Divide the first polynomial by the second in each of the following.

Solution:

$(i) \frac{3 x^{2}+4 x+5}{x-2}$

$=\frac{3 x(x-2)+10(x-2)+25}{(x-2)}$

$=\frac{(x-2)(3 x+10)+25}{(x-2)}$

$=(3 \mathrm{x}+10)+\frac{25}{(\mathrm{x}-2)}$

Therefore,

quotient $=3 x+10$ and remainder $=25$.

$(i i) \frac{10 x^{2}-7 x+8}{5 x-3}$

$=\frac{2 \mathrm{x}(5 \mathrm{x}-3)-\frac{1}{5}(5 \mathrm{x}-3)+\frac{47}{5}}{(5 \mathrm{x}-3)}$

$=\frac{(5 \mathrm{x}-3)\left(2 \mathrm{x}-\frac{1}{5}\right)+\frac{47}{5}}{(5 \mathrm{x}-3)}$

$=\left(2 \mathrm{x}-\frac{1}{5}\right)+\frac{\frac{47}{5}}{5 \mathrm{x}-3}$

Therefore,

$\mathrm{q}$ uot ient $=2 \mathrm{x}-\frac{1}{5}$ and remainder $=\frac{47}{5}$.

$($ iii $) \frac{5 \mathrm{y}^{3}-6 \mathrm{y}^{2}+6 \mathrm{y}-1}{5 \mathrm{y}-1}$

$=\frac{\mathrm{y}^{2}(5 \mathrm{y}-1)-\mathrm{y}(5 \mathrm{y}-1)+1(5 \mathrm{y}-1)}{(5 \mathrm{y}-1)}$

$=\frac{(5 \mathrm{y}-1)\left(\mathrm{y}^{2}-\mathrm{y}+1\right)}{(5 \mathrm{y}-1)}$

$=\left(\mathrm{y}^{2}-\mathrm{y}+1\right)$

Therefore,

Quotient $=\mathrm{y}^{2}-\mathrm{y}+1$ and remainder $=0$

(iv) $\frac{x^{4}-\mathrm{x}^{3}+5 \mathrm{x}}{\mathrm{x}-1}$

$=\frac{x^{3}(\mathrm{x}-1)+5(\mathrm{x}-1)+5}{\mathrm{x}-1}$

$=\frac{(\mathrm{x}-1)\left(\mathrm{x}^{3}+5\right)+5}{\mathrm{x}-1}$

$=\left(\mathrm{x}^{3}+5\right)+\frac{5}{x-1}$

Therefore, quotient $=x^{3}+5$ and remainder $=5$.

$(\mathrm{V}) \frac{\mathrm{y}^{4}+\mathrm{y}^{2}}{\mathrm{y}^{2}-2}$

$=\frac{y^{2}\left(y^{2}-2\right)+3\left(y^{2}-2\right)+6}{y^{2}-2}$

$=\frac{\left(y^{2}-2\right)\left(y^{2}+3\right)+6}{y^{2}-2}$

$=\left(y^{2}+3\right)+\frac{6}{y^{2}-2}$

Therefore, quotient $=\mathrm{y}^{2}+3$ and $\mathrm{r}$ emainder $=6$.