Let f(x) be a polynomial of degree 5. When f(x) is divided by (x–1)³ , the remainder 33, and when f(x) is divided by (x+1)³ , the remainder is –3. Then, equation to the tangent drawn to y=f(x) at x = 0 is

$$\begin{gathered} f\left(x\right)-33=0 has roots 1,1,1 and f\left(x\right)+3=0 has roots -1,-1,-1 \\ \therefore f\text{'}\left(x\right)=0 has roots 1,1,-1,-1 \\ \Rightarrow f\text{'}\left(x\right)=\lambda {\left(x^2-1\right)}^2=\lambda \left(x^4-2x^2+1\right) \\ \Rightarrow f\left(x\right)=\lambda \left(\frac{x^5}{5}-2\frac{x^3}{3}+x\right)+c,f\left(1\right)=33,f(-1)=-3 \\ \therefore \lambda =\frac{135}{4},c=15 \\ \Rightarrow f\left(x\right)=\frac{135}{4}\left(\frac{x^5}{5}-2\frac{x^3}{3}+x\right)+15=\frac{27{\mathrm{x}}^5}{4}-\frac{45{\mathrm{x}}^3}{2}+\frac{135\mathrm{x}}{4}+15 \end{gathered}$$

$\begin{array}{l}\mathrm{f}\left(\mathrm{x}\right)=\frac{27{\mathrm{x}}^5}{4}-\frac{45{\mathrm{x}}^3}{2}+\frac{135\mathrm{x}}{4}+15\text{ at }\\ \mathrm{x}=0,\mathrm{y}=15\Rightarrow {\mathrm{f}}^1\left(0\right)=\frac{135}{4}\\ \mathrm{Equation} \mathrm{of} \mathrm{tangent} \mathrm{is} \\ \mathrm{y}-15=\frac{135}{4}\left(\mathrm{x}-0\right)\\ 135\mathrm{x}-4\mathrm{y}+60=0\end{array}$