CALCULUS HELP!?

Since f(-3) = f(x) = f(6) = 0, then
(x + 3), x and (x - 6) are factors of f(x).

f(x) is cubic.
Let f(x) = a x (x + 3) (x - 6)

f(5) = 16 :
a × (5) × (5 + 3) × (5 - 6) = 160
-40a = 160
a = -4

Then, f(x) = 4 x (x + 3) (x - 6)
or f(x) = -4x³ + 12x² + 72x

Let f(x) = ax³ + bx² + cx + d

f(0) = 0
a(0)³ + b(0)² + c(0) + d
d = 0

Then, f(x) = ax³ + bx² + cx

f(5) = 160
a(5)³ + b(5)² + c(5) = 160
125a + 25b + 5c = 160
25a + 5b + c = 32 ...... [1]

f(-3) = 0
a(-3)³ + b(-3)² + c(-3) = 0
-27a + 9b - 3c = 0
9a - 3b + c = 0 ...... [2]

f(6) = 0
a(6)³ + b(6)² + c(6) = 0
216a + 36b + 6c = 0
36a + 6b + c = 0 ...... [3]

[3] - [2] :
27a + 9b = 0
3a + b = 0 ...... [4]

[1] - [2] :
16a + 8b = 32
2a + b = 4 ..... [5]

[4] - [5] :
a = -4

Put a = 4 into [4] :
3(-4) + b = 0
b = 12

Put a = -4 and b = 12 into [2] :
9(-4) - 3(12) + c = 0
c = 72

Hence, f(x) = -4x³ + 12x² + 72x

Find cubic f(x) such that f(-3) = f(0) = f(6) = 0 and f(5) = 160. Now since
x = -3,0 & 6 are zeros of f(x), we have g(x)=x(x+3)(x-6) as a factor.of f(x)
We have g(5) = 5*8*(- 1) = - 40. If we put f(x) = kg(x), then f(5) = kg(5) =
- 40k = 160 and k = - 4 so f(x) = - 4g(x) = - 4x(x+3)(x-6).

We are given three zeros for our f(x) function, we are trying to find:

x = -3, so x + 3 = 0 ; x = 0, and x = 6, so x - 6 = 0
Multiply them all together:

x[(x + 3)(x - 6)] =

x[x(x) + 3(x) - 6(x) + 3(-6)] = 0

x[x^2 - 3x - 18] = 0
Multiply in the x to each term:

x^3 - 3x^2 - 18x = 0
Now, if I input x = 5, I don't get 160, so multiply this by a constant to make it happen, as 0 = f(x), so it's our f(x) function, and we found the zeros, but we are not looking for the x-intercepts when y for f(x) = 0.

[x^3 - 3x^2 - 18x](a) = f(x)
Let f(x) = 160, and x = 5

[5^3 - 3((5)^2)) - 18(5)](a) = 160

-40(a) = 160
Divide both sides by -40:

a = -4

f(x) = -4(x^3 - 3x^2 - 18x)
Multiply in the -4 to make it look nice:

f(x) = -4(x)^3 + 12(x)^2 + 72(x)

*Up there! It's done*

f(x) = ax(x+3)(x-6)
160=-5a*8
a = -4
= -4x[x^2-3x-18]
= -4x^3+12x^2+72x

f = A ( x + 3) (x) ( x - 6), where 160 = A[8][5][-1]