Answer :

To find a cubic equation with [tex]\( x = -9 \)[/tex] as a solution, we can use the fact that if [tex]\( x = -9 \)[/tex] is a root of the equation, then [tex]\( (x + 9) \)[/tex] must be a factor of the cubic polynomial.

Let's start by representing the cubic equation in its factored form. If [tex]\( x = -9 \)[/tex] is a root, the polynomial can be expressed as:
[tex]\[ (x + 9) \cdot (Ax^2 + Bx + C) = 0 \][/tex]

To find the coefficients [tex]\( A \)[/tex], [tex]\( B \)[/tex], and [tex]\( C \)[/tex], we can expand the equation:

[tex]\[ (x + 9) \cdot (Ax^2 + Bx + C) = Ax^3 + Bx^2 + Cx + 9Ax^2 + 9Bx + 9C \][/tex]

Combine like terms:

[tex]\[ Ax^3 + (B + 9A)x^2 + (C + 9B)x + 9C = 0 \][/tex]

For simplicity, let’s assume [tex]\( A = 1 \)[/tex] and [tex]\( B = 0 \)[/tex]. This simplifies our polynomial to:

[tex]\[ x^3 + 9x^2 + Cx + 9C = 0 \][/tex]

Given [tex]\( B = 0 \)[/tex] and from the result:

[tex]\[ B + 9A = 9 + 0 \rightarrow 9 \][/tex]
[tex]\[ C = 0 \][/tex]
[tex]\[ 9C = 9(0) \rightarrow 0 \][/tex]

So, C must also be 0.

Therefore, the cubic equation simplifies further to:
[tex]\[ x^3 + 9x^2 = 0 \][/tex]

This is the cubic equation for which [tex]\( x = -9 \)[/tex] is one of its solutions.