There are infinitely many cubics that have those zeros, which is why your question reads "a possible equation" instead of "the equation".

`f(x)=ax(x+3)(x+2)`

is a solution for any *nonzero* constant .

You should remember that you may find the solutions to the equation if you know the x intercepts, hence, since the problem provides three x intercepts, thus, the cubic equation has three real solutions `x_1=-3, x_2=-2, x_3=0` .

You should remember that you may write the factored form of the equation if you know the solutions to the equation such that:

`f(x)=a(x-x_1)(x-x_2)(x-x_3)`

`f(x) = a(x-(-3))(x-(-2))(x-0)`

`f(x) = a(x+3)(x+2)x`

`x^3 + 5x^2 + 6x`

You may find the leading coefficient a using Vieta's relations for the equation `ax^3 + bx^2 + cx + d = 0` , such that:

`x_1+x_2+x_3 = -b/a `

`-3-2+0=-b/a => b/a=5 => a` =1

`x_1*x_2+x_1*x_3+x_2*x_3 = c/a`

`(-3)(-2) + (-3)*0 + (-2)*0 = c/a => c/a = 6 => a=1`

`x_1*x_2*x_3 = -d/a => d/a = 0 => d=0`

`x^3 + (-b/a)x^2 + (c/a)x - d/a = 0`

`x^3 + 5x^2 + 6x = 0`

**Hence, evaluating the cubic equation under the given conditions yields `x^3 + 5x^2 + 6x = 0` .**

If I have understood correctly cubic curve passes through x axis on points -3, 0 and -2. In that case the only possible equation is

`y = x(x+3)(x+2)` because `y=0` for `x = 0,-3,-2.` There is no other cubic curve that would satisfy those three conditions.

If you have `n` nul-points `x_0, ldots, x_(n-1)` then you can determine formula of any `n`-th degree polynomial in that way.

`P(x) = (x-x_0) cdots (x-x_(n-1))`

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