Give an equation for a polynomial function P with zeros 2, -3/2, and-6.?

Answer 1

There are zeros at 2, -3/2 and -6, so

f(x) = (x-2)*(x+3/2)*(x+6)

f(x) = x^3 + 11/2*x^2 - 6*x -18

Answer 2

The polynomial equation is zero at 2,-3/2, -6. Hence they become the roots of the equation P
suppose there is an equation in terms of x such that the values of x=2,-3/2,-6 satisfy the equation.
Hence x=2, x=-3/2, x=-6
x-2=0, 2x+3=0, x+6=0
Multiplying the left hand side terms=multiplying the Right terms
(x-2)(2x+3)(x+6)=0
2x^3+11x^2-12x-36=0 is the required equation(P) {^ means "power of"}

Answer 3

(x - 2)(2x + 3)(x + 6)
(x^2 + 6x - 2x - 12)(2x + 3)
(x^2 + 4x - 12)(2x + 3)
2x^3 + 3x^2 + 8x^2 + 12x - 24x - 36
2x^3 + 11x^2 - 12x - 36

Answer 4

You can get the answer by expanding the function
(x - p1) * (x - p2) * (x - p3)
= (x^2 - (p1 + p2)x +p1p2) * (x - p3)
= (x^3 - (p1 + p2)x^2 + p1p2x - p3x^2 + (p1 + p2)p3x - p1p2p3
= x^3 - (p1 + p2 + p3)x^2 + (p1p2 + (p1+p2)p3)x - p1p2p3

if you plug in
p1 = 2
p2 = -1.5
p3 = -6
you get
x^3 + 5.5x^2 - 6x - 18