Which polynomial function has x intercepts -1,0, and 2 and passes through the point (1,-6)?

Answer:f(x) = [tex]3x^3 - 3x^2 - 6x[/tex]Step-by-step explanation:Which polynomial function has x intercepts -1,0, and 2 and passes through the point (1,-6)?There are 3 distinct and real roots given in the question, which means that the  function must be a third degree polynomial. The roots are -1, 0, and 2. This means that f(x) = 0 at these points. The general form of the cubic equation is given by:f(x) = ax^3 + bx^2 + cx + d; where a, b, c, and d are arbitrary constants.From the given data:f(-1)=0 implies a*(-1)^3 + b*(-1)^2 + c(-1) + d = -a + b - c + d = 0. (Equation 1).f(0)=0 implies a*(0)^3 + b*(0)^2 + c(0) + d = 0a + 0b + 0c + d = 0. (Equation 2).f(2)=0 implies a*(2)^3 + b*(2)^2 + c(2) + d = 8a + 4b + 2c + d = 0. (Equation 3).f(1)=0 implies a*(1)^3 + b*(1)^2 + c(1) + d = a + b + c + d = -6. (Equation 4).Equation 2 shows that d = 0. So rest of the equations become:-a + b - c = 0; 8a + 4b + 2c = 0;  (Divide 2 on both sides of the equation to simplify).a + b + c = -6This system of equation can be solved using the Gaussian Elimination Method. Converting the system into the augmented matrix form:• 1 1 1 | -6  • -1 1 -1 | 0• 4 2 1 | 0Adding row 1 into row 3:• 1 1 1 | -6  • 0 2 0 | -6• 4 2 1 | 0Dividing row 2 with 2 and multiplying row 1 with -4 and add it into row 3:• 1 1        1 | -6• 0 1       0 | -3• 0 -2 -3 | 24Multiplying row 2 with 2 and add it into row 3:• 1 1       1 | -6• 0 1       0 | -3• 0 0 -3 | 18It can be seen that when this updated augmented matrix is converted into a system, it comes out to be:• a + b + c = -6• b  = -3• -3c = 18 (This implies that c = -6.)Put c = -6 and b = -3 in equation 1:• a + (-3) + (-6) = -6• a = -6 + 3 + 6• a = 3.So f(x) = [tex]3x^3 - 3x^2 - 6x[/tex] (All conditions are being satisfied)!!!