Question

Divide the first polynomial by the second in each of the following. Also, write the quotient and remainder:
(i) 3x² + 4x + 5, x − 2
(ii) 10x² − 7x + 8, 5x − 3
(iii) 5y³ − 6y² + 6y − 1, 5y − 1
(iv) x⁴ − x³ + 5x, x − 1
(v) y⁴ + y², y² − 2

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\frac{3{\mathrm{x}}^2+4\mathrm{x}+5}{\mathrm{x}-2} \\ =\frac{3\mathrm{x}(\mathrm{x}-2)+10(\mathrm{x}-2)+25}{(\mathrm{x}-2)} \\ =\frac{(\mathrm{x}-2)(3\mathrm{x}+10)+25}{(\mathrm{x}-2)} \\ =(3\mathrm{x}+10)+\frac{25}{(\mathrm{x}-2)} \\ \mathrm{Therefore}, \\ \text{q}\mathrm{uot}\text{ie}\mathrm{nt}=3\mathrm{x}+10\mathrm{and}\text{r}\mathrm{emainder}=25. \\ \left(\mathrm{ii}\right)\frac{10{\mathrm{x}}^2-7\mathrm{x}+8}{5\mathrm{x}-3} \\ =\frac{2\mathrm{x}(5\mathrm{x}-3)-{\displaystyle \frac{1}{5}}(5\mathrm{x}-3)+{\displaystyle \frac{47}{5}}}{(5\mathrm{x}-3)} \\ =\frac{(5\mathrm{x}-3)(2\mathrm{x}-{\displaystyle \frac{1}{5}})+{\displaystyle \frac{47}{5}}}{(5\mathrm{x}-3)} \\ =(2\mathrm{x}-\frac{1}{5})+\frac{{\displaystyle \frac{47}{5}}}{5\mathrm{x}-3} \\ \mathrm{Therefore}, \\ \text{q}\mathrm{uot}\text{ie}\mathrm{nt}=2\mathrm{x}-\frac{1}{5}\mathrm{and}\text{r}\mathrm{emainder}=\frac{47}{5}. \\ \left(\mathrm{iii}\right)\frac{5{\mathrm{y}}^3-6{\mathrm{y}}^2+6\mathrm{y}-1}{5\mathrm{y}-1} \\ =\frac{{\mathrm{y}}^2(5\mathrm{y}-1)-\mathrm{y}(5\mathrm{y}-1)+1(5\mathrm{y}-1)}{(5\mathrm{y}-1)} \\ =\frac{(5\mathrm{y}-1)({\mathrm{y}}^2-\mathrm{y}+1)}{(5\mathrm{y}-1)} \\ =({\mathrm{y}}^2-\mathrm{y}+1) \\ \mathrm{Therefore}, \\ \mathrm{Quotient}={\mathrm{y}}^2-\mathrm{y}+1\mathrm{and}\mathrm{remainder}=0 \end{gathered}$$

$$\begin{gathered} (iv)\frac{x^4{-x}^3+5x}{\mathrm{x-1}} \\ =\frac{x^3(x-1)+5(x-1)+5}{\mathrm{x-1}} \\ =\frac{{(x-1)(x}^3+5)+5}{\mathrm{x-1}} \\ {=(x}^3+5)+\frac{5}{x-1} \\ \mathrm{Therefore,}\text{q}\mathrm{uot}\text{ie}\mathrm{nt}=x^3+5\mathrm{and}\text{r}\mathrm{emainder}=5. \end{gathered}$$
$$\begin{gathered} \left(\mathrm{v}\right)\frac{{\mathrm{y}}^4+{\mathrm{y}}^2}{{\mathrm{y}}^2-2} \\ =\frac{{\mathrm{y}}^2({\mathrm{y}}^2-2)+3({\mathrm{y}}^2-2)+6}{{\mathrm{y}}^2-2} \\ =\frac{({\mathrm{y}}^2-2)({\mathrm{y}}^2+3)+6}{{\mathrm{y}}^2-2} \\ =({\mathrm{y}}^2+3)+\frac{6}{{\mathrm{y}}^2-2} \\ \mathrm{Therefore},\text{q}\mathrm{uotient}={\mathrm{y}}^2+3\mathrm{and}\text{r}\mathrm{emainder}=6. \end{gathered}$$