If 1, 3, 4, 0 are the roots of ${\mathrm{ax}}^4 + {\mathrm{bx}}^3 + {\mathrm{cx}}^2 + \mathrm{dx} +\mathrm{e} = 0$ then the roots of $\mathrm{a}{(\mathrm{x}+3)}^4+\mathrm{b}{(\mathrm{x}+ 3)}^3+\mathrm{c}{(\mathrm{x}+3)}^2+\mathrm{d}(\mathrm{x}+3)+\mathrm{e}=0$ are

$\begin{array}{l}f\left(x\right)=a{x}^4+b{x}^3+c{x}^2+dx+e=0\\ Nowa{(x+3)}^4+b{(x+3)}^3+c{(x+3)}^2+d(x+3)+e=0\\ Itisoftheformf(x+3)=0\\ Therootsoff(x+3)=\mathrm{Re}ducedby3oftherootsoff\left(x\right)\\ \therefore Therootsare1-3,3-3,4-3,0-3\\ \Rightarrow -2,0,1,-3\end{array}$