The curve $y=a{x}^3+b{x}^2+cx+5$ touches the x-axis at $P(-2, 0)$ and cuts the y-axis at the point Q where its gradient is 3, then the value of $|2a+4b+c|$ is 

We have,

$$\begin{gathered} y=a{x}^3+b{x}^2+cx+5 \\ \Rightarrow \frac{dy}{dx}=3a{x}^2+2bx+c \end{gathered}$$

Since the curve $y=a{x}^3+b{x}^2+cx+5$ touches the $x-$ axis at  $P(-2,0)$. This means that the curve passes through $P(-2,0)$ and  $x-$axis is the tangent at $P(-2,0)$
$$\begin{gathered} \therefore 0=8a+4b-2c+5 \\ \Rightarrow 8a-4b+2c=5 \\ And {\left(\frac{dy}{dx}\right)}_P=0 \end{gathered}$$

$$\begin{gathered} \Rightarrow 3a{(-2)}^2+2b\times (-2)+c=0 \\ \Rightarrow 12a-4b+c=0 \end{gathered}$$

The curve $y=a{x}^3+b{x}^2+cx+5$ meets  $y-$axis at Q. 
Putting $x=0$ in $y=a{x}^3+b{x}^2+cx+5,$ we get  $y=5$
Thus, the coordinates of Q are  $(0,5)$
It is given that the gradient of the curve at Q is 3.
$$\begin{gathered} \therefore {\left(\frac{dy}{dx}\right)}_Q=3 \\ \Rightarrow 3a\times 0+2b\times 0+c=3 \\ \Rightarrow c=3 \end{gathered}$$

Putting $c=3$ in (i) and (ii), we get 
$8a-4b=-1$ and $12a-4b=-3$
Solving these two equations, we get $a=-\frac{1}{2}$ and  $b=-\frac{3}{4}$
Substituting the value of a, b and c in the equation of the curve, we obtain 
$y=-\frac{1}{2}{x}^3-\frac{3}{4}{x}^2+3x+5$

As the equation of the curve.