NCERT Solutions Class 8 Maths Chapter 9

Q.1 Add the following.
(i) ab – bc, bc – ca, ca – ab
(ii) a – b + ab, b – c + bc, c – a + ac
(iii) 2p²q² – 3pq + 4, 5 + 7pq – 3p²q²
(iv) l² + m², m² + n², n² + l²,2lm + 2mn + 2nl

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{ab}–\mathrm{bc}\\ \mathrm{bc}–\mathrm{ca}\\ \text{+}\underset{¯}{–\mathrm{ab}\text{+}\mathrm{ca}}\\ \text{0}\\ \left(\mathrm{ii}\right)\mathrm{a}–\mathrm{b}+\mathrm{ab}\\ \mathrm{b}–\mathrm{c}+\mathrm{bc}\\ +\underset{¯}{–\mathrm{a}\text{+}\mathrm{c}+\mathrm{ac}}\\ \text{ab + bc + ac}\\ \left(\mathrm{iii}\right)2{\mathrm{p}}^2{\mathrm{q}}^2–3\mathrm{pq}+4\\ +\underset{¯}{–3{\mathrm{p}}^2{\mathrm{q}}^2+7\mathrm{pq}+5}\\ \text{–1}{\mathrm{p}}^2{\mathrm{q}}^2+4\mathrm{pq}+9\end{array}$

$\begin{array}{l}\left(\mathrm{iv}\right){\mathrm{l}}^2+{\mathrm{m}}^2\\ +\underset{¯}{\begin{array}{l}{\mathrm{m}}^2+{\mathrm{n}}^2\\ {\mathrm{l}}^2+{\mathrm{n}}^2\\ +2\mathrm{lm}+2\mathrm{mn}+2\mathrm{nl}\end{array}}\\ 2{\mathrm{l}}^2+2{\mathrm{m}}^2+2{\mathrm{n}}^2+2\mathrm{lm}+2\mathrm{mn}+2\mathrm{nl}\end{array}$

Q.2 (a) Subtract 4a – 7ab + 3b + 12 from 12a – 9ab + 5b – 3

(b) Subtract 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + 10xyz

(c) Subtract 4p²q – 3pq + 5pq² – 8p + 7q – 10 from 18 – 3p – 11q + 5pq – 2pq² + 5p²q

Ans

$\begin{array}{l}\text{(a) 12}a–\text{9}ab+\text{5}b–\text{3}\\ \\ \underset{¯}{\begin{array}{l}\text{4}a–\text{7}ab+\text{3}b+\text{12}\\ (–)\text{(+) (–) (–)}\end{array}}\\ \underset{¯}{\begin{array}{l}\\ \text{8}a–\text{2}ab+\text{2}b–\text{15}\\ \end{array}}\\ \\ \text{(b) 5}xy–\text{2}yz–\text{2}zx+\text{1}0xyz\\ \\ \underset{¯}{\begin{array}{l}\text{3}xy+\text{5}yz–\text{7}zx\\ (–)\text{(–) (+)}\end{array}}\\ \underset{¯}{\begin{array}{l}\\ \text{2xy – 7yz}+5zx+10xyz\\ \end{array}}\\ \left(c\right)18-3p-11q+5pq-2p{q}^2+5p^{2}q\\ \\ -10-8p+7q-3pq+5p{q}^2+4p^{2}q\\ \underset{¯}{(+)(+)(-)(+)(-)(-)}\\ \underset{¯}{\begin{array}{l}\\ \text{2}8+5p-18q+8pq-7p{q}^2+1p^{2}q\\ \end{array}}\end{array}$

Q.3 Find the product of the following pairs of monomials.

(I) 4, 7p

(ii) – 4p, 7p

(iii) – 4p, 7pq

(iv) 4p³, –3p

(v) 4p, 0

Ans

(i) 4 × 7p = 4 × 7 × p = 28p

(ii) –4p × 7p = (–4) × 7 × p × p = –28p²

(iii) –4p × 7pq =(–4) × 7 × p × p × q = –28p²q

(iv) 4p³ × (–3p) = 4 × (–3) × p³ × p = –12p⁴

(v) 4p × 0 = 4 × p × 0 = 0

Q.4 Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.

(p, q); (10m, 5n); (20x², 5y²); (4x, 3x²); (3mn, 4np)

Ans

$\begin{array}{c}\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Area of rectangle}=\text{Length × Breadth}\\ \text{Area of first rectangle}=\mathrm{p}\times \mathrm{q}\\ =\mathrm{pq}\\ \text{Area of second rectangle}=10\mathrm{m}\times \mathrm{}5\mathrm{n}\\ =10\times 5\times \mathrm{m}\times \mathrm{n}\\ =50\mathrm{mn}\\ \text{Area of third rectangle}\mathrm{}=20{\mathrm{x}}^2\times 5{\mathrm{y}}^2\\ =20\times 5\times {\mathrm{x}}^2\times {\mathrm{y}}^2\\ =100{\mathrm{x}}^2{\mathrm{y}}^2\\ \text{Area of fourth rectangle}\mathrm{}=4\mathrm{x}\times 3{\mathrm{x}}^2\\ =4\times 3\times \mathrm{x}\times {\mathrm{x}}^2\\ =12{\mathrm{x}}^3\\ \text{Area of fifth rectangle}\mathrm{}=3\mathrm{mn}\times 4\mathrm{np}\\ =3\times 4\times \mathrm{m}\times \mathrm{n}\times \mathrm{n}\times \mathrm{p}\\ =12{\mathrm{mn}}^2\mathrm{p}\end{array}$

Q.5 Complete the table of products.

$\begin{array}{|ccccccc|}\hline \frac{\mathrm{First}{\displaystyle }\mathrm{monomial}\to }{\mathrm{Second}{\displaystyle }\mathrm{monomial}\downarrow }& 2\mathrm{x}& -5\mathrm{y}& 3{\mathrm{x}}^2& -4\mathrm{xy}& 7{\mathrm{x}}^2\mathrm{y}& -9{\mathrm{x}}^2{\mathrm{y}}^2\\ 2\mathrm{x}& 4{\mathrm{x}}^2& ...& ...& ...& ...& ...\\ -5\mathrm{y}& ...& ...& -15{\mathrm{x}}^2\mathrm{y}& ...& ...& ...\\ 3{\mathrm{x}}^2& ...& ...& ...& ...& ...& ...\\ -4\mathrm{xy}& ...& ...& ...& ...& ...& ...\\ 7{\mathrm{x}}^2\mathrm{y}& ...& ...& ...& ...& ...& ...\\ -9{\mathrm{x}}^2{\mathrm{y}}^2& ...& ...& ...& ...& ...& ...\\ \hline\end{array}$

Ans

$\begin{array}{|ccccccc|}\hline \frac{\begin{array}{l}\mathrm{First}\\ \mathrm{monomial}\to \end{array}}{\begin{array}{l}\mathrm{Second}\\ \mathrm{monomial}\downarrow \end{array}}& 2\mathrm{x}& -5\mathrm{y}& 3{\mathrm{x}}^2& -4\mathrm{xy}& 7{\mathrm{x}}^2\mathrm{y}& -9{\mathrm{x}}^2{\mathrm{y}}^2\\ 2\mathrm{x}& 4{\mathrm{x}}^2& \underset{¯}{-10\mathrm{xy}}& \underset{¯}{6{\mathrm{x}}^3}& \underset{¯}{-8{\mathrm{x}}^2\mathrm{y}}& \underset{¯}{14{\mathrm{x}}^3\mathrm{y}}& \underset{¯}{-18{\mathrm{x}}^3{\mathrm{y}}^2}\\ -5\mathrm{y}& \underset{¯}{-10\mathrm{xy}}& \underset{¯}{25{\mathrm{y}}^2}& -15{\mathrm{x}}^2\mathrm{y}& \underset{¯}{20{\mathrm{xy}}^2}& \underset{¯}{-35{\mathrm{x}}^2{\mathrm{y}}^2}& \underset{¯}{45{\mathrm{x}}^2{\mathrm{y}}^3}\\ 3{\mathrm{x}}^2& \underset{¯}{6{\mathrm{x}}^3}& \underset{¯}{-15{\mathrm{x}}^2\mathrm{y}}& \underset{¯}{9{\mathrm{x}}^4}& \underset{¯}{-12{\mathrm{x}}^3\mathrm{y}}& \underset{¯}{21{\mathrm{x}}^4\mathrm{y}}& \underset{¯}{-27{\mathrm{x}}^4{\mathrm{y}}^2}\\ -4\mathrm{xy}& \underset{¯}{-8{\mathrm{x}}^2\mathrm{y}}& \underset{¯}{20{\mathrm{xy}}^2}& \underset{¯}{-12{\mathrm{x}}^3\mathrm{y}}& \underset{¯}{16{\mathrm{x}}^2{\mathrm{y}}^2}& \underset{¯}{-28{\mathrm{x}}^3{\mathrm{y}}^2}& \underset{¯}{36{\mathrm{x}}^3{\mathrm{y}}^3}\\ 7{\mathrm{x}}^2\mathrm{y}& \underset{¯}{14{\mathrm{x}}^3\mathrm{y}}& \underset{¯}{-35{\mathrm{x}}^2{\mathrm{y}}^2}& \underset{¯}{21{\mathrm{x}}^4\mathrm{y}}& \underset{¯}{-28{\mathrm{x}}^3{\mathrm{y}}^2}& \underset{¯}{49{\mathrm{x}}^4{\mathrm{y}}^2}& \underset{¯}{-63{\mathrm{x}}^4{\mathrm{y}}^3}\\ -9{\mathrm{x}}^2{\mathrm{y}}^2& \underset{¯}{-18{\mathrm{x}}^3{\mathrm{y}}^2}& \underset{¯}{45{\mathrm{x}}^2{\mathrm{y}}^3}& \underset{¯}{-27{\mathrm{x}}^4{\mathrm{y}}^2}& \underset{¯}{36{\mathrm{x}}^3{\mathrm{y}}^3}& \underset{¯}{-63{\mathrm{x}}^4{\mathrm{y}}^3}& \underset{¯}{81{\mathrm{x}}^4{\mathrm{y}}^4}\\ \hline\end{array}$

Q.6 Obtain the volume of rectangular boxes with the following length, breadth and height respectively.

(I) 5a, 3a², 7a⁴

(ii) 2p, 4q, 8r

(iii) xy, 2x²y, 2xy²

(iv) a, 2b, 3c

Ans

Volume of the the rectangular boxes = Length × Breadth × Height

(i) Volume = 5a × 3a² × 7a⁴

= 5 ×3 ×7 × a × a² × a⁴

= 105a⁷

(ii) Volume = 2p × 4q × 8r

= 2 × 4 × 8 × p × q × r

= 64pqr

(iii) Volume = xy × 2x²y × 2xy²

= 2 × 2 × x × x² × x × y × y × y²

= 4x⁴y⁴

(iv) Volume = a × 2b × 3c

= 2 × 3 × a × b × c

= 6abc

Q.7 Obtain the product of

(i) xy, yz, zx

(ii) a, – a², a³

(iii) 2, 4y, 8y², 16y³

(iv) a, 2b, 3c, 6abc

(v) m, – mn, mnp

Ans

$\begin{array}{l}\left(\text{i}\right)\mathrm{xy}\times \mathrm{yz}\times \mathrm{zx}=\mathrm{x}\times \mathrm{y}\times \mathrm{y}\times \mathrm{z}\times \mathrm{z}\times \mathrm{x}\\ ={\mathrm{x}}^2{\mathrm{y}}^2{\mathrm{z}}^2\\ \\ \mathrm{}\left(\text{ii}\right)\mathrm{a}\times \text{(}–{\mathrm{a}}^2)\times {\mathrm{a}}^3=-{\mathrm{a}}^6\\ \\ \left(\text{iii}\right)\text{2}\times \text{4}\mathrm{y}\times \text{8}{\mathrm{y}}^2\times \text{16}{\mathrm{y}}^3=\text{2}\times \text{4}\times \text{8}\times \text{16 ×}\mathrm{y}\times {\mathrm{y}}^2\times {\mathrm{y}}^3\\ =1024{\mathrm{y}}^6\\ \\ \left(\text{iv}\right)\mathrm{a}\times \text{2}\mathrm{b}\times \text{3}\mathrm{c}\times \text{6}\mathrm{abc}=2\times 3\times 6\times \mathrm{a}\times \mathrm{b}\times \mathrm{c}\times \mathrm{abc}\\ =36{\mathrm{a}}^2{\mathrm{b}}^2{\mathrm{c}}^2\\ \\ \mathrm{}\left(\text{v}\right)\mathrm{m}\times (–\mathrm{mn})\times \mathrm{mnp}=\mathrm{m}\times (–\mathrm{m})\times \mathrm{n}\times \mathrm{m}\times \mathrm{n}\times \mathrm{p}\\ =-{\mathrm{m}}^3{\mathrm{n}}^2\mathrm{p}\end{array}$

Q.8 Carry out the multiplication of the expressions in each of the following pairs.
(i) 4p, q + r
(ii) ab, a – b
(iii) a + b, 7a²b²
(iv) a² – 9, 4a
(v) pq + qr + rp, 0

Ans

$\begin{array}{l}\left(\text{i}\right)\text{4}\mathrm{p}\times (\mathrm{q}+\mathrm{r})\\ =(\text{4}\mathrm{p}\times \mathrm{q})+(\text{4}\mathrm{p}\times \mathrm{r})\\ =4\mathrm{pq}+4\mathrm{pr}\\ \\ \left(\text{ii}\right)\mathrm{ab}\times (\mathrm{a}–\mathrm{b})\\ =\mathrm{ab}\times \mathrm{a}-\mathrm{ab}\times \mathrm{b}\\ ={\mathrm{a}}^2\mathrm{b}-{\mathrm{ab}}^2\\ \mathrm{}\\ \mathrm{}\left(\text{iii}\right)(\mathrm{a}+\mathrm{b})\times 7{\mathrm{a}}^2{\mathrm{b}}^2\\ =\mathrm{a}\times 7{\mathrm{a}}^2{\mathrm{b}}^2+\mathrm{b}\times 7{\mathrm{a}}^2{\mathrm{b}}^2\\ =7{\mathrm{a}}^3{\mathrm{b}}^2+7{\mathrm{a}}^2{\mathrm{b}}^3\\ {}^{}\\ \mathrm{}\left(\text{iv}\right)\left({\mathrm{a}}^2–9\right)\times \text{4}\mathrm{a}={\mathrm{a}}^2\times \text{4}\mathrm{a}–9\times 4\mathrm{a}\\ =4{\mathrm{a}}^3-36\mathrm{a}\mathrm{}\\ \mathrm{}\\ \mathrm{}\left(\text{v}\right)\text{(}\mathrm{pq}+\mathrm{qr}+\mathrm{rp})\times \text{0}\\ =(\mathrm{pq}\times 0)+(\mathrm{qr}\times 0)+(\mathrm{rp}\times \text{0})\\ =\text{0}\end{array}$

Q.9 Complete the table.

$\begin{array}{|cccc|}\hline & \mathrm{First}\mathrm{Expression}& \mathrm{Second}\mathrm{Expression}& \mathrm{Product}\\ \left(\mathrm{i}\right)& \mathrm{a}& \mathrm{b}+\mathrm{c}+\mathrm{d}& ...\\ \left(\mathrm{ii}\right)& \mathrm{x}+\mathrm{y}-5& 5\mathrm{xy}& ...\\ \left(\mathrm{iii}\right)& \mathrm{p}& 6{\mathrm{p}}^2-7\mathrm{p}+5& ...\\ \left(\mathrm{iv}\right)& 4{\mathrm{p}}^2{\mathrm{q}}^2& {\mathrm{p}}^2-{\mathrm{q}}^2& ...\\ \left(\mathrm{v}\right)& \mathrm{a}+\mathrm{b}+\mathrm{c}& \mathrm{abc}& ...\\ \hline\end{array}$

Ans

$\begin{array}{|cccc|}\hline & \text{First Expression}& \text{Second Expression}& \text{Product}\\ \left(\mathrm{i}\right)& \mathrm{a}& \mathrm{b}+\mathrm{c}+\mathrm{d}& \mathrm{ab}+\mathrm{ac}+\mathrm{ad}\\ \left(\mathrm{ii}\right)& \mathrm{x}+\mathrm{y}-5& 5\mathrm{xy}& 5{\mathrm{x}}^2\mathrm{y}+5{\mathrm{xy}}^2-25\mathrm{xy}\\ \left(\mathrm{iii}\right)& \mathrm{p}& 6{\mathrm{p}}^2-7\mathrm{p}+5& 6{\mathrm{p}}^3-7{\mathrm{p}}^2+5\mathrm{p}\\ \left(\mathrm{iv}\right)& 4{\mathrm{p}}^2{\mathrm{q}}^2& {\mathrm{p}}^2-{\mathrm{q}}^2& 4{\mathrm{p}}^4{\mathrm{q}}^2-4{\mathrm{p}}^2{\mathrm{q}}^4\\ \left(\mathrm{v}\right)& \mathrm{a}+\mathrm{b}+\mathrm{c}& \mathrm{abc}& {\mathrm{a}}^2\mathrm{bc}+{\mathrm{ab}}^2\mathrm{c}+{\mathrm{abc}}^2\\ \hline\end{array}$

Q.10

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{product}.\\ \left(\mathrm{i}\right)\left({\mathrm{a}}^2\right)\times \left(2{\mathrm{a}}^{22}\right)\times \left(4{\mathrm{a}}^{26}\right)\\ \left(\mathrm{ii}\right)\left(\frac{2}{3}\mathrm{xy}\right)\times \left(\frac{–9}{10}{\mathrm{x}}^2{\mathrm{y}}^2\right)\\ \left(\mathrm{iii}\right)\left(\frac{–10}{3}{\mathrm{pq}}^3\right)\times \left(\frac{6}{5}{\mathrm{p}}^3\mathrm{q}\right)\\ \left(\mathrm{iv}\right)\mathrm{x}\times {\mathrm{x}}^2\times {\mathrm{x}}^3\times {\mathrm{x}}^4\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\left({\mathrm{a}}^2\right)\times \left(2{\mathrm{a}}^{22}\right)\times \left(4{\mathrm{a}}^{26}\right)=2\times 4\times {\mathrm{a}}^2\times {\mathrm{a}}^{22}\times {\mathrm{a}}^{26}\\ =8{\mathrm{a}}^{50}\\ \\ \left(\mathrm{ii}\right)\left(\frac{2}{3}\mathrm{xy}\right)\times \left(\frac{-9}{10}{\mathrm{x}}^2{\mathrm{y}}^2\right)=\frac{2}{3}\times \frac{-9}{10}\times \mathrm{x}\times {\mathrm{x}}^2\times \mathrm{y}\times {\mathrm{y}}^2\\ =-\frac{3}{5}{\mathrm{x}}^3{\mathrm{y}}^3\\ \\ \left(\mathrm{iii}\right)\left(\frac{-10}{3}{\mathrm{pq}}^3\right)\times \left(\frac{6}{5}{\mathrm{p}}^3\mathrm{q}\right)=\frac{-10}{3}\times \frac{6}{5}\times \mathrm{p}\times {\mathrm{q}}^3\times {\mathrm{p}}^3\times \mathrm{q}\\ =-4{\mathrm{p}}^4{\mathrm{q}}^4\\ \\ \left(\mathrm{iv}\right)\mathrm{x}\times {\mathrm{x}}^2\times {\mathrm{x}}^3\times {\mathrm{x}}^4={\mathrm{x}}^{10}\end{array}$

Q.11

$\begin{array}{l}\left(\mathrm{a}\right)\mathrm{Simplify}3\mathrm{x}(4\mathrm{x}-5)+3\mathrm{and}\mathrm{find}\mathrm{its}\mathrm{values}\mathrm{for}\\ \left(\mathrm{i}\right)\mathrm{x}=3\left(\mathrm{ii}\right)\mathrm{x}=\frac{1}{2}\\ \\ \left(\mathrm{b}\right)\mathrm{Simplify}\mathrm{a}({\mathrm{a}}^2+\mathrm{a}+1)+5\mathrm{and}\mathrm{find}\mathrm{its}\mathrm{value}\mathrm{for}\\ \left(\mathrm{i}\right)\mathrm{a}=0,\left(\mathrm{ii}\right)\mathrm{a}=1,\left(\mathrm{iii}\right)\mathrm{a}=-1\end{array}$

Ans

$\begin{array}{l}\text{(a)}\\ \text{3x (4x – 5) + 3}\\ \text{=\hspace{0.33em}3x×4x-3x×5+3}\\ {\text{=\hspace{0.33em}12x}}^{\text{2}}\text{-15x+3}\\ \\ {\text{(i) For x = 3, 12x}}^{\text{2}}\text{-15x+3=\hspace{0.33em}12}{\left(\text{3}\right)}^{\text{2}}\text{-15×3+3}\\ \text{=\hspace{0.33em}108-45+3}\\ \text{=\hspace{0.33em}66}\end{array}$

$\begin{array}{l}\left(\mathrm{ii}\right)\text{For}\mathrm{x}=\frac{1}{2},12{\mathrm{x}}^2-15\mathrm{x}+3=12{\left(\frac{1}{2}\right)}^2-15\times \frac{1}{2}+3\\ =12\times \frac{1}{4}-\frac{15}{2}+3\\ =3-\frac{15}{2}+3\\ =6-\frac{15}{2}=\frac{12-15}{2}=\frac{-3}{2}\\ \left(\mathrm{b}\right)\\ \mathrm{a}({\mathrm{a}}^2+\mathrm{a}+1)+5=\mathrm{a}\times {\mathrm{a}}^2+\mathrm{a}\times \mathrm{a}+\mathrm{a}\times 1+5\\ ={\mathrm{a}}^3+{\mathrm{a}}^2+\mathrm{a}+5\\ \\ \left(\mathrm{i}\right)\text{For}\mathrm{a}=0,{\mathrm{a}}^3+{\mathrm{a}}^2+\mathrm{a}+5=0+0+0+5\\ =5\\ \\ \left(\mathrm{ii}\right)\text{For}\mathrm{a}=1,{\mathrm{a}}^3+{\mathrm{a}}^2+\mathrm{a}+5=1^3+1^2+1+5\\ =8\\ \\ \left(\mathrm{iii}\right)\text{For}\mathrm{a}=-1,{\mathrm{a}}^3+{\mathrm{a}}^2+\mathrm{a}+5={(-1)}^3+{(-1)}^2+(-1)+5\\ =(-1)+1+(-1)+5=4\end{array}$

Q.12

$\begin{array}{l}\left(\mathrm{a}\right)\mathrm{Add}:\mathrm{p}(\mathrm{p}-\mathrm{q}),\mathrm{q}(\mathrm{q}-\mathrm{r})\mathrm{and}\mathrm{r}(\mathrm{r}-\mathrm{p})\\ \left(\mathrm{b}\right)\mathrm{Add}:2\mathrm{x}(\mathrm{z}-\mathrm{x}-\mathrm{y})\mathrm{and}2\mathrm{y}(\mathrm{z}-\mathrm{y}-\mathrm{x})\\ \left(\mathrm{c}\right)\mathrm{Subtract}:3\mathrm{l}(\mathrm{l}-4\mathrm{m}+5\mathrm{n})\mathrm{from}4\mathrm{l}(10\mathrm{n}-3\mathrm{m}+2\mathrm{l})\\ \left(\mathrm{d}\right)\mathrm{Subtract}:3\mathrm{a}(\mathrm{a}+\mathrm{b}+\mathrm{c})-2\mathrm{b}(\mathrm{a}-\mathrm{b}+\mathrm{c})\\ \mathrm{from}4\mathrm{c}(-\mathrm{a}+\mathrm{b}+\mathrm{c})\end{array}$

Ans

$\begin{array}{l}\text{First we will simplify the given expressions and then add or subtract them}.\\ \left(\mathrm{a}\right)\mathrm{p}(\mathrm{p}-\mathrm{q})={\mathrm{p}}^2-\mathrm{pq}\\ \mathrm{q}(\mathrm{q}-\mathrm{r})={\mathrm{q}}^2-\mathrm{qr}\\ \mathrm{r}(\mathrm{r}-\mathrm{p})={\mathrm{r}}^2-\mathrm{rp}\\ \mathrm{Now},\text{adding all the expressions}\\ {\mathrm{p}}^2-\mathrm{pq}\\ {\mathrm{q}}^2-\mathrm{qr}\\ \text{+}\underset{¯}{{\mathrm{r}}^2-\mathrm{rp}}\\ \underset{¯}{\begin{array}{l}{\mathrm{p}}^2-\mathrm{pq}+{\mathrm{q}}^2-\mathrm{qr}+{\mathrm{r}}^2-\mathrm{rp}\\ \end{array}}\\ \left(\mathrm{b}\right)2\mathrm{x}(\mathrm{z}-\mathrm{x}-\mathrm{y})=2\mathrm{xz}-2{\mathrm{x}}^2-2\mathrm{xy}\\ 2\mathrm{y}(\mathrm{z}-\mathrm{y}-\mathrm{x})=2\mathrm{yz}-2{\mathrm{y}}^2-2\mathrm{yx}\\ \\ \mathrm{Now},\text{adding all the expressions}\\ 2\mathrm{xz}-2{\mathrm{x}}^2-2\mathrm{xy}\\ +\underset{¯}{-2\mathrm{yx}+2\mathrm{yz}-2{\mathrm{y}}^2}\\ \underset{¯}{2\mathrm{xz}-2{\mathrm{x}}^2-4\mathrm{xy}+2\mathrm{yz}-2{\mathrm{y}}^2}\\ \\ \\ \left(\mathrm{c}\right)3\mathrm{l}(\mathrm{l}-4\mathrm{m}+5\mathrm{n})=3{\mathrm{l}}^2-12\mathrm{lm}+15\ln \\ 4\mathrm{l}(10\mathrm{n}-3\mathrm{m}+2\mathrm{l})=40\ln -12\mathrm{lm}+8{\mathrm{l}}^2\\ \\ \mathrm{On}\text{subtracting them, we get}\\ 40\ln -12\mathrm{lm}+8{\mathrm{l}}^2\\ \underset{¯}{\begin{array}{l}15\ln -12\mathrm{lm}+3{\mathrm{l}}^2\\ (-\text{)\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}(+)\text{(}-\text{)}\end{array}}\\ {\text{25ln +5l}}^{\text{2}}\end{array}$

$\begin{array}{l}\left(\mathrm{d}\right)3\mathrm{a}(\mathrm{a}+\mathrm{b}+\mathrm{c})-2\mathrm{b}(\mathrm{a}-\mathrm{b}+\mathrm{c})=3{\mathrm{a}}^2+\mathrm{ab}+3\mathrm{ac}+2{\mathrm{b}}^2-2\mathrm{bc}\\ 4\mathrm{c}(-\mathrm{a}+\mathrm{b}+\mathrm{c})=-4\mathrm{ac}+4\mathrm{cb}+4{\mathrm{c}}^2\\ \mathrm{On}\text{subtracting them, we get}\\ -4\mathrm{ac}+4\mathrm{cb}+4{\mathrm{c}}^2\\ \underset{¯}{\begin{array}{l}+3\mathrm{ac}-2\mathrm{bc}\text{+}3{\mathrm{a}}^2+\mathrm{ab}+2{\mathrm{b}}^2\\ (-)\text{(+) (}-\text{) (}-\text{) (}-\text{)}\end{array}}\\ \underset{¯}{\begin{array}{l}-{\text{7ac + 6bc + 4c}}^{\text{2}}-3{\mathrm{a}}^2-\mathrm{ab}-2{\mathrm{b}}^2\\ \end{array}}\\ \\ \end{array}$

Q.13

$\begin{array}{l}\mathrm{Multiply}\mathrm{the}\mathrm{binomials}.\\ \left(\mathrm{i}\right)(2\mathrm{x}+5)\mathrm{and}(4\mathrm{x}-3)\left(\mathrm{ii}\right)(\mathrm{y}-8)\mathrm{and}(3\mathrm{y}-4)\\ \left(\mathrm{iii}\right)(2.5\mathrm{l}-0.5\mathrm{m})\mathrm{and}(2.5\mathrm{l}+0.5\mathrm{m})\left(\mathrm{iv}\right)(\mathrm{a}+3\mathrm{b})\mathrm{and}(\mathrm{x}+5)\\ \left(\mathrm{v}\right)(2\mathrm{pq}+3{\mathrm{q}}^2)\mathrm{and}(3\mathrm{pq}-2{\mathrm{q}}^2)\\ \left(\mathrm{vi}\right)\frac{3}{4}{\mathrm{a}}^2+3{\mathrm{b}}^2\mathrm{and}4({\mathrm{a}}^2-\frac{2}{3}{\mathrm{b}}^2)\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)(2\mathrm{x}+5)\mathrm{and}(4\mathrm{x}-3)=2\mathrm{x}\times (4\mathrm{x}-3)+5\times (4\mathrm{x}-3)\\ =8{\mathrm{x}}^2-6\mathrm{x}+20\mathrm{x}-15\\ =8{\mathrm{x}}^2+14\mathrm{x}-15\\ \\ \left(\mathrm{ii}\right)(\mathrm{y}-8)\mathrm{and}(3\mathrm{y}-4)=\mathrm{y}\times (3\mathrm{y}-4)-8\times (3\mathrm{y}-4)\\ =3{\mathrm{y}}^2-4\mathrm{y}-24\mathrm{y}+32\\ =3{\mathrm{y}}^2+32-28\mathrm{y}\\ \\ \left(\mathrm{iii}\right)(2.5\mathrm{l}-0.5\mathrm{m})\mathrm{and}(2.5\mathrm{l}+0.5\mathrm{m})=2.5\mathrm{l}\times (2.5\mathrm{l}+0.5\mathrm{m})-0.5\mathrm{m}\times (2.5\mathrm{l}+0.5\mathrm{m})\\ =6.25{\mathrm{l}}^2+1.25\mathrm{lm}-1.25\mathrm{lm}-0.25{\mathrm{m}}^2\\ =6.25{\mathrm{l}}^2-0.25{\mathrm{m}}^2\\ \\ \left(\mathrm{iv}\right)(\mathrm{a}+3\mathrm{b})\mathrm{and}(\mathrm{x}+5)=\mathrm{a}\times (\mathrm{x}+5)+3\mathrm{b}\times (\mathrm{x}+5)\\ =\mathrm{ax}+5\mathrm{a}+3\mathrm{bx}+15\mathrm{b}\\ \\ \left(\mathrm{v}\right)(2\mathrm{pq}+3{\mathrm{q}}^2)\mathrm{and}(3\mathrm{pq}-2{\mathrm{q}}^2)=2\mathrm{pq}\times (3\mathrm{pq}-2{\mathrm{q}}^2)+3{\mathrm{q}}^2\times (3\mathrm{pq}-2{\mathrm{q}}^2)\\ =6{\mathrm{p}}^2{\mathrm{q}}^2-4{\mathrm{pq}}^3+9{\mathrm{pq}}^3-6{\mathrm{q}}^4\\ =6{\mathrm{p}}^2{\mathrm{q}}^2+5{\mathrm{pq}}^3-6{\mathrm{q}}^4\\ \\ \left(\mathrm{vi}\right)\frac{3}{4}{\mathrm{a}}^2+3{\mathrm{b}}^2\mathrm{and}4({\mathrm{a}}^2-\frac{2}{3}{\mathrm{b}}^2)=\left(\frac{3}{4}{\mathrm{a}}^2\right)\times 4({\mathrm{a}}^2-\frac{2}{3}{\mathrm{b}}^2)+\left(3{\mathrm{b}}^2\right)\times 4({\mathrm{a}}^2-\frac{2}{3}{\mathrm{b}}^2)\\ =3{\mathrm{a}}^4-2{\mathrm{a}}^2{\mathrm{b}}^2+12{\mathrm{a}}^2{\mathrm{b}}^2-8{\mathrm{b}}^4\\ =3{\mathrm{a}}^4+10{\mathrm{a}}^2{\mathrm{b}}^2-8{\mathrm{b}}^4\end{array}$

Q.14 Find the product.

(i) (5 – 2x) (3 + x)

(ii) (x + 7y) (7x – y)

(iii) (a² + b) (a + b²)

(iv) (p² – q²) (2p + q)

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\left(5–2\mathrm{x}\right)\left(3+\mathrm{x}\right)=5\times \left(3+\mathrm{x}\right)-2\mathrm{x}\left(3+\mathrm{x}\right)\\ =15+5\mathrm{x}-6\mathrm{x}-2{\mathrm{x}}^2\\ =15-\mathrm{x}-2{\mathrm{x}}^2\\ \\ \left(\mathrm{ii}\right)\left(\mathrm{x}+7\mathrm{y}\right)\left(7\mathrm{x}–\mathrm{y}\right)=\mathrm{x}\times \left(7\mathrm{x}–\mathrm{y}\right)+7\mathrm{y}\times \left(7\mathrm{x}–\mathrm{y}\right)\\ =7{\mathrm{x}}^2-\mathrm{xy}+49\mathrm{xy}-7{\mathrm{y}}^2\\ =7{\mathrm{x}}^2+48\mathrm{xy}-7{\mathrm{y}}^2\\ \\ \left(\mathrm{iii}\right)\left({\mathrm{a}}^2+\mathrm{b}\right)\left(\mathrm{a}+{\mathrm{b}}^2\right)={\mathrm{a}}^2\times \left(\mathrm{a}+{\mathrm{b}}^2\right)+\mathrm{b}\times \left(\mathrm{a}+{\mathrm{b}}^2\right)\\ ={\mathrm{a}}^3+{\mathrm{a}}^2{\mathrm{b}}^2+\mathrm{ba}+{\mathrm{b}}^3\\ \\ \left(\mathrm{iv}\right)\left({\mathrm{p}}^2–{\mathrm{q}}^2\right)\left(2\mathrm{p}+\mathrm{q}\right)={\mathrm{p}}^2\times \left(2\mathrm{p}+\mathrm{q}\right)–{\mathrm{q}}^2\times \left(2\mathrm{p}+\mathrm{q}\right)\\ =2{\mathrm{p}}^3+{\mathrm{p}}^2\mathrm{q}-2{\mathrm{q}}^3-{\mathrm{q}}^3\\ =2{\mathrm{p}}^3+{\mathrm{p}}^2\mathrm{q}-3{\mathrm{q}}^3\end{array}$

Q.15 Simplify.

(i) (x² – 5) (x + 5) + 25

(ii) (a² + 5) (b³ + 3) + 5

(iii) (t + s²) (t² – s)

(iv) (a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd)

(v) (x + y)(2x + y) + (x + 2y)(x – y)

(vi) (x + y)(x² – xy + y²)

(vii) (1.5x – 4y)(1.5x + 4y + 3) – 4.5x + 12y

(viii) (a + b + c)(a + b – c)

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{}({\mathrm{x}}^2–5)(\mathrm{x}+5)+25\\ ={\text{x}}^{\text{2}}(\mathrm{x}+5)-5(\mathrm{x}+5)+25\\ ={\mathrm{x}}^3+5{\mathrm{x}}^2-5\mathrm{x}-25+25\\ ={\mathrm{x}}^3+5{\mathrm{x}}^2-5\mathrm{x}\\ \left(\mathrm{ii}\right)\mathrm{}({\mathrm{a}}^2+5)({\mathrm{b}}^3+3)+5\\ ={\mathrm{a}}^2({\mathrm{b}}^3+3)+5({\mathrm{b}}^3+3)+5\\ ={\mathrm{a}}^2{\mathrm{b}}^3+3{\mathrm{a}}^2+5{\mathrm{b}}^3+15+5\\ ={\mathrm{a}}^2{\mathrm{b}}^3+3{\mathrm{a}}^2+5{\mathrm{b}}^3+20\\ \\ \left(\mathrm{iii}\right)\mathrm{}(\mathrm{t}+{\mathrm{s}}^2)({\mathrm{t}}^2–\mathrm{s})\\ =\mathrm{t}({\mathrm{t}}^2–\mathrm{s})+{\mathrm{s}}^2({\mathrm{t}}^2–\mathrm{s})\\ ={\mathrm{t}}^3-\mathrm{ts}+{\mathrm{s}}^2{\mathrm{t}}^2-{\mathrm{s}}^3\\ \\ \left(\mathrm{iv}\right)(\mathrm{a}+\mathrm{b})(\mathrm{c}–\mathrm{d})+(\mathrm{a}–\mathrm{b})(\mathrm{c}+\mathrm{d})+2(\mathrm{ac}+\mathrm{bd})\\ =\mathrm{a}(\mathrm{c}–\mathrm{d})+\mathrm{b}(\mathrm{c}–\mathrm{d})+\mathrm{a}(\mathrm{c}+\mathrm{d})-\mathrm{b}(\mathrm{c}+\mathrm{d})+2(\mathrm{ac}+\mathrm{bd})\\ =\mathrm{ac}-\mathrm{ad}+\mathrm{bc}-\mathrm{bd}+\mathrm{ac}+\mathrm{ad}-\mathrm{bc}-\mathrm{bd}+2\mathrm{ac}+2\mathrm{bd}\\ =4\mathrm{ac}\\ \left(\mathrm{v}\right)\mathrm{}(\mathrm{x}+\mathrm{y})(2\mathrm{x}+\mathrm{y})+(\mathrm{x}+2\mathrm{y})(\mathrm{x}–\mathrm{y})\\ =\mathrm{x}(2\mathrm{x}+\mathrm{y})+\mathrm{y}(2\mathrm{x}+\mathrm{y})+\mathrm{x}(\mathrm{x}–\mathrm{y})+2\mathrm{y}(\mathrm{x}–\mathrm{y})\\ =2{\mathrm{x}}^2+\mathrm{xy}+2\mathrm{xy}+{\mathrm{y}}^2+{\mathrm{x}}^2-\mathrm{xy}+2\mathrm{xy}-2{\mathrm{y}}^2\\ =3{\mathrm{x}}^2-{\mathrm{y}}^2+4\mathrm{xy}\\ \\ \mathrm{}\left(\mathrm{vi}\right)(\mathrm{x}+\mathrm{y})({\mathrm{x}}^2–\mathrm{xy}+{\mathrm{y}}^2)\\ =\mathrm{x}({\mathrm{x}}^2–\mathrm{xy}+{\mathrm{y}}^2)+\mathrm{y}({\mathrm{x}}^2–\mathrm{xy}+{\mathrm{y}}^2)\\ ={\mathrm{x}}^3-{\mathrm{x}}^2\mathrm{y}+{\mathrm{xy}}^2+{\mathrm{yx}}^2-{\mathrm{xy}}^2+{\mathrm{y}}^3\\ ={\mathrm{x}}^3+{\mathrm{y}}^3\\ \\ \left(\mathrm{vii}\right)(1.5\mathrm{x}–4\mathrm{y})(1.5\mathrm{x}+4\mathrm{y}+3)–4.5\mathrm{x}+12\mathrm{y}\\ =1.5\mathrm{x}(1.5\mathrm{x}+4\mathrm{y}+3)-4\mathrm{y}(1.5\mathrm{x}+4\mathrm{y}+3)–4.5\mathrm{x}+12\mathrm{y}\\ =2.25{\mathrm{x}}^2+6\mathrm{xy}+4.5\mathrm{x}-6\mathrm{xy}-16{\mathrm{y}}^2-12\mathrm{y}-4.5\mathrm{x}+12\mathrm{y}\\ =2.25{\mathrm{x}}^2-16{\mathrm{y}}^2\\ \\ \left(\mathrm{viii}\right)(\mathrm{a}+\mathrm{b}+\mathrm{c})(\mathrm{a}+\mathrm{b}–\mathrm{c})\mathrm{}\\ =\mathrm{a}(\mathrm{a}+\mathrm{b}–\mathrm{c})\mathrm{}+\mathrm{b}(\mathrm{a}+\mathrm{b}–\mathrm{c})+\mathrm{c}\mathrm{}(\mathrm{a}+\mathrm{b}–\mathrm{c})\mathrm{}\\ ={\mathrm{a}}^2+\mathrm{ab}-\mathrm{ac}+\mathrm{ab}+{\mathrm{b}}^2-\mathrm{bc}+\mathrm{ac}+\mathrm{cb}-{\mathrm{c}}^2\\ ={\mathrm{a}}^2+{\mathrm{b}}^2-{\mathrm{c}}^2+2\mathrm{ab}\end{array}$

Q.16

$\begin{array}{l}\mathrm{Use}\mathrm{a}\mathrm{suitable}\mathrm{identity}\mathrm{to}\mathrm{get}\mathrm{each}\mathrm{of}\mathrm{the}\mathrm{following}\mathrm{products}.\\ \left(\mathrm{i}\right)(\mathrm{x}+3)(\mathrm{x}+3)\left(\mathrm{ii}\right)(2\mathrm{y}+5)(2\mathrm{y}+5)\\ \left(\mathrm{iii}\right)\left(2\mathrm{a}-7\right)\left(2\mathrm{a}-7\right)\left(\mathrm{iv}\right)\left(3\mathrm{a}-\frac{1}{2}\right)\left(3\mathrm{a}-\frac{1}{2}\right)\\ \left(\mathrm{v}\right)(1.1\mathrm{m}-0.4)(1.1\mathrm{m}+0.4)\left(\mathrm{vi}\right)({\mathrm{a}}^2+{\mathrm{b}}^2)(-{\mathrm{a}}^2+{\mathrm{b}}^2)\\ \left(\mathrm{vii}\right)(6\mathrm{x}-7)(6\mathrm{x}+7)\left(\mathrm{viii}\right)(-\mathrm{a}+\mathrm{c})(-\mathrm{a}+\mathrm{c})\\ \left(\mathrm{ix}\right)\left(\frac{\mathrm{x}}{2}+\frac{3\mathrm{y}}{4}\right)\left(\frac{\mathrm{x}}{2}+\frac{3\mathrm{y}}{4}\right)\left(\mathrm{x}\right)\left(7\mathrm{a}-9\mathrm{b}\right)\left(7\mathrm{a}-9\mathrm{b}\right)\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\left(\mathrm{x}+3\right)\left(\mathrm{x}+3\right)={(\mathrm{x}+3)}^2\\ ={\left(\mathrm{x}\right)}^2+2\times 3\times \mathrm{x}+{\left(3\right)}^2\left\{\begin{array}{l}\text{By using the identity}\\ {\text{(a+b)}}^{\text{2}}={\mathrm{a}}^2+{\mathrm{b}}^2+2\mathrm{ab}\end{array}\right\}\\ ={\mathrm{x}}^2+6\mathrm{x}+9\\ \\ \left(\mathrm{ii}\right)\left(2\mathrm{y}+5\right)\left(2\mathrm{y}+5\right)={(2\mathrm{y}+5)}^2\\ ={\left(2\mathrm{y}\right)}^2+2\times 5\times 2\mathrm{y}+{\left(5\right)}^2\left\{\begin{array}{l}\text{By using the identity}\\ {\text{(a+b)}}^{\text{2}}={\mathrm{a}}^2+{\mathrm{b}}^2+2\mathrm{ab}\end{array}\right\}\\ =4{\mathrm{y}}^2+20\mathrm{y}+25\\ \\ \left(\mathrm{iii}\right)\left(2\mathrm{a}-7\right)\left(2\mathrm{a}-7\right)={(2\mathrm{a}-7)}^2\\ ={\left(2\mathrm{a}\right)}^2-2\times 2\mathrm{a}\times 7+{\left(7\right)}^2\left\{\begin{array}{l}\text{By using the identity}\\ {(\text{a}-\text{b})}^{\text{2}}={\mathrm{a}}^2+{\mathrm{b}}^2-2\mathrm{ab}\end{array}\right\}\\ =4{\mathrm{a}}^2-28\mathrm{a}+49\\ \\ \left(\mathrm{iv}\right)\left(3\mathrm{a}-\frac{1}{2}\right)\left(3\mathrm{a}-\frac{1}{2}\right)={(3\mathrm{a}-\frac{1}{2})}^2\\ ={\left(3\mathrm{a}\right)}^2-2\times 3\mathrm{a}\times \frac{1}{2}+{\left(\frac{1}{2}\right)}^2\left\{\begin{array}{l}\text{By using the identity}\\ {(\text{a}-\text{b})}^{\text{2}}={\mathrm{a}}^2+{\mathrm{b}}^2-2\mathrm{ab}\end{array}\right\}\\ =9{\mathrm{a}}^2-3\mathrm{a}+\frac{1}{4}\\ \left(\mathrm{v}\right)(1.1\mathrm{m}-0.4)(1.1\mathrm{m}+0.4)\\ ={(1.1\mathrm{m})}^2-{(0.4)}^2\left\{\begin{array}{l}\text{By using the identity}\\ \text{(a+b)(a}-{\text{b)=a}}^{\text{2}}-{\text{b}}^{\text{2}}\end{array}\right\}\\ =1.21{\mathrm{m}}^2-0.16\\ \\ \left(\mathrm{vi}\right)({\mathrm{a}}^2+{\mathrm{b}}^2)(-{\mathrm{a}}^2+{\mathrm{b}}^2)\\ ={\left({\mathrm{b}}^2\right)}^2-{\left({\mathrm{a}}^2\right)}^2\left\{\begin{array}{l}\text{By using the identity}\\ \text{(a+b)(a}-{\text{b)=a}}^{\text{2}}-{\text{b}}^{\text{2}}\end{array}\right\}\\ ={\mathrm{b}}^4-{\mathrm{a}}^4\\ \\ \left(\mathrm{vii}\right)(6\mathrm{x}-7)(6\mathrm{x}+7)\\ ={\left(6\mathrm{x}\right)}^2-{\left(7\right)}^2\left\{\begin{array}{l}\text{By using the identity}\\ \text{(a+b)(a}-{\text{b)=a}}^{\text{2}}-{\text{b}}^{\text{2}}\end{array}\right\}\\ =36{\mathrm{x}}^2-49\\ \\ \left(\mathrm{viii}\right)(-\mathrm{a}+\mathrm{c})(-\mathrm{a}+\mathrm{c})\\ ={(-\mathrm{a}+\mathrm{c})}^2\left\{\begin{array}{l}\text{By using the identity}\\ {\text{(a+b)}}^{\text{2}}={\mathrm{a}}^2+{\mathrm{b}}^2+2\mathrm{ab}\end{array}\right\}\\ ={\mathrm{c}}^2+{\mathrm{a}}^2-2\mathrm{ac}\end{array}$

$\begin{array}{l}\left(\mathrm{ix}\right)(\frac{\mathrm{x}}{2}+\frac{3\mathrm{y}}{4})(\frac{\mathrm{x}}{2}+\frac{3\mathrm{y}}{4})={(\frac{\mathrm{x}}{2}+\frac{3\mathrm{y}}{4})}^2\\ ={\left(\frac{\mathrm{x}}{2}\right)}^2+{\left(\frac{3\mathrm{y}}{4}\right)}^2+\text{2}\left(\frac{\mathrm{x}}{2}\right)\left(\frac{3\mathrm{y}}{4}\right)\left\{\begin{array}{l}\text{By using the identity}\\ {\text{(a + b)}}^{\text{2}}={\mathrm{a}}^2+{\mathrm{b}}^2+2\mathrm{ab}\end{array}\right\}\\ \text{=}\frac{{\mathrm{x}}^2}{4}+\frac{9{\mathrm{y}}^2}{16}+\frac{3\mathrm{xy}}{4}\\ \\ \left(\mathrm{x}\right)\left(7\mathrm{a}-9\mathrm{b}\right)\left(7\mathrm{a}-9\mathrm{b}\right)={(7\mathrm{a}-9\mathrm{b})}^2\\ ={\left(7\mathrm{a}\right)}^2+{\left(9\mathrm{b}\right)}^2-\text{2}\left(7\mathrm{a}\right)\left(9\mathrm{b}\right)\left\{\begin{array}{l}\text{By using the identity}\\ {(\text{a}-\text{b})}^{\text{2}}={\mathrm{a}}^2+{\mathrm{b}}^2-2\mathrm{ab}\end{array}\right\}\\ =49{\mathrm{a}}^2+81{\mathrm{b}}^2-126\mathrm{ab}\end{array}$

Q.17

$\begin{array}{l}\mathrm{Use}\mathrm{the}\mathrm{identity}(\mathrm{x}+\mathrm{a})(\mathrm{x}+\mathrm{b})={\mathrm{x}}^2+(\mathrm{a}+\mathrm{b})\mathrm{x}+\mathrm{ab}\\ \mathrm{to}\mathrm{find}\mathrm{the}\mathrm{following}\mathrm{products}.\\ \left(\mathrm{i}\right)(\mathrm{x}+3)(\mathrm{x}+7)\\ \left(\mathrm{ii}\right)(4\mathrm{x}+5)(4\mathrm{x}+1)\\ \left(\mathrm{iii}\right)(4\mathrm{x}-5)(4\mathrm{x}-1)\\ \left(\mathrm{iv}\right)(4\mathrm{x}+5)(4\mathrm{x}-1)\\ \left(\mathrm{v}\right)(2\mathrm{x}+5\mathrm{y})(2\mathrm{x}+3\mathrm{y})\\ \left(\mathrm{vi}\right)(2{\mathrm{a}}^2+9)(2{\mathrm{a}}^2+5)\\ \left(\mathrm{vii}\right)(\mathrm{xyz}-4)(\mathrm{xyz}-2)\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)(\mathrm{x}+3)(\mathrm{x}+7)={\mathrm{x}}^2+(3+7)\mathrm{x}+3\times 7\\ ={\mathrm{x}}^2+10\mathrm{x}+21\\ \\ \left(\mathrm{ii}\right)\left(4\mathrm{x}+5\right)\left(4\mathrm{x}+1\right)={\left(4\mathrm{x}\right)}^2+\left(5+1\right)\left(4\mathrm{x}\right)+5\times 1\\ =16{\mathrm{x}}^2+24\mathrm{x}+5\\ \\ \left(\mathrm{iii}\right)\left(4\mathrm{x}-5\right)\left(4\mathrm{x}-1\right)={\left(4\mathrm{x}\right)}^2+[(-5)+(–1)]\left(4\mathrm{x}\right)+(–5)\times (-1)\\ =16{\mathrm{x}}^2-24\mathrm{x}+5\\ \\ \left(\mathrm{iv}\right)\left(4\mathrm{x}+5\right)\left(4\mathrm{x}-1\right)={\left(4\mathrm{x}\right)}^2+[5+(-1)]\left(4\mathrm{x}\right)+5\times (-1)\\ =16{\mathrm{x}}^2+16\mathrm{x}-5\\ \\ \left(\mathrm{v}\right)\left(2\mathrm{x}+5\mathrm{y}\right)\left(2\mathrm{x}+3\mathrm{y}\right)={\left(2\mathrm{x}\right)}^2+\left(5\mathrm{y}+3\mathrm{y}\right)\left(2\mathrm{x}\right)+5\mathrm{y}\times 3\mathrm{y}\\ =4{\mathrm{x}}^2+16\mathrm{xy}+15{\mathrm{y}}^2\\ \\ \left(\mathrm{vi}\right)\left(2{\mathrm{a}}^2+9\right)\left(2{\mathrm{a}}^2+5\right)={\left(2{\mathrm{a}}^2\right)}^2+\left(9+5\right)\left(2{\mathrm{a}}^2\right)+9\times 5\\ =4{\mathrm{a}}^4+28{\mathrm{a}}^2+45\\ \\ \left(\mathrm{vii}\right)\left(\mathrm{xyz}-4\right)\left(\mathrm{xyz}-2\right)={\left(\mathrm{xyz}\right)}^2+[(-4)+(–2)]\left(\mathrm{xyz}\right)+(–4)\times (-2)\\ ={\mathrm{x}}^2{\mathrm{y}}^2{\mathrm{z}}^2-6\mathrm{xyz}+8\end{array}$