Polynomial is passes through 4 collinear points.

So least degree polynomial be of degree 3.

Let P(x) = Ax³ + Bx² + Cx +D be required polynomial.

∵ P(x) passes through (0, -1).

∴ -1 = D        ........(1)

∵ P(x) passes through (1, -1).

∴ 1 = A + B + C + D

⇒ A + B + C = 2          ........(2)     (From (1))

∵ P(x) passes through (2, 1)

∴ 1 = 8A + 4B + 2C + D

⇒ 8A + 4B + 2C = 2     ........(3)     (From (1))

By Applying equation (3) - 2 x equation (2), we obtain

(8A + 4B + 2C) - (2A + 2B + 2C) = 2 - 4

⇒ 6A + 2B = -2

⇒ 3A + B = -1        .......(4)

∵ P(x) passes through (3, -2)

∴ -2 = 27A + 9B + 3C + D

⇒ 27A + 9B + 3C = -1      ........(5)     (From (1))

By applying equation (5) - \(\frac32\) x equation (3), we obtain 

(27A + 9B + 3C) - (12A + 6B + 3C) = -1 - 3

⇒ 15A + 3B = -4        ........(6)

By applying equation (6) - 3 x equation (4), we obtain 

(15A + 3B) - (9A + 3B) = -4 - (-3)

⇒ 6A = -1

⇒ A = \(\frac{-1}6\)

From equation (4), we get

\(\frac{-3}{6} + B = -1\)

⇒ \(B = -1 + \frac12 = \frac{-1}{2}\)

Put A = \(\frac{-1}6\) & B = \(\frac{-1}2\) in equation (2), we get

\(-\frac{1}{6} -\frac{1}{2} + C = 2\)

⇒ \(C = 2 + \frac16+ \frac12 = \frac{12 + 1 + 3}{6} = \frac{16}{6} = \frac83\)

∴ \(P(x) = \frac{-x^3}{6}-\frac{x^2}{2} + \frac{8}{3}x - 1\)

is the least degree polynomial passing through the points (0, -1) , (1, 1) , (2, 1) & (3, -2).