The curve `y=a x^3+b x^2+c x+5` touches the x-axis at `P(-2,0)` and cuts the y-axis at the point `Q` where its gradient is 3. Find the equation of the curve completely.

Explanation

To find the equation of the curve \( y = ax^3 + bx^2 + cx + 5 \) that touches the x-axis at \( P(-2, 0) \) and cuts the y-axis at point \( Q \) where its gradient is 3, we will follow these steps:

### Step 1: Use the point where the curve touches the x-axis
Since the curve touches the x-axis at \( P(-2, 0) \), we know that this point must satisfy the equation of the curve. Thus, we substitute \( x = -2 \) and \( y = 0 \):

\[ 0 = a(-2)^3 + b(-2)^2 + c(-2) + 5 \]
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