If the curve $\mathrm{y}={\mathrm{ax}}^3+{\mathrm{bx}}^2+\mathrm{cx}+5$ touches x axis at P(-2,0) and cuts y axis at point Q where its gradient is 3. Find $\mathrm{c}+2\mathrm{b}+\mathrm{a}$

$$\begin{gathered} \frac{dy}{dx}=3a{x}^2+2bx+c \\ \text{ Since, }x\text{-axis is tangent at }(-2,0)\text{, } \\ \therefore {\left|\frac{dy}{dx}\right|}_{x=-2}=0 \\ \begin{array}{r}\Rightarrow 0=3a(-2{)}^2+2b(-2)+c\\ \Rightarrow 0=12a-4b+c\dots \left(\mathrm{i}\right)\end{array} \\ \text{ Again, slope of tangent at }(0,5)\text{ is }3 \\ \begin{array}{r}\Rightarrow {\left|\frac{dy}{dx}\right|}_{(0,5)}=3\\ \Rightarrow 3=3a(0{)}^2+2b(0)+c\\ \Rightarrow 3=c.....\left(ii\right)\\ \begin{array}{r}0=a(-2{)}^3+b(-2{)}^2+c(-2)+5\\ \mathbf{0}=-8a+4b-2c+5......\left(iii\right)\end{array}\\ solving \left(i\right),\left(ii\right)\Rightarrow 2a-4b=-3\\ solving \left(iii\right),\left(ii\right)\Rightarrow -8a+4b=1\\ \end{array} \end{gathered}$$
$\begin{array}{l} \\ \mathrm{c}=3,\mathrm{a}=-\frac{1}{2},\mathrm{b}=-\frac{3}{4}.\\ \mathrm{c}+2\mathrm{b}+\mathrm{a}=3-\frac{3}{2}-\frac{1}{2}=1\end{array}$