Question 1155338
Find the function ​f(x)=ax^3+bx^2+cx+d for which, f(-3)=-112, f(-1)=-2, f(1)=4, f(2)=13.

I know the answer is f(x) 3x^3-4x^2+5

But, i would like to know how to solve this
<pre>{{{matrix(1,3, f(x), "=", Ax^3 + Bx^2 + Cx + D)}}}
Given: f(- 3) = - 112, or (- 3, - 112)
Substituting (- 3, - 112) in {{{matrix(1,3, f(x), "=", Ax^3 + Bx^2 + Cx + D)}}}, we get: {{{matrix(2,3, f(- 3), "=", A(- 3)^3 + B(- 3)^2 + C(- 3) + D, 
- 112, "=", - 27A + 9B - 3C + D)}}} ---- eq (i)

{{{matrix(1,3, f(x), "=", Ax^3 + Bx^2 + Cx + D)}}} 
Given: f(- 1) = - 2, or (- 1, - 2)
Substituting (- 1, - 2) in {{{matrix(1,3, f(x), "=", Ax^3 + Bx^2 + Cx + D)}}}, we get: {{{matrix(2,3, f(- 1), "=", A(- 1)^3 + B(- 1)^2 + C(- 1) + D, 
- 2, "=", - A + B - C + D)}}} ------ eq (ii)

{{{matrix(1,3, f(x), "=", Ax^3 + Bx^2 + Cx + D)}}} 
Given: f(1) = 4, or (1, 4)
Substituting (1, 4) in {{{matrix(1,3, f(x), "=", Ax^3 + Bx^2 + Cx + D)}}}, we get: {{{matrix(2,3, f(1), "=", A(1)^3 + B(1)^2 + C(1) + D, 
4, "=", A + B + C + D)}}} ----------------- eq (iii)

{{{matrix(1,3, f(x), "=", Ax^3 + Bx^2 + Cx + D)}}}
Given:  f(2) = 13, or (2, 13)
Substituting (2, 13) in {{{matrix(1,3, f(x), "=", Ax^3 + Bx^2 + Cx + D)}}}, we get: {{{matrix(2,3, f(2), "=", A(2)^3 + B(2)^2 + C(2) + D, 
13, "=", 8A + 4B + 2C + D)}}} ---------------- eq (iv)

We now have: {{{system(matrix(1,6, - 27A + 9B - 3C + D, "=", - 112, "--", "eq", "(i)"),
matrix(1,6, - A + B - C + D, "=", - 2, "--------", "eq", "(ii)"),
matrix(1,6, A + B + C + D, "=", 4, "-----------", "eq", "(iii)"),
matrix(1,6, 8A + 4B + 2C + D, "=", 13, "-------", "eq", "(iv)"))}}} 

35A - 5B + 5C = 125____7A -  B + C = 25 ---- Subtracting eq (i) from eq (iv) -------- eq (v)
                       7A + 3B + C = 9 ----- Subtracting eq (iii) from eq (iv) ------ eq (vi)
                            4B = - 16 ------ Subtracting eq (v) from eq (vi)
                            {{{matrix(1,5, B, "=", (- 16)/4, "=", highlight(- 4))}}} 
 
        2B + 2D = 2____2(B + D) = 2(1) -- Adding eq (ii) & (iii) 
                         B + D = 1 ------ eq (vii)
                       - 4 + D = 1 ------ Substituting - 4 for B in eq (vii)
                            {{{matrix(1,5, D, "=",  1 + 4, "=", highlight(5))}}}

A + B + C + D = 4
A - 4 + C + 5 = 4 -------- Substituting - 4 for B, and 5 for D in eq (iii)
                                      A + C = 3 ------ eq (viii)

8A + 4B + 2C + D = 13 ------- eq (iv)
8A + 4(- 4) + 2C + 5 = 13 --- Substituting - 4 for B, and 5 for D in eq (iv)
8A - 16 + 2C + 5 = 13
8A + 2C = 24____2(4A + C) = 2(12)____4A + C = 12 ----- eq (ix)
                                     3A = 9 ------ Subtracting eq (viii) from eq (ix)
                                     {{{matrix(1,5, A, "=", 9/3, "=", highlight(3))}}}

                                  A + C = 3 ----- eq (viii)
                                  3 + C = 3 ----- Substituting 3 for A in eq (viii)
                                     {{{matrix(1,5, C, "=", 3 - 3, "=", highlight(0))}}}

With {{{system(matrix(4,3, A, being, 3, B, being, - 4, C, being, 0, D, being, 5))}}}, {{{matrix(1,8, f(x), "=", Ax^3 + Bx^2 + Cx + D, becomes, f(x), "=", "=", 3x^3 - 4x^2 + 5))}}}</pre>