Suppose $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^4+{\mathrm{ax}}^3+{\mathrm{bx}}^2+\mathrm{cx}+\mathrm{d}$, and $\mathrm{f}\left(1\right)=\mathrm{f}\left(2\right)=\mathrm{f}\left(3\right)=\mathrm{f}\left(4\right)=0$. Then the value of ‘b’ is

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^4+{\mathrm{ax}}^3+{\mathrm{bx}}^2+\mathrm{cx}+\mathrm{d} \\ \mathrm{Since} \mathrm{f}\left(1\right)=\mathrm{f}\left(2\right)=\mathrm{f}\left(3\right)=\mathrm{f}\left(4\right)=0\mathrm{, } \end{gathered}$$

Therefore $(\mathrm{x}-1),(\mathrm{x}-2),(\mathrm{x}-3),(\mathrm{x}-4)$ are factors of f(x) 

Degree of f(x) = 4

Degree of $(\mathrm{x}-1)(\mathrm{x}-2)(\mathrm{x}-3)(\mathrm{x}-4)=4$

And coefficient of x⁴ is one

$\begin{array}{l}\mathrm{f}\left(\mathrm{x}\right)=1(\mathrm{x}-1)(\mathrm{x}-2)(\mathrm{x}-3)(\mathrm{x}-4)\\ {\mathrm{x}}^4+{\mathrm{ax}}^3+{\mathrm{bx}}^2+\mathrm{cx}+\mathrm{d}=(\mathrm{x}-1)(\mathrm{x}-2)(\mathrm{x}-3)(\mathrm{x}-4)\\ =\left({\mathrm{x}}^2-3\mathrm{x}+2\right)\left({\mathrm{x}}^2-7\mathrm{x}+12\right)\\ ={\mathrm{x}}^4-10{\mathrm{x}}^3+35{\mathrm{x}}^2-50\mathrm{x}+24\\ {\mathrm{x}}^4+{\mathrm{ax}}^3+{\mathrm{bx}}^2+\mathrm{cx}+\mathrm{d}={\mathrm{x}}^4-10{\mathrm{x}}^3+35{\mathrm{x}}^2-50\mathrm{x}+24\end{array}$

$\therefore$By comparison b = 35

   (or)

If $\mathrm{f}\left(1\right)=\mathrm{f}\left(2\right)=\mathrm{f}\left(3\right)=\mathrm{f}\left(4\right)=0$, the roots of 

$$\begin{gathered} {\mathrm{x}}^4+{\mathrm{ax}}^3+{\mathrm{bx}}^2+\mathrm{cx}+\mathrm{d}=0 \mathrm{are} 1, 2, 3 \mathrm{and} 4 \\ \begin{array}{l}\mathrm{\Sigma \alpha \beta }=\frac{\mathrm{b}}{\mathrm{a}}\\ \left(1\right)\left(2\right)+(2\times 3)+(3\times 4)+(4\times 1)+(1\times 3)+(2\times 4)=\frac{\mathrm{b}}{1}\\ 2+6+12+4+3+8=\mathrm{b}\\ \mathrm{b}=35\end{array} \end{gathered}$$