The curve y= ax³+ bx²+cx + 5 touches the x-axis at P(-2, 0) and cuts the y-axis at a point Q where its gradient is 3, then the value of -10 a- 100b+ 1000 c must be

$\because \frac{\mathrm{dy}}{\mathrm{dx}}=3{\mathrm{ax}}^2+2\mathrm{bx}+\mathrm{c}----\left(\mathrm{i}\right)$

Since, the curve y = ax³ + bx² + cx +5 touches X-axis at P(-2 0) then X-axis is the tangent at (-2, 0) the curve
meets Y-axis in (0, 5).

$\begin{array}{l}{\left.\therefore \frac{\mathrm{dy}}{\mathrm{dx}}\right|}_{(0,5)}=0+0+\mathrm{c}=3\text{ (given) ----}\left(\mathrm{ii}\right)\\ \text{ and }{\left.\frac{\mathrm{dy}}{\mathrm{dx}}\right|}_{(-2,0)}=0\Rightarrow 12\mathrm{a}-4\mathrm{b}+\mathrm{c}=0\\ \Rightarrow 12\mathrm{a}-4\mathrm{b}+3=0 [\mathrm{from} \mathrm{Eq}. (\mathrm{ii}\left)\right] ..........\left(\mathrm{iii}\right)\end{array}$

and (-2, 0) lies on the curve, then

$\begin{array}{l}0=-8\mathrm{a}+4\mathrm{b}-2\mathrm{c}+5\\ \Rightarrow -8\mathrm{a}+4\mathrm{b}-1=0\\ \Rightarrow 8\mathrm{a}-4\mathrm{b}+1=0----\left(\mathrm{iv}\right)\end{array}$

From Egs. (ii) and (iv) we get

$\begin{array}{l}\mathrm{a}=-\frac{1}{2},\mathrm{b}=-\frac{3}{4}\\ \therefore -10\mathrm{a}-100\mathrm{b}+1000\mathrm{c}=-10\times \frac{-1}{2}-100\times \frac{-3}{4}+1000\times 3\\ =5+75+3000=3080\end{array}$