RD Sharma solution class 8 chapter 8 Division of Algebraic Expression Exercise 8.5

Exercise 8.5

Page-8.15

Question 1:

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:

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

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

Question 2:

Find whether the first polynomial is a factor of the second.
(i) x + 1, 2x² + 5x + 4
(ii) y − 2, 3y³ + 5y² + 5y + 2
(iii) 4x² − 5, 4x⁴ + 7x² + 15
(iv) 4 − z, 3z² − 13z + 4
(v) 2a − 3, 10a² − 9a − 5
(vi) 4y + 1, 8y² − 2y + 1

Answer 2:

$$\begin{gathered} \left(\mathrm{i}\right) \frac{2{\mathrm{x}}^2+5\mathrm{x}+4}{\mathrm{x}+1} \\ =\frac{2\mathrm{x}(\mathrm{x}+1)+3(\mathrm{x}+1)+1}{\mathrm{x}+1} \\ =\frac{(\mathrm{x}+1)(2\mathrm{x}+3)+1}{(\mathrm{x}+1)} \\ =(2\mathrm{x}+3)+\frac{1}{\mathrm{x}+1} \\ \because \mathrm{Remainder}=1 \\ \mathrm{Therefore}, (\mathrm{x}+1) \mathrm{is} \mathrm{not} \mathrm{a} \mathrm{factor} \mathrm{of} 2{\mathrm{x}}^2+5\mathrm{x}+4 \end{gathered}$$

$$\begin{gathered} (\mathrm{ii)} \frac{{3y}^3{+5y}^2+5y+2}{\mathrm{y-2}} \\ =\frac{{\mathrm{3y}}^2(y-2)+11y(y-2)+27(y-2)+56}{\mathrm{y-2}} \\ =\frac{{(y-2)(3y}^2+11y+27)+56}{\mathrm{y-2}} \\ {=(3y}^2+11y+27)+\frac{56}{\mathrm{y-2}} \\ \because \mathrm{Remainder} = 56 \\ \therefore (y-2) is \mathrm{not} a \mathrm{factor} of {\mathrm{3y}}^3{+5y}^2+5y+2. \end{gathered}$$


$$\begin{gathered} \left(\mathrm{iii}\right) \frac{4{\mathrm{x}}^4{+}^2+15}{{\mathrm{4x}}^2-5} \\ = \frac{{\mathrm{x}}^2{(4x}^2{-5)+3(4x}^2-5)+30}{{\mathrm{4x}}^2-5} \\ = \frac{({\mathrm{4x}}^2{-5)(x}^2+3)+30}{{\mathrm{4x}}^2-5} \\ {=(x}^2+3)+\frac{30}{{\mathrm{4x}}^2-5} \\ \because \mathrm{Remainder} = 30 \\ \mathrm{T}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{e}, (4{\mathrm{x}}^2-5) \mathrm{i}\mathrm{s} \mathrm{n}\mathrm{o}\mathrm{t} \mathrm{a} \mathrm{f}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{r} \mathrm{o}\mathrm{f} 4{\mathrm{x}}^4+7{\mathrm{x}}^2+15 \end{gathered}$$

$$\begin{gathered} (iv) \frac{{\mathrm{3z}}^2-13z+4}{\mathrm{4-z}} \\ =\frac{{\mathrm{3z}}^2-12z-z+4}{\mathrm{4-z}} \\ =\frac{\mathrm{3z(z-4)-1(z-4)}}{\mathrm{4-z}} \\ =\frac{(z-4)(3z-1)}{\mathrm{4-z}} \\ =\frac{(\mathrm{4-z)(1-3z)}}{\mathrm{4-z}} \\ =1-3z \\ \because \mathrm{Remainder} = 0 \\ \therefore (4-z) \mathrm{is} a \mathrm{factor} of {\mathrm{3z}}^2-13z+4. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{V}\right) \frac{{\mathrm{10a}}^2-9a-5}{\mathrm{2a-3}} \\ =\frac{\mathrm{5a(2a-3)+3(2a-3)+4}}{\mathrm{2a-3}} \\ =\frac{(2a-3)(5a+3)+4}{\mathrm{2a-3}} \\ =(5a+3)+\frac{4}{\mathrm{2a-3}} \\ \because \mathrm{Remainder} = 4 \\ \therefore (\; 2a-3) \mathrm{is} \mathrm{not} a \mathrm{factor} \mathrm{of} {\mathrm{10a}}^2-9a-5. \end{gathered}$$

$$\begin{gathered} (vi) \frac{{\mathrm{8y}}^2-2y+1}{\mathrm{4y+1}} \\ =\frac{\mathrm{2y}(4y+1)-1(4y+1)+2}{\mathrm{4y+1}} \\ =\frac{(4y+1)(2y-1)+2}{\mathrm{4y+1}} \\ =(2y-1)+\frac{2}{\mathrm{4y+1}} \\ \because \mathrm{Remainder} = 2 \\ \therefore (4y+1) \mathrm{is} \mathrm{not} a \mathrm{factor} \mathrm{of} {\mathrm{8y}}^2-2y+1. \end{gathered}$$