Find a polynomial that passes through the points below (0,-12)(1,0)(3,5)(4,12) using Lagrange's interpolating polynomial method ?

To find the Lagrange interpolating polynomial that passes through the given points, we can use the formula:

P(x) = L0(x)y0 + L1(x)y1 + L2(x)y2 + L3(x)y3

where Li(x) are the Lagrange basis polynomials and yi are the corresponding y-values of the points.

First, let's calculate the Lagrange basis polynomials:

L0(x) = (x-1)(x-3)(x-4) / (0-1)(0-3)(0-4) = x^3 - 7x^2 + 12x
L1(x) = (x-0)(x-3)(x-4) / (1-0)(1-3)(1-4) = -2x^3 + 15x^2 - 18x
L2(x) = (x-0)(x-1)(x-4) / (3-0)(3-1)(3-4) = 3x^3 - 16x^2 + 21x
L3(x) = (x-0)(x-1)(x-3) / (4-0)(4-1)(4-3) = -x^3 + 7x^2 - 12x

Now, substitute the Lagrange basis polynomials and y-values into the formula:

P(x) = (-12)(x^3 - 7x^2 + 12x) + (0)(-2x^3 + 15x^2 - 18x) + (5)(3x^3 - 16x^2 + 21x) + (12)(-x^3 + 7x^2 - 12x)

Simplifying the expression, we get:

P(x) = -12x^3 + 84x^2 - 144x + 0 + 0 + 15x^3 - 80x^2 + 105x - 0 - 12x^3 + 84x^2 - 144x

P(x) = 15x^3 - 80x^2 + 105x

Therefore, the Lagrange interpolating polynomial that passes through the given points is:

P(x) = 15x^3 - 80x^2 + 105x