 b + 2 c = -2    

4 b + 2 c = -14 (eq. 1)

f(4) = 2(4)^3 + b(4)^2 + c(4) - 4 = -4

      128 + 16 b + 4 c - 4 = -4

16 b + 4 c = -128 (eq. 2)

Multiplying eq. 1 by 2 and then subtracting it to eq. 2 gives

8 b = -100

b = -12.5

Replacing in eq. 1

4(-12.5) + 2 c = -14

2 c = 36

c = 18

The polynomial is 2x^3 - 12.5x^2 + 18x - 4  

(2)

My polynomial: x^3 + x^2 + x + 1

In table attached, x and f(x) values are shown

a = 3rd difference/3! =  6/6 = 1

f(0) = d = 1

f(1) =  (1)^3 + b(1)^2 + c(1) + 1 = 4

1 + b + c + 1 = 4

b + c = 2 (eq. 1)  

f(2) =  (2)^3 + b(2)^2 + c(2) + 1 = 15  

8 + 4 b + 2 c + 1 = 15

4 b + 2 c = 6 (eq. 2)

Multiplying eq. 1 by 2 and then subtracting it to eq. 2 gives

2 b = 2

b = 1

Replacing in eq. 1

(1) + c = 2

c = 1

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