Question

Find a polynomial of the form f(x) = ax^3 + bx^2 + cx + d such that f(0) = −6, f(1) = 4, f(3) = 10, and f(4) = 0.

Answer 1

`f(x) = -(x^(3))/(2) - (x^(2))/(3) + (65x)/(6) - 6`

Explanation:

We use a system of equations to solve this question.

f(0) = −6

This means that when
`x = 0, f(x) = -6`. So

`f(x) = ax^(3) + bx^(2) + cx + d`

`-6 = a*0^(3) + b*0^(2) + c*0 + d`

`d = -6`

So

`f(x) = ax^(3) + bx^(2) + cx - 6`

f(1) = 4

This means that when
`x = 1, f(x) = 4`

So

`f(x) = ax^(3) + bx^(2) + cx - 6`

`4 = a*1^(3) + b*1^(2) + c*1 - 6`

`a + b + c = 10`

f(3) = 10

This means that when
`x = 3, f(x) = 10`

So

`f(x) = ax^(3) + bx^(2) + cx - 6`

`10 = a*3^(3) + b*3^(2) + c*3 - 6`

`27a + 9b + 3c = 16`

f(4) = 0

This means that when
`x = 4, f(x) = 0`

So

`f(x) = ax^(3) + bx^(2) + cx - 6`

`0 = a*4^(3) + b*4^(2) + c*4 - 6`

`64a + 16b + 4c = 6`

So we have to solve the following system of equations:

a + b + c = 10

27a + 9b + 3c = 16

64a + 16b + 4c = 6

From the first equation:

a = 10 - b - c

Replacing on the second, and on the third:

`27(10 - b - c) + 9b + 3c = 16`

`270 - 27b - 27c + 9b + 3c = 16`

`18b + 24c = 254`

And

`64(10 - b - c) + 16b + 4c = 6`

`640 - 64b - 64c + 16b + 4c = 6`

`48b + 60c = 634`

So

18b + 24c = 254

48b + 60c = 634

Multiplying the first one by 5, and the second one by -2, to eliminate c

90b + 120c = 1270

-96b - 120c = 1268

`-6b = 2`

`6b = -2`

`b = -(1)/(3)`

Then

`18b + 24c = 254`

`24c = 254 - 18b`

`24c = 254 - 18(-1)/(3)`

`24c = 260`

`c = (65)/(6)`

And

`a = 10 - b - c = 10 -(-(1)/(3)) - (65)/(6) = -(1)/(2)`

So

`f(x) = -(x^(3))/(2) - (x^(2))/(3) + (65x)/(6) - 6`