Page 69

Think and Reflect

Question. Try and find other patterns like this one. For example, you could consider 4 consecutive squares and see if you can find a pattern.

Answer— Consider four consecutive numbers $n,n+1,n+2,n+3$. Their squares are:

$$n^2,{(n+1)}^2,{(n+2)}^2,{(n+3)}^2$$

Add the first and fourth squares, and subtract the second and third squares:

$$\begin{array}{cc}\hfill \hfill & \hfill n^2+{(n+3)}^2-[{(n+1)}^2+{(n+2)}^2]\hfill \\ \hfill \hfill & \hfill =n^2+n^2+6n+9-(n^2+2n+1+n^2+4n+4)\hfill \\ \hfill \hfill & \hfill =4\hfill \end{array}$$

Thus, for any four consecutive square numbers,

$$\mathrm{\text{first square}}+\mathrm{\text{fourth square}}-\mathrm{\text{second square}}-\mathrm{\text{third square}}=4$$

For example,

$$1+16-4-9=4.$$

Page 70

Question 1. Think of numbers $𝐚$ and $𝐛$ where $𝐚$ and $𝐛$ do not represent lengths of line segments. What if $𝐚$ and $𝐛$ are negative numbers? Does ${(𝐚\mathbf{+}𝐛)}^{\mathrm{𝟐}}\mathbf{=}{𝐚}^{\mathrm{𝟐}}\mathbf{+}\mathrm{𝟐}𝐚𝐛\mathbf{+}{𝐛}^{\mathrm{𝟐}}$ still work?

Answer— Yes. For example, let $a=-2$ and $b=-3$.

$${(a+b)}^2={(-2-3)}^2={(-5)}^2=25$$

Also,

$$a^2+2ab+b^2=4+2(-2)(-3)+9=4+12+9=25.$$

Therefore, the identity is also true for negative values.

Question 2. What is the difference between an equation and an identity?

Answer—

• An equation is true only for one or more particular values of its variables. For example, $x^2-1=24$ is true only for $x=5$ or $x=-5$.
• An identity is true for every permissible value of its variables. For example, ${(x+y)}^2=x^2+2xy+y^2$ is true for all values of $x$ and $y$.

Page 71

Question 1. By now you must have observed that ${(𝐚\mathbf{+}𝐛)}^{\mathrm{𝟐}}\mathbf{\ne }{𝐚}^{\mathrm{𝟐}}\mathbf{+}{𝐛}^{\mathrm{𝟐}}$. But can you find out which one of them will be greater?

Answer—

$${(a+b)}^2-(a^2+b^2)=2ab.$$

Therefore:

• if $ab>0$, then ${(a+b)}^2>a^2+b^2$;
• if $ab<0$, then ${(a+b)}^2<a^2+b^2$;
• if $ab=0$, then ${(a+b)}^2=a^2+b^2$.

Thus, which expression is greater depends upon the sign of $ab$.

Question 2. If $𝐚\mathbf{=}\mathrm{𝟏𝟎}$ and $𝐛\mathbf{=}\mathrm{𝟐}$, what are the values of ${(𝐚\mathbf{+}𝐛)}^{\mathrm{𝟐}}$ and ${𝐚}^{\mathrm{𝟐}}\mathbf{+}{𝐛}^{\mathrm{𝟐}}$?

Answer—

$${(a+b)}^2={(10+2)}^2={12}^2=144$$

and

$$a^2+b^2={10}^2+2^2=100+4=104.$$

Hence,

$${(a+b)}^2=144,a^2+b^2=104$$

and ${(a+b)}^2>a^2+b^2$.

Question 3. Is ${(𝐚\mathbf{+}𝐛)}^{\mathrm{𝟐}}\mathbf{>}{𝐚}^{\mathrm{𝟐}}\mathbf{+}{𝐛}^{\mathrm{𝟐}}$ true for all numbers $𝐚$ and $𝐛$?

Answer— No. For example, take $a=2$ and $b=-1$.

$${(a+b)}^2={(2-1)}^2=1$$

but

$$a^2+b^2=2^2+{(-1)}^2=5.$$

Therefore, ${(a+b)}^2<a^2+b^2$ in this case.

Think and Reflect

1. What can you say about $𝐚$ and $𝐛$ if ${(𝐚\mathbf{+}𝐛)}^{\mathrm{𝟐}}\mathbf{<}{𝐚}^{\mathrm{𝟐}}\mathbf{+}{𝐛}^{\mathrm{𝟐}}$?

Answer—

$$a^2+2ab+b^2<a^2+b^2$$

$$2ab<0\Rightarrow ab<0.$$

Therefore, $a$ and $b$ have opposite signs.

2. What can you say about $𝐚$ and $𝐛$ if ${(𝐚\mathbf{+}𝐛)}^{\mathrm{𝟐}}\mathbf{>}{𝐚}^{\mathrm{𝟐}}\mathbf{+}{𝐛}^{\mathrm{𝟐}}$?

Answer—

$$a^2+2ab+b^2>a^2+b^2$$

$$2ab>0\Rightarrow ab>0.$$

Therefore, $a$ and $b$ have the same non-zero sign.

3. When will ${(𝐚\mathbf{+}𝐛)}^{\mathrm{𝟐}}$ be equal to ${𝐚}^{\mathrm{𝟐}}\mathbf{+}{𝐛}^{\mathrm{𝟐}}$?

Answer—

$$a^2+2ab+b^2=a^2+b^2$$

$$2ab=0\Rightarrow ab=0.$$

Therefore, equality holds when $a=0$, $b=0$, or both are zero.

4. What term will decide which is larger: ${(𝐚\mathbf{+}𝐛)}^{\mathrm{𝟐}}$ or ${𝐚}^{\mathrm{𝟐}}\mathbf{+}{𝐛}^{\mathrm{𝟐}}$? Use the expansion of ${(𝐚\mathbf{+}𝐛)}^{\mathrm{𝟐}}$ to decide.

Answer— Since

$${(a+b)}^2=a^2+b^2+2ab,$$

the term $2ab$ decides which expression is larger.

Exercise Set 4.1 — Pages 71–72

1. Using the identity ${(𝐚\mathbf{+}𝐛)}^{\mathrm{𝟐}}\mathbf{=}{𝐚}^{\mathrm{𝟐}}\mathbf{+}\mathrm{𝟐}𝐚𝐛\mathbf{+}{𝐛}^{\mathrm{𝟐}}$, expand the following:

(i) ${(\mathrm{𝟕}𝐱\mathbf{+}\mathrm{𝟒}𝐲)}^{\mathrm{𝟐}}$

Answer—

$$\begin{array}{cc}\hfill \hfill & \hfill {(7x+4y)}^2={\left(7x\right)}^2+2\left(7x\right)\left(4y\right)+{\left(4y\right)}^2\hfill \\ \hfill \hfill & \hfill =49x^2+56xy+16y^2\hfill \end{array}$$

(ii) ${(\frac{\mathrm{𝟕}}{\mathrm{𝟓}}𝐱\mathbf{+}\frac{\mathrm{𝟑}}{\mathrm{𝟐}}𝐲)}^{\mathrm{𝟐}}$

Answer—

$$\begin{array}{cc}\hfill \hfill & \hfill {(\frac{7}{5}x+\frac{3}{2}y)}^2={\left(\frac{7}{5}x\right)}^2+2\left(\frac{7}{5}x\right)\left(\frac{3}{2}y\right)+{\left(\frac{3}{2}y\right)}^2\hfill \\ \hfill \hfill & \hfill =\frac{49}{25}{x}^2+\frac{21}{5}xy+\frac{9}{4}{y}^2\hfill \end{array}$$

(iii) ${(\mathbf{2.5}𝐩\mathbf{+}\mathbf{1.5}𝐪)}^{\mathrm{𝟐}}$

Answer—

$$\begin{array}{cc}\hfill \hfill & \hfill {(2.5p+1.5q)}^2={\left(2.5p\right)}^2+2\left(2.5p\right)\left(1.5q\right)+{\left(1.5q\right)}^2\hfill \\ \hfill \hfill & \hfill =6.25p^2+7.5pq+2.25q^2\hfill \end{array}$$

(iv) ${(\frac{\mathrm{𝟑}}{\mathrm{𝟒}}𝐬\mathbf{+}\mathrm{𝟖}𝐭)}^{\mathrm{𝟐}}$

Answer—

$$\begin{array}{cc}\hfill \hfill & \hfill {(\frac{3}{4}s+8t)}^2={\left(\frac{3}{4}s\right)}^2+2\left(\frac{3}{4}s\right)\left(8t\right)+{\left(8t\right)}^2\hfill \\ \hfill \hfill & \hfill =\frac{9}{16}{s}^2+12st+64t^2\hfill \end{array}$$

(v) ${(𝐱\mathbf{+}\frac{\mathrm{𝟏}}{\mathrm{𝟐}𝐲})}^{\mathrm{𝟐}}$

Answer—

$$\begin{array}{cc}\hfill \hfill & \hfill {(x+\frac{1}{2y})}^2=x^2+2\left(x\right)\left(\frac{1}{2y}\right)+{\left(\frac{1}{2y}\right)}^2\hfill \\ \hfill \hfill & \hfill =x^2+\frac{x}{y}+\frac{1}{4y^2},y\ne 0\hfill \end{array}$$

(vi) ${(\frac{\mathrm{𝟏}}{𝐱}\mathbf{+}\frac{\mathrm{𝟏}}{𝐲})}^{\mathrm{𝟐}}$

Answer—

$$\begin{array}{cc}\hfill \hfill & \hfill {(\frac{1}{x}+\frac{1}{y})}^2=\frac{1}{x^2}+2\left(\frac{1}{x}\right)\left(\frac{1}{y}\right)+\frac{1}{y^2}\hfill \\ \hfill \hfill & \hfill =\frac{1}{x^2}+\frac{2}{xy}+\frac{1}{y^2},x,y\ne 0\hfill \end{array}$$

2. Using the same identity, find the values of the following:

(i) ${\left(\mathrm{𝟔𝟒}\right)}^{\mathrm{𝟐}}$

Answer—

$$\begin{array}{cc}\hfill \hfill & \hfill {64}^2={(60+4)}^2\hfill \\ \hfill \hfill & \hfill ={60}^2+2\left(60\right)\left(4\right)+4^2\hfill \\ \hfill \hfill & \hfill =3600+480+16\hfill \\ \hfill \hfill & \hfill =4096\hfill \end{array}$$

(ii) ${\left(\mathrm{𝟏𝟎𝟓}\right)}^{\mathrm{𝟐}}$

Answer—

$$\begin{array}{cc}\hfill \hfill & \hfill {105}^2={(100+5)}^2\hfill \\ \hfill \hfill & \hfill ={100}^2+2\left(100\right)\left(5\right)+5^2\hfill \\ \hfill \hfill & \hfill =10000+1000+25\hfill \\ \hfill \hfill & \hfill =11025\hfill \end{array}$$

(iii) ${\left(\mathrm{𝟐𝟎𝟓}\right)}^{\mathrm{𝟐}}$

Answer—

$$\begin{array}{cc}\hfill \hfill & \hfill {205}^2={(200+5)}^2\hfill \\ \hfill \hfill & \hfill ={200}^2+2\left(200\right)\left(5\right)+5^2\hfill \\ \hfill \hfill & \hfill =40000+2000+25\hfill \\ \hfill \hfill & \hfill =42025\hfill \end{array}$$

Page 72

Question 1. Let us try to factor $\mathrm{𝟓𝟎}{𝐩}^{\mathrm{𝟐}}\mathbf{+}\mathrm{𝟔𝟎}𝐩𝐪\mathbf{+}\mathrm{𝟏𝟖}{𝐪}^{\mathrm{𝟐}}$. What will $𝐚$ and $𝐛$ be in this case?

Answer— First, take $2$ as a common factor:

$$50p^2+60pq+18q^2=2(25p^2+30pq+9q^2).$$

Now,

$$25p^2={\left(5p\right)}^2,9q^2={\left(3q\right)}^2$$

and

$$30pq=2\left(5p\right)\left(3q\right).$$

Therefore, for the expression inside the brackets,

$$a=5p,b=3q.$$

Hence, $50p^2+60pq+18q^2=2{(5p+3q)}^2$.

Question 2. Think of a term whose square is $\mathrm{𝟓𝟎}{𝐩}^{\mathrm{𝟐}}$.

Answer—

$${\left(5\sqrt{2}p\right)}^2=50p^2.$$

Therefore, one such term is $5\sqrt{2}p$, which is the same as $\sqrt{50}p$.

Page 73

Question 1. Think of a term whose square is $\mathrm{𝟐𝟓}{𝐩}^{\mathrm{𝟐}}$. What about $\mathrm{𝟗}{𝐪}^{\mathrm{𝟐}}$?

Answer—

$${\left(5p\right)}^2=25p^2,{\left(3q\right)}^2=9q^2.$$

Therefore, the required terms are $5p$ and $3q$.

Think and Reflect

Question. What if we replace $b$ by $-b$ in ${(a+b)}^2=a^2+2ab+b^2$?

Answer— Replacing $b$ by $-b$, we get:

$$\begin{array}{cc}\hfill \hfill & \hfill {[a+(-b)]}^2=a^2+2a(-b)+{(-b)}^2\hfill \\ \hfill \hfill & \hfill {(a-b)}^2=a^2-2ab+b^2.\hfill \end{array}$$

Thus,

$${(a-b)}^2=a^2-2ab+b^2.$$

Exercise Set 4.2 — Pages 74–75

1. Factor completely:

(i) $\mathrm{𝟗}{𝐱}^{\mathrm{𝟐}}\mathbf{+}\mathrm{𝟐𝟒}𝐱𝐲\mathbf{+}\mathrm{𝟏𝟔}{𝐲}^{\mathrm{𝟐}}$

Answer—

$$\begin{array}{cc}\hfill \hfill & \hfill 9x^2+24xy+16y^2={\left(3x\right)}^2+2\left(3x\right)\left(4y\right)+{\left(4y\right)}^2\hfill \\ \hfill \hfill & \hfill ={(3x+4y)}^2\hfill \end{array}$$

(ii) $\mathrm{𝟒}{𝐬}^{\mathrm{𝟐}}\mathbf{+}\mathrm{𝟐𝟎}𝐬𝐭\mathbf{+}\mathrm{𝟐𝟓}{𝐭}^{\mathrm{𝟐}}$

Answer—

$$\begin{array}{cc}\hfill \hfill & \hfill 4s^2+20st+25t^2={\left(2s\right)}^2+2\left(2s\right)\left(5t\right)+{\left(5t\right)}^2\hfill \\ \hfill \hfill & \hfill ={(2s+5t)}^2\hfill \end{array}$$

(iii) $\mathrm{𝟒𝟗}{𝐱}^{\mathrm{𝟐}}\mathbf{+}\mathrm{𝟐𝟖}𝐱𝐲\mathbf{+}\mathrm{𝟒}{𝐲}^{\mathrm{𝟐}}$

Answer—

$$\begin{array}{cc}\hfill \hfill & \hfill 49x^2+28xy+4y^2={\left(7x\right)}^2+2\left(7x\right)\left(2y\right)+{\left(2y\right)}^2\hfill \\ \hfill \hfill & \hfill ={(7x+2y)}^2\hfill \end{array}$$

(iv) $\mathrm{𝟔𝟒}{𝐩}^{\mathrm{𝟐}}\mathbf{+}\frac{\mathrm{𝟑𝟐}}{\mathrm{𝟑}}𝐩𝐪\mathbf{+}\frac{\mathrm{𝟒}}{\mathrm{𝟗}}{𝐪}^{\mathrm{𝟐}}$

Answer—

$$\begin{array}{cc}\hfill \hfill & \hfill 64p^2+\frac{32}{3}pq+\frac{4}{9}{q}^2={\left(8p\right)}^2+2\left(8p\right)\left(\frac{2}{3}q\right)+{\left(\frac{2}{3}q\right)}^2\hfill \\ \hfill \hfill & \hfill ={(8p+\frac{2}{3}q)}^2\hfill \end{array}$$

(v) $\mathrm{𝟑}{𝐚}^{\mathrm{𝟐}}\mathbf{+}\mathrm{𝟒}𝐚𝐛\mathbf{+}\frac{\mathrm{𝟒}}{\mathrm{𝟑}}{𝐛}^{\mathrm{𝟐}}$

Answer—

$$\begin{array}{cc}\hfill \hfill & \hfill 3a^2+4ab+\frac{4}{3}{b}^2=\frac{1}{3}(9a^2+12ab+4b^2)\hfill \\ \hfill \hfill & \hfill =\frac{1}{3}[{\left(3a\right)}^2+2\left(3a\right)\left(2b\right)+{\left(2b\right)}^2]\hfill \\ \hfill \hfill & \hfill =\frac{1}{3}{(3a+2b)}^2\hfill \end{array}$$

(vi) $\frac{\mathrm{𝟗}}{\mathrm{𝟓}}{𝐬}^{\mathrm{𝟐}}\mathbf{+}\mathrm{𝟔}𝐬𝐯\mathbf{+}\mathrm{𝟓}{𝐯}^{\mathrm{𝟐}}$

Answer—

$$\begin{array}{cc}\hfill \hfill & \hfill \frac{9}{5}{s}^2+6sv+5v^2=\frac{1}{5}(9s^2+30sv+25v^2)\hfill \\ \hfill \hfill & \hfill =\frac{1}{5}[{\left(3s\right)}^2+2\left(3s\right)\left(5v\right)+{\left(5v\right)}^2]\hfill \\ \hfill \hfill & \hfill =\frac{1}{5}{(3s+5v)}^2\hfill \end{array}$$

Hint Question. $\mathrm{𝟐}$ was taken out as a common factor in Example 7. Is it possible to do something similar in Exercises (v) and (vi) above?

Answer— Yes. In part (v), take $\frac{1}{3}$ as a common factor, and in part (vi), take $\frac{1}{5}$ as a common factor. The remaining expressions become perfect-square trinomials.

2. Find the values of the following using the identity ${(𝐚\mathbf{-}𝐛)}^{\mathrm{𝟐}}\mathbf{=}{𝐚}^{\mathrm{𝟐}}\mathbf{-}\mathrm{𝟐}𝐚𝐛\mathbf{+}{𝐛}^{\mathrm{𝟐}}$.

(i) ${\left(\mathrm{𝟕𝟗}\right)}^{\mathrm{𝟐}}$

Answer—

$$\begin{array}{cc}\hfill \hfill & \hfill {79}^2={(80-1)}^2\hfill \\ \hfill \hfill & \hfill ={80}^2-2\left(80\right)\left(1\right)+1^2\hfill \\ \hfill \hfill & \hfill =6400-160+1\hfill \\ \hfill \hfill & \hfill =6241\hfill \end{array}$$

(ii) ${\left(\mathrm{𝟏𝟗𝟑}\right)}^{\mathrm{𝟐}}$

Answer—

$$\begin{array}{cc}\hfill \hfill & \hfill {193}^2={(200-7)}^2\hfill \\ \hfill \hfill & \hfill ={200}^2-2\left(200\right)\left(7\right)+7^2\hfill \\ \hfill \hfill & \hfill =40000-2800+49\hfill \\ \hfill \hfill & \hfill =37249\hfill \end{array}$$

(iii) ${\left(\mathrm{𝟐𝟗𝟗}\right)}^{\mathrm{𝟐}}$

Answer—

$$\begin{array}{cc}\hfill \hfill & \hfill {299}^2={(300-1)}^2\hfill \\ \hfill \hfill & \hfill ={300}^2-2\left(300\right)\left(1\right)+1^2\hfill \\ \hfill \hfill & \hfill =90000-600+1\hfill \\ \hfill \hfill & \hfill =89401\hfill \end{array}$$

Page 75

Question 1. What will happen if we want to find the square of the sum of three numbers $𝐚\mathbf{,}𝐛$ and $𝐜$, that is, ${(𝐚\mathbf{+}𝐛\mathbf{+}𝐜)}^{\mathrm{𝟐}}$?

Answer— Put $d=b+c$. Then:

$$\begin{array}{cc}\hfill \hfill & \hfill {(a+b+c)}^2={(a+d)}^2\hfill \\ \hfill \hfill & \hfill =a^2+2ad+d^2\hfill \\ \hfill \hfill & \hfill =a^2+2a(b+c)+{(b+c)}^2\hfill \\ \hfill \hfill & \hfill =a^2+2ab+2ac+b^2+2bc+c^2.\hfill \end{array}$$

Therefore,

$${(a+b+c)}^2=a^2+b^2+c^2+2ab+2bc+2ca.$$

Question 2. What if I put $𝐝\mathbf{=}𝐛\mathbf{+}𝐜$?

Answer— If $d=b+c$, then:

$$a+b+c=a+d.$$

Therefore,

$${(a+b+c)}^2={(a+d)}^2.$$

Page 76 — Think and Reflect

Question. Label the squares and rectangles in Fig. 4.4 so that it represents the identity ${(𝐚\mathbf{+}𝐛\mathbf{+}𝐜)}^{\mathrm{𝟐}}\mathbf{=}{𝐚}^{\mathrm{𝟐}}\mathbf{+}{𝐛}^{\mathrm{𝟐}}\mathbf{+}{𝐜}^{\mathrm{𝟐}}\mathbf{+}\mathrm{𝟐}𝐚𝐛\mathbf{+}\mathrm{𝟐}𝐛𝐜\mathbf{+}\mathrm{𝟐}𝐜𝐚$.

Answer: The rows and columns have widths $a,b,c$. Therefore, label the nine parts as follows:

Thus, the total area is

$$a^2+b^2+c^2+ab+ab+bc+bc+ac+ac=a^2+b^2+c^2+2ab+2bc+2ca.$$

Pages 76–77 — Exercise Set 4.3

1. Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier.

(i) ${\mathrm{𝟏𝟏𝟕}}^{\mathrm{𝟐}}$

Using ${(a-b)}^2=a^2-2ab+b^2$,

$${117}^2={(120-3)}^2=14400-720+9=13689.$$

(ii) ${\mathrm{𝟕𝟖}}^{\mathrm{𝟐}}$

$${78}^2={(80-2)}^2=6400-320+4=6084.$$

(iii) ${\mathrm{𝟏𝟗𝟖}}^{\mathrm{𝟐}}$

$${198}^2={(200-2)}^2=40000-800+4=39204.$$

(iv) ${\mathrm{𝟐𝟏𝟒}}^{\mathrm{𝟐}}$

$${214}^2={(200+14)}^2=40000+5600+196=45796.$$

(v) ${\mathrm{𝟏𝟏𝟎𝟒}}^{\mathrm{𝟐}}$

Using the three-term identity,

$$\begin{array}{cc}\hfill \hfill & \hfill {1104}^2={(1000+100+4)}^2\hfill \\ \hfill \hfill & \hfill =1000000+10000+16+200000+8000+800\hfill \\ \hfill \hfill & \hfill =1218816.\hfill \end{array}$$

(vi) ${\mathrm{𝟏𝟏𝟐𝟎}}^{\mathrm{𝟐}}$

$$\begin{array}{cc}\hfill \hfill & \hfill {1120}^2={(1000+100+20)}^2\hfill \\ \hfill \hfill & \hfill =1000000+10000+400+200000+40000+4000\hfill \\ \hfill \hfill & \hfill =1254400.\hfill \end{array}$$

2. Factor using suitable identities:

(i) $\mathrm{𝟏𝟔}{𝐲}^{\mathrm{𝟐}}\mathbf{-}\mathrm{𝟐𝟒}𝐲\mathbf{+}\mathrm{𝟗}$

$$16y^2-24y+9={\left(4y\right)}^2-2\left(4y\right)\left(3\right)+3^2={(4y-3)}^2.$$

(ii) $\frac{\mathrm{𝟗}}{\mathrm{𝟒}}{𝐬}^{\mathrm{𝟐}}\mathbf{+}\mathrm{𝟔}𝐬𝐭\mathbf{+}\mathrm{𝟒}{𝐭}^{\mathrm{𝟐}}$

$$\frac{9}{4}{s}^2+6st+4t^2={\left(\frac{3s}{2}\right)}^2+2\left(\frac{3s}{2}\right)\left(2t\right)+{\left(2t\right)}^2={(\frac{3s}{2}+2t)}^2.$$

(iii) $\frac{{𝐦}^{\mathrm{𝟐}}}{\mathrm{𝟗}}\mathbf{+}\frac{𝐦𝐤}{\mathrm{𝟑}}\mathbf{+}\frac{{𝐤}^{\mathrm{𝟐}}}{\mathrm{𝟒}}\mathbf{+}\mathrm{𝟑}𝐧𝐤\mathbf{+}\mathrm{𝟐}𝐦𝐧\mathbf{+}\mathrm{𝟗}{𝐧}^{\mathrm{𝟐}}$

$$\frac{m^2}{9}+\frac{mk}{3}+\frac{k^2}{4}+3nk+2mn+9n^2={(\frac{m}{3}+\frac{k}{2}+3n)}^2.$$

Here, $\frac{mk}{3}=2\left(\frac{m}{3}\right)\left(\frac{k}{2}\right)$, $3nk=2\left(\frac{k}{2}\right)\left(3n\right)$, and $2mn=2\left(3n\right)\left(\frac{m}{3}\right)$.

(iv) $\frac{{𝐩}^{\mathrm{𝟐}}}{\mathrm{𝟏𝟔}}\mathbf{-}\mathrm{𝟐}\mathbf{+}\frac{\mathrm{𝟏𝟔}}{{𝐩}^{\mathrm{𝟐}}}$

$$\frac{p^2}{16}-2+\frac{16}{p^2}={\left(\frac{p}{4}\right)}^2-2\left(\frac{p}{4}\right)\left(\frac{4}{p}\right)+{\left(\frac{4}{p}\right)}^2={(\frac{p}{4}-\frac{4}{p})}^2,p\ne 0.$$

(v) $\mathrm{𝟗}{𝐚}^{\mathrm{𝟐}}\mathbf{+}\mathrm{𝟒}{𝐛}^{\mathrm{𝟐}}\mathbf{+}{𝐜}^{\mathrm{𝟐}}\mathbf{-}\mathrm{𝟏𝟐}𝐚𝐛\mathbf{+}\mathrm{𝟔}𝐚𝐜\mathbf{-}\mathrm{𝟒}𝐛𝐜$

$$9a^2+4b^2+c^2-12ab+6ac-4bc={(3a-2b+c)}^2.$$

3. Expand the following using the identity ${(𝐚\mathbf{+}𝐛\mathbf{+}𝐜)}^{\mathrm{𝟐}}\mathbf{=}{𝐚}^{\mathrm{𝟐}}\mathbf{+}{𝐛}^{\mathrm{𝟐}}\mathbf{+}{𝐜}^{\mathrm{𝟐}}\mathbf{+}\mathrm{𝟐}𝐚𝐛\mathbf{+}\mathrm{𝟐}𝐛𝐜\mathbf{+}\mathrm{𝟐}𝐜𝐚$:

(i) ${(𝐩\mathbf{+}\mathrm{𝟑}𝐪\mathbf{+}\mathrm{𝟕}𝐫)}^{\mathrm{𝟐}}$

$${(p+3q+7r)}^2=p^2+9q^2+49r^2+6pq+42qr+14pr.$$

(ii) ${(\mathrm{𝟑}𝐱\mathbf{-}\mathrm{𝟐}𝐲\mathbf{+}\mathrm{𝟒}𝐳)}^{\mathrm{𝟐}}$

$${(3x-2y+4z)}^2=9x^2+4y^2+16z^2-12xy-16yz+24xz.$$

4. Is this an identity? ${(𝐚\mathbf{+}𝐛\mathbf{-}𝐜)}^{\mathrm{𝟐}}\mathbf{+}{(𝐚\mathbf{-}𝐛\mathbf{+}𝐜)}^{\mathrm{𝟐}}\mathbf{+}{(𝐚\mathbf{-}𝐛\mathbf{-}𝐜)}^{\mathrm{𝟐}}\mathbf{=}\mathrm{𝟐}{𝐚}^{\mathrm{𝟐}}\mathbf{+}\mathrm{𝟐}{𝐛}^{\mathrm{𝟐}}\mathbf{+}\mathrm{𝟐}{𝐜}^{\mathrm{𝟐}}$.

Answer: No, it is not an identity. For $a=b=c=1$,

$$\mathrm{\text{LHS}}=1^2+1^2+{(-1)}^2=3,\mathrm{\text{RHS}}=2+2+2=6.$$

Since the equality is not true for all values, it is not an identity.

Page 77 — Prompt Based on Fig. 4.5

Question. Look at the following figure. Justify the identity ${𝐚}^{\mathrm{𝟐}}\mathbf{=}(𝐚\mathbf{+}𝐛)(𝐚\mathbf{-}𝐛)\mathbf{+}{𝐛}^{\mathrm{𝟐}}$ for yourself.

Answer: A square of side $a$ has area $a^2$. If a square of side $b$ is removed, the remaining area is $a^2-b^2$. The remaining pieces can be rearranged into a rectangle of length $a+b$ and breadth $a-b$. Hence,

$$a^2-b^2=(a+b)(a-b).$$

Adding $b^2$ to both sides gives

$$a^2=(a+b)(a-b)+b^2.$$

Page 78 — Think and Reflect

1. Try to evaluate the following using a suitable identity: (i) ${\mathrm{𝟑𝟓}}^{\mathrm{𝟐}}$ (ii) ${\mathrm{𝟔𝟓}}^{\mathrm{𝟐}}$ (iii) ${\mathrm{𝟖𝟓}}^{\mathrm{𝟐}}$ (iv) ${\mathrm{𝟏𝟎𝟓}}^{\mathrm{𝟐}}$. Do you observe any interesting pattern?

Using $a^2=(a+b)(a-b)+b^2$ with $b=5$:

$${35}^2=(35+5)(35-5)+5^2=40\times 30+25=1225,$$

$${65}^2=70\times 60+25=4225,$$

$${85}^2=90\times 80+25=7225,$$

$${105}^2=110\times 100+25=11025.$$

Pattern: To square a number ending in $5$, multiply the number before $5$ by its successor and write $25$ at the end. In general,

$${(10n+5)}^2=100n(n+1)+25.$$

2. Observe the two rows of figures below. They represent an algebraic identity. Try to identify it.

Answer: The figures in the upper row have areas

$${(a+b+c)}^2,{(a+b-c)}^2,{(a-b+c)}^2,{(a-b-c)}^2.$$

The figures in the lower row have areas ${\left(2a\right)}^2,{\left(2b\right)}^2,{\left(2c\right)}^2$. Therefore,

$${(a+b+c)}^2+{(a+b-c)}^2+{(a-b+c)}^2+{(a-b-c)}^2=4(a^2+b^2+c^2).$$

Page 79 — Think and Reflect

Question. Suppose $\mathrm{𝟕}𝐱$ is split as $\mathrm{𝟐}𝐱\mathbf{+}\mathrm{𝟓}𝐱$; can a similar rectangular arrangement be formed? Consider other possibilities and check.

Answer: No. Splitting $7x$ as $2x+5x$ would give possible side lengths $x+2$ and $x+5$. The required number of unit tiles would then be $2\times 5=10$, but the expression contains $12$ unit tiles. Hence, two unit tiles would be left over.

The non-negative splits of $7$ are:

$$0+7,1+6,2+5,3+4.$$

Their products are $0,6,10,12$, respectively. Only $3+4$ (or $4+3$) has product $12$. Therefore, only this split forms the required rectangle.

Algebra tiles can be used to represent products and find factors.

1. Figure out the product of $𝐱\mathbf{+}\mathrm{𝟐}$ and $𝐱\mathbf{+}\mathrm{𝟑}$ using algebra tiles.

Answer: Arrange one $x^2$-tile, five $x$-tiles and six unit tiles into a rectangle with sides $x+2$ and $x+3$. Thus,

$$(x+2)(x+3)=x^2+5x+6.$$

2. Lay out algebra tiles for ${𝐱}^{\mathrm{𝟐}}\mathbf{+}\mathrm{𝟏𝟏}𝐱\mathbf{+}\mathrm{𝟑𝟎}$ in such a way that you will see its factors.

Answer: Since $5+6=11$ and $5\times 6=30$, split $11x$ as $5x+6x$. One $x^2$-tile, eleven $x$-tiles and thirty unit tiles form a rectangle with sides $x+5$ and $x+6$. Hence,

$$x^2+11x+30=(x+5)(x+6).$$

Page 80 — Think and Reflect

Question. We have seen that $(𝐱\mathbf{+}\mathrm{𝟑})(𝐱\mathbf{+}\mathrm{𝟒})\mathbf{=}{𝐱}^{\mathrm{𝟐}}\mathbf{+}\mathrm{𝟕}𝐱\mathbf{+}\mathrm{𝟏𝟐}$. Also $(𝐱\mathbf{+}\mathrm{𝟔})(𝐱\mathbf{+}\mathrm{𝟕})\mathbf{=}{𝐱}^{\mathrm{𝟐}}\mathbf{+}\mathrm{𝟏𝟑}𝐱\mathbf{+}\mathrm{𝟒𝟐}$. Generalise the pattern to get an expression for $(𝐱\mathbf{+}𝐚)(𝐱\mathbf{+}𝐛)$.

Answer:

$$\begin{array}{cc}\hfill \hfill & \hfill (x+a)(x+b)=x(x+b)+a(x+b)\hfill \\ \hfill \hfill & \hfill =x^2+bx+ax+ab.\hfill \end{array}$$

Therefore,

$$(x+a)(x+b)=x^2+(a+b)x+ab.$$

Question. Now consider the case where we have a rectangle of sidelengths $\mathrm{𝟐}𝐱\mathbf{+}\mathrm{𝟑}$ and $\mathrm{𝟑}𝐱\mathbf{+}\mathrm{𝟏}$, as shown in Fig. 4.8. What can you say about its area $(\mathrm{𝟐}𝐱\mathbf{+}\mathrm{𝟑})(\mathrm{𝟑}𝐱\mathbf{+}\mathrm{𝟏})$?

Answer: Fig. 4.8 contains six $x^2$-tiles, eleven $x$-tiles and three unit tiles. Therefore,

$$\begin{array}{cc}\hfill \hfill & \hfill (2x+3)(3x+1)=6x^2+2x+9x+3\hfill \\ \hfill \hfill & \hfill =6x^2+11x+3.\hfill \end{array}$$

Question. Fill in the blanks with the appropriate expressions to make the equation true. $(𝐩𝐱\mathbf{+}𝐚)(𝐪𝐱\mathbf{+}𝐛)\mathbf{=}\left(\mathbf{\_}\mathbf{\_}\mathbf{\_}\mathbf{\_}\mathbf{\_}\right){𝐱}^{\mathrm{𝟐}}\mathbf{+}\left(\mathbf{\_}\mathbf{\_}\mathbf{\_}\mathbf{\_}\mathbf{\_}\right)𝐱\mathbf{+}\mathbf{\_}\mathbf{\_}\mathbf{\_}\mathbf{\_}\mathbf{\_}\mathbf{.}$ Also, verify your answer using the distributive property.

Answer: The blanks are $pq$, $pb+aq$, and $ab$.

$$(px+a)(qx+b)=\left(pq\right)x^2+(pb+aq)x+ab.$$

Verification:

$$\begin{array}{cc}\hfill \hfill & \hfill (px+a)(qx+b)=px(qx+b)+a(qx+b)\hfill \\ \hfill \hfill & \hfill =pq{x}^2+pbx+aqx+ab\hfill \\ \hfill \hfill & \hfill =pq{x}^2+(pb+aq)x+ab.\hfill \end{array}$$

Question. But how do we achieve this ‘splitting of the $𝐱$ term’ without using tiles?

Answer: Compare $x^2+7x+12$ with $x^2+(a+b)x+ab$. We need two numbers whose sum is $7$ and product is $12$. These numbers are $3$ and $4$. Therefore,

$$\begin{array}{cc}\hfill \hfill & \hfill x^2+7x+12=x^2+3x+4x+12\hfill \\ \hfill \hfill & \hfill =x(x+3)+4(x+3)\hfill \\ \hfill \hfill & \hfill =(x+3)(x+4).\hfill \end{array}$$

Page 81 — Example 11 Check

Question. What if we choose $𝐚\mathbf{=}\mathrm{𝟐}$ and $𝐛\mathbf{=}\mathrm{𝟏𝟓}$? Or $𝐚\mathbf{=}\mathrm{𝟑}$ and $𝐛\mathbf{=}\mathrm{𝟏𝟎}$?

Answer: Neither pair works. Although both pairs have product $30$,

$$2+15=17\ne 11,3+10=13\ne 11.$$

The correct pair is $5,6$, because $5+6=11$ and $5\times 6=30$. Thus,

$$x^2+11x+30=(x+5)(x+6).$$

Pages 81–82 — Exercise Set 4.4

1. Fill in the blanks to complete the following identities:

(i) ${𝐬}^{\mathrm{𝟐}}\mathbf{-}\mathrm{𝟏𝟏}𝐬\mathbf{+}\mathrm{𝟐𝟒}\mathbf{=}\left(\mathbf{\_}\mathbf{\_}\mathbf{\_}\mathbf{\_}\mathbf{\_}\mathbf{\_}\mathbf{\_}\mathbf{\_}\right)\left(\mathbf{\_}\mathbf{\_}\mathbf{\_}\mathbf{\_}\mathbf{\_}\mathbf{\_}\mathbf{\_}\mathbf{\_}\right)$

$$s^2-11s+24=(s-3)(s-8).$$

(ii) $\left(\mathbf{\_}\mathbf{\_}\mathbf{\_}\mathbf{\_}\mathbf{\_}\mathbf{\_}\mathbf{\_}\mathbf{\_}\right)(𝐱\mathbf{+}\mathrm{𝟏})\mathbf{=}\mathrm{𝟑}{𝐱}^{\mathrm{𝟐}}\mathbf{-}\mathrm{𝟒}𝐱\mathbf{-}\mathrm{𝟕}$

$$(3x-7)(x+1)=3x^2-4x-7.$$

(iii) $\mathrm{𝟏𝟎}{𝐱}^{\mathrm{𝟐}}\mathbf{-}\mathrm{𝟏𝟏}𝐱\mathbf{-}\mathrm{𝟔}\mathbf{=}\left(\mathrm{𝟐}𝐱\mathbf{-}\mathbf{\_}\mathbf{\_}\mathbf{\_}\right)\left(\mathbf{\_}\mathbf{\_}\mathbf{\_}\mathbf{+}\mathrm{𝟐}\right)$

$$10x^2-11x-6=(2x-3)(5x+2).$$

(iv) $\mathrm{𝟔}{𝐱}^{\mathrm{𝟐}}\mathbf{+}\mathrm{𝟕}𝐱\mathbf{+}\mathrm{𝟐}\mathbf{=}\left(\mathbf{\_}\mathbf{\_}\mathbf{\_}\mathbf{\_}\mathbf{\_}\mathbf{\_}\mathbf{\_}\mathbf{\_}\mathbf{\_}\mathbf{\_}\mathbf{\_}\mathbf{\_}\right)\left(\mathbf{\_}\mathbf{\_}\mathbf{\_}\mathbf{\_}\mathbf{\_}\mathbf{\_}\mathbf{\_}\mathbf{\_}\mathbf{\_}\mathbf{\_}\mathbf{\_}\right)$

$$6x^2+7x+2=(3x+2)(2x+1).$$

2. Select and use the identity that will help you to find the following products without multiplying directly:

(i) ${\left(\mathrm{𝟒𝟏}\right)}^{\mathrm{𝟐}}$

$${41}^2={(40+1)}^2=1600+80+1=1681.$$

(ii) ${\left(\mathrm{𝟐𝟕}\right)}^{\mathrm{𝟐}}$

$${27}^2={(30-3)}^2=900-180+9=729.$$

(iii) $(\mathrm{𝟐𝟑}\mathbf{\times }\mathrm{𝟏𝟕})$

$$23\times 17=(20+3)(20-3)={20}^2-3^2=400-9=391.$$

(iv) ${\left(\mathrm{𝟏𝟑𝟓}\right)}^{\mathrm{𝟐}}$

$${135}^2=(135+5)(135-5)+5^2=140\times 130+25=18225.$$

(v) ${\left(\mathrm{𝟗𝟕}\right)}^{\mathrm{𝟐}}$

$${97}^2={(100-3)}^2=10000-600+9=9409.$$

(vi) $(\mathrm{𝟏𝟖}\mathbf{\times }\mathrm{𝟐𝟗})$

Using $mn=\frac{{(m+n)}^2-{(m-n)}^2}{4}$,

$$18\times 29=\frac{{47}^2-{11}^2}{4}=\frac{2209-121}{4}=522.$$

(vii) $(\mathrm{𝟑𝟒}\mathbf{\times }\mathrm{𝟒𝟑})$

$$34\times 43=\frac{{77}^2-9^2}{4}=\frac{5929-81}{4}=1462.$$

(viii) ${\left(\mathrm{𝟐𝟎𝟓}\right)}^{\mathrm{𝟐}}$

$${205}^2=(205+5)(205-5)+5^2=210\times 200+25=42025.$$

3. Factor the following:

(i) $\mathrm{𝟗}{𝐚}^{\mathrm{𝟐}}\mathbf{+}{𝐛}^{\mathrm{𝟐}}\mathbf{+}\mathrm{𝟒}{𝐜}^{\mathrm{𝟐}}\mathbf{-}\mathrm{𝟔}𝐚𝐛\mathbf{+}\mathrm{𝟏𝟐}𝐚𝐜\mathbf{-}\mathrm{𝟒}𝐛𝐜$

$$9a^2+b^2+4c^2-6ab+12ac-4bc={(3a-b+2c)}^2.$$

(ii) $\mathrm{𝟏𝟔}{𝐬}^{\mathrm{𝟐}}\mathbf{+}\mathrm{𝟐𝟓}{𝐭}^{\mathrm{𝟐}}\mathbf{-}\mathrm{𝟒𝟎}𝐬𝐭$

$$16s^2+25t^2-40st={(4s-5t)}^2.$$

(iii) ${𝐫}^{\mathrm{𝟐}}\mathbf{-}𝐫\mathbf{-}\mathrm{𝟒𝟐}$

Since $-7+6=-1$ and $(-7)\left(6\right)=-42$,

$$r^2-r-42=(r-7)(r+6).$$

(iv) $\mathrm{𝟒𝟗}{𝐠}^{\mathrm{𝟐}}\mathbf{+}\mathrm{𝟏𝟒}𝐠𝐡\mathbf{+}{𝐡}^{\mathrm{𝟐}}$

$$49g^2+14gh+h^2={(7g+h)}^2.$$

(v) $\mathrm{𝟔𝟒}{𝐮}^{\mathrm{𝟐}}\mathbf{+}\mathrm{𝟏𝟐𝟏}{𝐯}^{\mathrm{𝟐}}\mathbf{+}\mathrm{𝟒}{𝐰}^{\mathrm{𝟐}}\mathbf{-}\mathrm{𝟏𝟕𝟔}𝐮𝐯\mathbf{-}\mathrm{𝟑𝟐}𝐮𝐰\mathbf{+}\mathrm{𝟒𝟒}𝐯𝐰$

$$64u^2+121v^2+4w^2-176uv-32uw+44vw={(8u-11v-2w)}^2.$$

Page 82 — Think and Reflect: James and Reshma

Question. James and Reshma were talking about algebraic identities they learnt in school.

James: ${(a-b)}^2(a+b)=(a^2-2ab+b^2)(a+b)$

Reshma: I have a different idea. ${(a-b)}^2(a+b)=(a-b)\left[(a-b)(a+b)\right]=(a-b)(a^2-b^2)$. I will find this product to get the answer.

According to you, who is correct and why?

Answer: Both James and Reshma are correct. James first uses the identity for ${(a-b)}^2$, while Reshma first uses $(a-b)(a+b)=a^2-b^2$. Both methods give

$$\begin{array}{cc}\hfill \hfill & \hfill {(a-b)}^2(a+b)=(a^2-2ab+b^2)(a+b)\hfill \\ \hfill \hfill & \hfill =a^3-a^{2}b-a{b}^2+b^3,\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill \hfill & \hfill (a-b)(a^2-b^2)=a^3-a{b}^2-a^{2}b+b^3\hfill \\ \hfill \hfill & \hfill =a^3-a^{2}b-a{b}^2+b^3.\hfill \end{array}$$

Question. Try to combine more such identities and find new results.

Answer: Some examples are:

$${(a+b)}^2+{(a-b)}^2=2(a^2+b^2),$$

$${(a+b)}^2-{(a-b)}^2=4ab,$$

and

$${(a+b)}^2(a-b)=a^3+a^{2}b-a{b}^2-b^3.$$

Page 82 — Section 4.7 Opening Prompts

Question. What do you think ${(𝐚\mathbf{+}𝐛)}^{\mathrm{𝟑}}$ will look like?

Answer:

$$\begin{array}{cc}\hfill \hfill & \hfill {(a+b)}^3=(a+b){(a+b)}^2\hfill \\ \hfill \hfill & \hfill =(a+b)(a^2+2ab+b^2)\hfill \\ \hfill \hfill & \hfill =a^3+3a^{2}b+3a{b}^2+b^3.\hfill \end{array}$$

Question. What if we have a cube of edge $𝐚\mathbf{+}𝐛$? Can we divide a cube of edge $(𝐚\mathbf{+}𝐛)$ into smaller cubes and cuboids and represent this new identity?

Answer: Yes. Divide each edge into lengths $a$ and $b$. The cube is divided into:

• one cube of volume $a^3$,
• three cuboids of volume $a^{2}b$ each,
• three cuboids of volume $a{b}^2$ each, and
• one cube of volume $b^3$.

Their total volume is

$$a^3+3a^{2}b+3a{b}^2+b^3={(a+b)}^3,$$

which represents the new identity geometrically.

Printed Page 83

Activity: We know that the volume of this cube is ${(𝐚\mathbf{+}𝐛)}^{\mathrm{𝟑}}$. Let us split this cube into smaller cubes and cuboids.

Answer:

The cube of side $a+b$ can be divided into:

• one cube of volume $a^3$,
• one cube of volume $b^3$,
• three cuboids of volume $a^{2}b$ each,
• three cuboids of volume $a{b}^2$ each.

Therefore,

$${(a+b)}^3=a^3+3a^{2}b+3a{b}^2+b^3$$

Printed Page 84

1. What happens when we replace $𝐛$ with $\mathbf{-}𝐛$ in this new identity?

Answer:

Replace $b$ by $-b$ in

$${(a+b)}^3=a^3+3a^{2}b+3a{b}^2+b^3.$$

Then,

$${[a+(-b)]}^3=a^3+3a^2(-b)+3a{(-b)}^2+{(-b)}^3.$$

Therefore,

$${(a-b)}^3=a^3-3a^{2}b+3a{b}^2-b^3$$

The positive and negative terms occur alternately.

2. Example 13: What is the side of the cube whose volume is ${𝐩}^{\mathrm{𝟑}}\mathbf{+}\mathrm{𝟔}{𝐩}^{\mathrm{𝟐}}𝐪\mathbf{+}\mathrm{𝟏𝟐}𝐩{𝐪}^{\mathrm{𝟐}}\mathbf{+}\mathrm{𝟖}{𝐪}^{\mathrm{𝟑}}$ cubic units?

Answer:

Using

$${(a+b)}^3=a^3+3a^{2}b+3a{b}^2+b^3,$$

$$\begin{array}{cc}\hfill \hfill & \hfill p^3+6p^{2}q+12p{q}^2+8q^3=p^3+3p^2\left(2q\right)+3p{\left(2q\right)}^2+{\left(2q\right)}^3\hfill \\ \hfill \hfill & \hfill ={(p+2q)}^3.\hfill \end{array}$$

Hence, the side of the cube is

$$p+2q\mathrm{\text{units}}.$$

3. Example 14: Now consider the expression $\mathrm{𝟖}{𝐧}^{\mathrm{𝟑}}\mathbf{-}\mathrm{𝟔𝟎}{𝐧}^{\mathrm{𝟐}}𝐦\mathbf{+}\mathrm{𝟏𝟓𝟎}𝐧{𝐦}^{\mathrm{𝟐}}\mathbf{-}\mathrm{𝟏𝟐𝟓}{𝐦}^{\mathrm{𝟑}}$. If you write it in the form ${(𝐚\mathbf{-}𝐛)}^{\mathrm{𝟑}}$, what will be $𝐚$ and $𝐛$?

Answer:

$$\begin{array}{cc}\hfill \hfill & \hfill 8n^3-60n^{2}m+150n{m}^2-125m^3={\left(2n\right)}^3-3{\left(2n\right)}^2\left(5m\right)\hfill \\ \hfill \hfill & \hfill +3\left(2n\right){\left(5m\right)}^2-{\left(5m\right)}^3\hfill \\ \hfill \hfill & \hfill ={(2n-5m)}^3.\hfill \end{array}$$

Therefore,

$$a=2n,b=5m.$$

4. Try to multiply the following using the distributive property.

(i) $(𝐱\mathbf{-}𝐲)({𝐱}^{\mathrm{𝟐}}\mathbf{+}𝐱𝐲\mathbf{+}{𝐲}^{\mathrm{𝟐}})$

Answer:

$$\begin{array}{cc}\hfill \hfill & \hfill (x-y)(x^2+xy+y^2)=x^3+x^{2}y+x{y}^2-x^{2}y-x{y}^2-y^3\hfill \\ \hfill \hfill & \hfill =x^3-y^3.\hfill \end{array}$$

Thus,

$$x^3-y^3=(x-y)(x^2+xy+y^2).$$

(ii) $(𝐱\mathbf{+}𝐲)({𝐱}^{\mathrm{𝟐}}\mathbf{-}𝐱𝐲\mathbf{+}{𝐲}^{\mathrm{𝟐}})$

Answer:

$$\begin{array}{cc}\hfill \hfill & \hfill (x+y)(x^2-xy+y^2)=x^3-x^{2}y+x{y}^2+x^{2}y-x{y}^2+y^3\hfill \\ \hfill \hfill & \hfill =x^3+y^3.\hfill \end{array}$$

Thus,

$$x^3+y^3=(x+y)(x^2-xy+y^2).$$

Printed Page 85

1. Verify $(𝐱\mathbf{-}𝐲)({𝐱}^{\mathrm{𝟐}}\mathbf{+}𝐱𝐲\mathbf{+}{𝐲}^{\mathrm{𝟐}})\mathbf{=}{𝐱}^{\mathrm{𝟑}}\mathbf{-}{𝐲}^{\mathrm{𝟑}}$ for yourself by choosing different values of $𝐱$ and $𝐲$.

Answer:

Take $x=3$ and $y=2$.

$$\mathrm{\text{LHS}}=(3-2)(3^2+3\cdot 2+2^2)=1(9+6+4)=19$$

$$\mathrm{\text{RHS}}=3^3-2^3=27-8=19$$

Thus, LHS $=$ RHS.

Again, take $x=-1$ and $y=4$.

$$\mathrm{\text{LHS}}=(-1-4)[{(-1)}^2+(-1)\left(4\right)+4^2]=(-5)\left(13\right)=-65$$

$$\mathrm{\text{RHS}}={(-1)}^3-4^3=-1-64=-65$$

Hence, the identity is verified.

2. Predict what $(𝐱\mathbf{+}𝐲)({𝐱}^{\mathrm{𝟐}}\mathbf{-}𝐱𝐲\mathbf{+}{𝐲}^{\mathrm{𝟐}})$ will be.

Answer:

$$\begin{array}{cc}\hfill \hfill & \hfill (x+y)(x^2-xy+y^2)=x^3-x^{2}y+x{y}^2+x^{2}y-x{y}^2+y^3\hfill \\ \hfill \hfill & \hfill =x^3+y^3.\hfill \end{array}$$

Think and Reflect

3. Do you think $𝐱\mathbf{-}𝐲$ is also a factor of ${𝐱}^{\mathrm{𝟒}}\mathbf{-}{𝐲}^{\mathrm{𝟒}}$? Can you see how $𝐱\mathbf{-}𝐲$ is a factor of ${𝐱}^{\mathrm{𝟒}}\mathbf{-}{𝐲}^{\mathrm{𝟒}}$?

Answer:

Yes.

$$\begin{array}{cc}\hfill \hfill & \hfill x^4-y^4={\left(x^2\right)}^2-{\left(y^2\right)}^2\hfill \\ \hfill \hfill & \hfill =(x^2-y^2)(x^2+y^2)\hfill \\ \hfill \hfill & \hfill =(x-y)(x+y)(x^2+y^2).\hfill \end{array}$$

Therefore,

$$x-y\mathrm{\text{is a factor of}}{x}^4-y^4.$$

4. How about ${𝐱}^{\mathrm{𝟓}}\mathbf{-}{𝐲}^{\mathrm{𝟓}}$? Does this also have $𝐱\mathbf{-}𝐲$ as a factor?

Answer:

Yes.

$$x^5-y^5=(x-y)(x^4+x^{3}y+x^2{y}^2+x{y}^3+y^4)$$

Therefore, $x-y$ is also a factor of $x^5-y^5$. In general,

$$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\cdots +x{y}^{n-2}+y^{n-1}).$$

5. Example 15: The sum of three numbers is 10 and their product is 25. The sum of their squares is 38. Try to use the previous identity to find the sum of the cubes of these three numbers.

Answer:

Let the numbers be $x,y,z$. Then,

$$x+y+z=10,xyz=25,x^2+y^2+z^2=38.$$

First, using

$${(x+y+z)}^2=x^2+y^2+z^2+2(xy+xz+yz),$$

$$100=38+2(xy+xz+yz).$$

Therefore,

$$xy+xz+yz=31.$$

Now use

$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-xz-yz).$$

$$x^3+y^3+z^3-75=10(38-31)=70.$$

Hence,

$$x^3+y^3+z^3=145.$$

Printed Page 86

Example 16: Simplify the rational expression $\frac{{𝐱}^{\mathrm{𝟐}}\mathbf{-}\mathrm{𝟕}𝐱\mathbf{+}\mathrm{𝟏𝟐}}{\mathrm{𝟓}{𝐱}^{\mathrm{𝟐}}\mathbf{+}\mathrm{𝟓}𝐱\mathbf{-}\mathrm{𝟏𝟎𝟎}}$, assuming that $\mathrm{𝟓}{𝐱}^{\mathrm{𝟐}}\mathbf{+}\mathrm{𝟓}𝐱\mathbf{-}\mathrm{𝟏𝟎𝟎}\mathbf{\ne }\mathrm{𝟎}$.

Answer:

For the numerator, the two required numbers are $-3$ and $-4$:

$$x^2-7x+12=(x-3)(x-4).$$

For the denominator, take $5$ common:

$$\begin{array}{cc}\hfill \hfill & \hfill 5x^2+5x-100=5(x^2+x-20)\hfill \\ \hfill \hfill & \hfill =5(x-4)(x+5).\hfill \end{array}$$

$$\begin{array}{cc}\hfill \hfill & \hfill \frac{x^2-7x+12}{5x^2+5x-100}=\frac{(x-3)(x-4)}{5(x-4)(x+5)}\hfill \\ \hfill \hfill & \hfill =\frac{x-3}{5(x+5)}.\hfill \end{array}$$

The restrictions from the original denominator are

$$x\ne 4,-5.$$

Printed Page 87

Think and Reflect: Try to simplify the following rational expression:

$$\frac{36s^2-12st+t^2}{t^2+2ts-48s^2}=\frac{{(6s-t)}^2}{(\_\_\_+\_\_\_)(\_\_\_+\_\_\_)}.$$

Assume that $t^2+2ts-48s^2\ne 0$.

Answer:

$$36s^2-12st+t^2={(6s-t)}^2={(t-6s)}^2.$$

For the denominator,

$$t^2+2ts-48s^2=(t+8s)(t-6s).$$

Thus, the blanks are

$$(t+8s)(t-6s).$$

$$\begin{array}{cc}\hfill \hfill & \hfill \frac{36s^2-12st+t^2}{t^2+2ts-48s^2}=\frac{{(t-6s)}^2}{(t+8s)(t-6s)}\hfill \\ \hfill \hfill & \hfill =\frac{t-6s}{t+8s}.\hfill \end{array}$$

Restrictions:

$$t\ne 6s,-8s.$$

Exercise Set 4.5

1. Simplify the following rational expressions assuming that the expressions in the denominators are not equal to zero.

(i) $\frac{\mathrm{𝟑}{𝐩}^{\mathrm{𝟐}}\mathbf{-}\mathrm{𝟑}𝐩𝐪\mathbf{-}\mathrm{𝟏𝟖}{𝐪}^{\mathrm{𝟐}}}{{𝐩}^{\mathrm{𝟐}}\mathbf{+}\mathrm{𝟑}𝐩𝐪\mathbf{-}\mathrm{𝟏𝟎}{𝐪}^{\mathrm{𝟐}}}$

Answer:

$$3p^2-3pq-18q^2=3(p-3q)(p+2q)$$

and

$$p^2+3pq-10q^2=(p+5q)(p-2q).$$

Hence,

$$\frac{3(p-3q)(p+2q)}{(p+5q)(p-2q)}.$$

There is no common factor to cancel. Here,

$$p\ne -5q,p\ne 2q.$$

(ii) $\frac{{𝐧}^{\mathrm{𝟑}}\mathbf{-}\mathrm{𝟑}{𝐧}^{\mathrm{𝟐}}𝐦\mathbf{+}\mathrm{𝟑}𝐧{𝐦}^{\mathrm{𝟐}}\mathbf{-}{𝐦}^{\mathrm{𝟑}}}{\mathrm{𝟓}{𝐦}^{\mathrm{𝟐}}\mathbf{-}\mathrm{𝟏𝟎}𝐦𝐧\mathbf{+}\mathrm{𝟓}{𝐧}^{\mathrm{𝟐}}}$

Answer:

$$n^3-3n^{2}m+3n{m}^2-m^3={(n-m)}^3$$

and

$$5m^2-10mn+5n^2=5{(m-n)}^2=5{(n-m)}^2.$$

Therefore,

$$\frac{n-m}{5},n\ne m.$$

(iii) $\frac{{𝐰}^{\mathrm{𝟑}}\mathbf{-}{𝐯}^{\mathrm{𝟑}}\mathbf{+}{𝐱}^{\mathrm{𝟑}}\mathbf{+}\mathrm{𝟑}𝐰𝐯𝐱}{{𝐰}^{\mathrm{𝟐}}\mathbf{+}{𝐯}^{\mathrm{𝟐}}\mathbf{+}{𝐱}^{\mathrm{𝟐}}\mathbf{-}\mathrm{𝟐}𝐰𝐯\mathbf{-}\mathrm{𝟐}𝐯𝐱\mathbf{+}\mathrm{𝟐}𝐰𝐱}$

Answer:

Taking $a=w,b=x,c=-v$ in the three-cube identity,

$$\begin{array}{cc}\hfill \hfill & \hfill w^3-v^3+x^3+3wvx=(w+x-v)\hfill \\ \hfill \hfill & \hfill \times (w^2+x^2+v^2-wx+xv+wv).\hfill \end{array}$$

Also,

$$w^2+v^2+x^2-2wv-2vx+2wx={(w+x-v)}^2.$$

Therefore,

$$\frac{w^2+x^2+v^2-wx+xv+wv}{w+x-v},$$

where $w+x-v\ne 0$.

(iv) $\frac{\mathrm{𝟒}{𝐲}^{\mathrm{𝟐}}\mathbf{-}\mathrm{𝟐𝟎}𝐲𝐳\mathbf{+}\mathrm{𝟐𝟓}{𝐳}^{\mathrm{𝟐}}}{\mathrm{𝟐𝟓}{𝐳}^{\mathrm{𝟐}}\mathbf{-}\mathrm{𝟒}{𝐲}^{\mathrm{𝟐}}}$

Answer:

$$4y^2-20yz+25z^2={(5z-2y)}^2$$

and

$$25z^2-4y^2=(5z-2y)(5z+2y).$$

Thus,

$$\frac{5z-2y}{5z+2y},$$

where $5z-2y\ne 0$ and $5z+2y\ne 0$.

(v) $\frac{({𝐱}^{\mathrm{𝟐}}\mathbf{+}𝐱\mathbf{-}\mathrm{𝟔})({𝐱}^{\mathrm{𝟐}}\mathbf{-}\mathrm{𝟕}𝐱\mathbf{+}\mathrm{𝟏𝟐})}{({𝐱}^{\mathrm{𝟐}}\mathbf{-}\mathrm{𝟔}𝐱\mathbf{+}\mathrm{𝟖})({𝐱}^{\mathrm{𝟐}}\mathbf{-}\mathrm{𝟗})}$

Answer:

$$\begin{array}{cc}\hfill \hfill & \hfill x^2+x-6=(x+3)(x-2),\hfill \\ \hfill \hfill & \hfill x^2-7x+12=(x-3)(x-4),\hfill \\ \hfill \hfill & \hfill x^2-6x+8=(x-2)(x-4),\hfill \\ \hfill \hfill & \hfill x^2-9=(x-3)(x+3).\hfill \end{array}$$

Therefore, the simplified value is $1$, where $x\ne -3,2,3,4$.

(vi) $\frac{{𝐩}^{\mathrm{𝟒}}\mathbf{-}\mathrm{𝟏𝟔}}{{𝐩}^{\mathrm{𝟐}}\mathbf{-}\mathrm{𝟒}𝐩\mathbf{+}\mathrm{𝟒}}$

Answer:

$$\begin{array}{cc}\hfill \hfill & \hfill p^4-16=(p^2-4)(p^2+4)=(p-2)(p+2)(p^2+4),\hfill \\ \hfill \hfill & \hfill p^2-4p+4={(p-2)}^2.\hfill \end{array}$$

Therefore,

$$\frac{(p+2)(p^2+4)}{p-2},p\ne 2.$$

Example 17: Saira has arranged a square of side $𝐱$ units, 8 rectangular strips of sides $𝐱$ units and width 1 unit, and 15 squares of side 1 unit to form a bigger rectangle. Find the length and breadth of the rectangle in terms of $𝐱$.

Answer:

The total area is

$$x^2+8x+15.$$

We need two numbers whose sum is $8$ and product is $15$. They are $3$ and $5$.

$$x^2+8x+15=(x+3)(x+5).$$

Hence, possible dimensions are

$$\mathrm{\text{Length}}=x+5\mathrm{\text{units}}$$

and

$$\mathrm{\text{Breadth}}=x+3\mathrm{\text{units}}.$$

Printed Page 88

1. Now draw Saira’s rectangle using these pieces.

Answer:

Draw a rectangle of dimensions $(x+5)\times (x+3)$ and divide each side into $x$ and its remaining constant part. The four regions will be:

• an $x\times x$ square,
• an $x\times 3$ region made from three $x\times 1$ strips,
• a $5\times x$ region made from five $x\times 1$ strips, and
• a $5\times 3$ region made from fifteen $1\times 1$ squares.

Thus, all the given pieces form a rectangle of dimensions

$$(x+5)\mathrm{\text{units}}\times (x+3)\mathrm{\text{units}}.$$

2. Example 18: A rectangular pool is such that its breadth is 4 metres less than its length and its area is 96 sq. metres. Find the length and breadth of the pool.

Answer:

Let the length be $x$ metres. Then the breadth is $x-4$ metres.

$$x(x-4)=96$$

$$x^2-4x-96=0$$

$$x^2-12x+8x-96=0$$

$$x(x-12)+8(x-12)=0$$

$$(x-12)(x+8)=0.$$

Thus,

$$x=12\mathrm{\text{or}}x=-8.$$

Since a physical length cannot be negative, $x=-8$ is rejected.

$$\mathrm{\text{Length}}=12\mathrm{\text{m}},\mathrm{\text{Breadth}}=8\mathrm{\text{m}}.$$

Source correction: The printed statement $(-12)\times 8=96$ is incorrect. It should be

$$(-12)\times 8=-96.$$

End-of-Chapter Exercises

1. Use suitable identities to find the following products.

(i) ${(\mathbf{-}\mathrm{𝟑}𝐱\mathbf{+}\mathrm{𝟒})}^{\mathrm{𝟐}}$

Answer:

$$\begin{array}{cc}\hfill \hfill & \hfill {(-3x+4)}^2={(-3x)}^2+2(-3x)\left(4\right)+4^2\hfill \\ \hfill \hfill & \hfill =9x^2-24x+16.\hfill \end{array}$$

(ii) $(\mathrm{𝟐}𝐬\mathbf{+}\mathrm{𝟕})(\mathrm{𝟐}𝐬\mathbf{-}\mathrm{𝟕})$

Answer:

$${\left(2s\right)}^2-7^2=4s^2-49.$$

(iii) $({𝐩}^{\mathrm{𝟐}}\mathbf{+}\frac{\mathrm{𝟏}}{\mathrm{𝟐}})({𝐩}^{\mathrm{𝟐}}\mathbf{-}\frac{\mathrm{𝟏}}{\mathrm{𝟐}})$

Answer:

$$p^4-\frac{1}{4}.$$

(iv) $(\mathrm{𝟐}𝐧\mathbf{+}\mathrm{𝟕})(\mathrm{𝟐}𝐧\mathbf{-}\mathrm{𝟕})$

Answer:

$$4n^2-49.$$

(v) $(𝐬\mathbf{-}\mathrm{𝟐}𝐭)({𝐬}^{\mathrm{𝟐}}\mathbf{+}\mathrm{𝟐}𝐬𝐭\mathbf{+}\mathrm{𝟒}{𝐭}^{\mathrm{𝟐}})$

Answer:

Using $a^3-b^3=(a-b)(a^2+ab+b^2)$,

$$s^3-8t^3.$$

(vi) ${(\frac{\mathrm{𝟏}}{\mathrm{𝟐}𝐫}\mathbf{-}\mathrm{𝟒}𝐫)}^{\mathrm{𝟐}}$

Answer:

$$\begin{array}{cc}\hfill \hfill & \hfill {(\frac{1}{2r}-4r)}^2=\frac{1}{4r^2}-2\left(\frac{1}{2r}\right)\left(4r\right)+16r^2\hfill \\ \hfill \hfill & \hfill =\frac{1}{4r^2}-4+16r^2,\hfill \end{array}$$

where $r\ne 0$.

(vii) ${(\mathbf{-}\mathrm{𝟑}𝐦\mathbf{+}\mathrm{𝟒}𝐤\mathbf{-}𝐥)}^{\mathrm{𝟐}}$

Answer:

$$9m^2+16k^2+l^2-24mk-8kl+6ml.$$

(viii) ${(𝐱\mathbf{-}\frac{\mathrm{𝟏}}{\mathrm{𝟑}}𝐲)}^{\mathrm{𝟑}}$

Answer:

$$x^3-x^{2}y+\frac{1}{3}x{y}^2-\frac{1}{27}{y}^3.$$

(ix) ${(\frac{\mathrm{𝟕}}{\mathrm{𝟐}}𝐤\mathbf{-}\frac{\mathrm{𝟐}}{\mathrm{𝟑}}𝐦)}^{\mathrm{𝟑}}$

Answer:

$$\frac{343}{8}{k}^3-\frac{49}{2}{k}^{2}m+\frac{14}{3}k{m}^2-\frac{8}{27}{m}^3.$$

Printed Page 89

2. Find the values using suitable identities.

(i) $\mathrm{𝟏𝟕}\mathbf{\times }\mathrm{𝟐𝟏}$

Answer:

$$17\times 21=(19-2)(19+2)={19}^2-2^2$$

$$17\times 21=357.$$

(ii) $\mathrm{𝟏𝟎𝟒}\mathbf{\times }\mathrm{𝟗𝟔}$

Answer:

$$(100+4)(100-4)={100}^2-4^2$$

$$104\times 96=9984.$$

(iii) $\mathrm{𝟐𝟒}\mathbf{\times }\mathrm{𝟏𝟔}$

Answer:

$$(20+4)(20-4)={20}^2-4^2$$

$$24\times 16=384.$$

(iv) ${\mathrm{𝟏𝟒𝟕}}^{\mathrm{𝟑}}$

Answer:

$$\begin{array}{cc}\hfill \hfill & \hfill {147}^3={(150-3)}^3\hfill \\ \hfill \hfill & \hfill ={150}^3-3{\left(150\right)}^2\left(3\right)+3\left(150\right){\left(3\right)}^2-3^3\hfill \\ \hfill \hfill & \hfill =3,176,523.\hfill \end{array}$$

(v) ${\mathrm{𝟏𝟗𝟗}}^{\mathrm{𝟑}}$

Answer:

$$\begin{array}{cc}\hfill \hfill & \hfill {199}^3={(200-1)}^3\hfill \\ \hfill \hfill & \hfill ={200}^3-3{\left(200\right)}^2+3\left(200\right)-1\hfill \\ \hfill \hfill & \hfill =7,880,599.\hfill \end{array}$$

(vi) ${\mathrm{𝟏𝟐𝟕}}^{\mathrm{𝟑}}$

Answer:

$$\begin{array}{cc}\hfill \hfill & \hfill {127}^3={(120+7)}^3\hfill \\ \hfill \hfill & \hfill ={120}^3+3{\left(120\right)}^2\left(7\right)+3\left(120\right){\left(7\right)}^2+7^3\hfill \\ \hfill \hfill & \hfill =2,048,383.\hfill \end{array}$$

(vii) ${(\mathbf{-}\mathrm{𝟏𝟎𝟕})}^{\mathrm{𝟑}}$

Answer:

$$\begin{array}{cc}\hfill \hfill & \hfill {(-107)}^3=-{(100+7)}^3\hfill \\ \hfill \hfill & \hfill =-1,225,043.\hfill \end{array}$$

(viii) ${(\mathbf{-}\mathrm{𝟐𝟗𝟗})}^{\mathrm{𝟑}}$

Answer:

$$\begin{array}{cc}\hfill \hfill & \hfill {(-299)}^3=-{(300-1)}^3\hfill \\ \hfill \hfill & \hfill =-26,730,899.\hfill \end{array}$$

3. Factor the following algebraic expressions.

(i) $\mathrm{𝟒}{𝐲}^{\mathrm{𝟐}}\mathbf{+}\mathrm{𝟏}\mathbf{+}\frac{\mathrm{𝟏}}{\mathrm{𝟏𝟔}{𝐲}^{\mathrm{𝟐}}}$

Answer:

$${(2y+\frac{1}{4y})}^2,y\ne 0.$$

(ii) $\mathrm{𝟗}{𝐦}^{\mathrm{𝟐}}\mathbf{-}\frac{\mathrm{𝟏}}{\mathrm{𝟐𝟓}{𝐧}^{\mathrm{𝟐}}}$

Answer:

$$(3m-\frac{1}{5n})(3m+\frac{1}{5n}),n\ne 0.$$

(iii) $\mathrm{𝟐𝟕}{𝐛}^{\mathrm{𝟑}}\mathbf{-}\frac{\mathrm{𝟏}}{\mathrm{𝟔𝟒}{𝐛}^{\mathrm{𝟑}}}$

Answer:

$$\begin{array}{cc}\hfill \hfill & \hfill 27b^3-\frac{1}{64b^3}={\left(3b\right)}^3-{\left(\frac{1}{4b}\right)}^3\hfill \\ \hfill \hfill & \hfill =(3b-\frac{1}{4b})(9b^2+\frac{3}{4}+\frac{1}{16b^2}),\hfill \end{array}$$

where $b\ne 0$.

(iv) ${𝐱}^{\mathrm{𝟐}}\mathbf{+}\frac{\mathrm{𝟓}𝐱}{\mathrm{𝟔}}\mathbf{+}\frac{\mathrm{𝟏}}{\mathrm{𝟔}}$

Answer:

$$(x+\frac{1}{2})(x+\frac{1}{3}).$$

(v) $\mathrm{𝟐𝟕}{𝐮}^{\mathrm{𝟑}}\mathbf{-}\frac{\mathrm{𝟏}}{\mathrm{𝟏𝟐𝟓}}\mathbf{-}\frac{\mathrm{𝟐𝟕}{𝐮}^{\mathrm{𝟐}}}{\mathrm{𝟓}}\mathbf{+}\frac{\mathrm{𝟗}𝐮}{\mathrm{𝟐𝟓}}$

Answer:

Rearranging the terms,

$$27u^3-\frac{27u^2}{5}+\frac{9u}{25}-\frac{1}{125}={(3u-\frac{1}{5})}^3.$$

(vi) $\mathrm{𝟔𝟒}{𝐲}^{\mathrm{𝟑}}\mathbf{+}\frac{\mathrm{𝟏}}{\mathrm{𝟏𝟐𝟓}}{𝐳}^{\mathrm{𝟑}}$

Answer:

$$(4y+\frac{z}{5})(16y^2-\frac{4}{5}yz+\frac{z^2}{25}).$$

(vii) ${𝐩}^{\mathrm{𝟑}}\mathbf{+}\mathrm{𝟐𝟕}{𝐪}^{\mathrm{𝟑}}\mathbf{+}{𝐫}^{\mathrm{𝟑}}\mathbf{-}\mathrm{𝟗}𝐩𝐪𝐫$

Answer:

Taking the three terms as $p,3q,r$,

$$(p+3q+r)(p^2+9q^2+r^2-3pq-pr-3qr).$$

(viii) $\mathrm{𝟗}{𝐦}^{\mathrm{𝟐}}\mathbf{-}\mathrm{𝟏𝟐}𝐦\mathbf{+}\mathrm{𝟒}$

Answer:

$${(3m-2)}^2.$$

(ix) $\mathrm{𝟗}{𝐱}^{\mathrm{𝟑}}\mathbf{-}\frac{\mathrm{𝟖}}{\mathrm{𝟑}}{𝐲}^{\mathrm{𝟑}}\mathbf{+}\frac{{𝐳}^{\mathrm{𝟑}}}{\mathrm{𝟑}}\mathbf{+}\mathrm{𝟔}𝐱𝐲𝐳$

Answer:

$$\begin{array}{cc}\hfill \hfill & \hfill 9x^3-\frac{8}{3}{y}^3+\frac{z^3}{3}+6xyz\hfill \\ \hfill \hfill & \hfill =\frac{1}{3}[{\left(3x\right)}^3+{(-2y)}^3+z^3-3\left(3x\right)(-2y)z].\hfill \end{array}$$

Therefore,

$$\frac{1}{3}(3x-2y+z)(9x^2+4y^2+z^2+6xy-3xz+2yz).$$

(x) $\mathrm{𝟒}{𝐱}^{\mathrm{𝟐}}\mathbf{+}\mathrm{𝟗}{𝐲}^{\mathrm{𝟐}}\mathbf{+}\mathrm{𝟑𝟔}{𝐳}^{\mathrm{𝟐}}\mathbf{+}\mathrm{𝟏𝟐}𝐱𝐳\mathbf{+}\mathrm{𝟑𝟔}𝐲𝐳\mathbf{+}\mathrm{𝟐𝟒}𝐱𝐲$

Answer:

Source correction: The two printed mixed terms $12xz$ and $24xy$ appear to have been interchanged. As printed, the expression does not factorise through the standard identities taught in the chapter. The intended expression is evidently

$$4x^2+9y^2+36z^2+12xy+24xz+36yz,$$

which gives

$${(2x+3y+6z)}^2.$$

(xi) $\mathrm{𝟐𝟕}{𝐮}^{\mathrm{𝟑}}\mathbf{-}\frac{\mathrm{𝟏}}{\mathrm{𝟐𝟏𝟔}}\mathbf{-}\frac{\mathrm{𝟗}{𝐮}^{\mathrm{𝟐}}}{\mathrm{𝟐}}\mathbf{+}\frac{𝐮}{\mathrm{𝟒}}$

Answer:

Rearranging the terms,

$$27u^3-\frac{9u^2}{2}+\frac{u}{4}-\frac{1}{216}={(3u-\frac{1}{6})}^3.$$

4. Simplify the following.

(i) $\frac{\mathrm{𝟒}{𝐱}^{\mathrm{𝟐}}\mathbf{+}\mathrm{𝟒}𝐱\mathbf{+}\mathrm{𝟏}}{\mathrm{𝟒}{𝐱}^{\mathrm{𝟐}}\mathbf{-}\mathrm{𝟏}}$

Answer:

$$\begin{array}{cc}\hfill \hfill & \hfill \frac{4x^2+4x+1}{4x^2-1}=\frac{{(2x+1)}^2}{(2x+1)(2x-1)}\hfill \\ \hfill \hfill & \hfill =\frac{2x+1}{2x-1},\hfill \end{array}$$

where $x\ne \pm \frac{1}{2}$.

(ii) $\frac{\mathrm{𝟗}(\mathrm{𝟑}{𝐚}^{\mathrm{𝟑}}\mathbf{-}\mathrm{𝟐𝟒}{𝐛}^{\mathrm{𝟑}})}{\mathrm{𝟗}{𝐚}^{\mathrm{𝟐}}\mathbf{-}\mathrm{𝟑𝟔}{𝐛}^{\mathrm{𝟐}}}$

Answer:

$$\begin{array}{cc}\hfill \hfill & \hfill \frac{9(3a^3-24b^3)}{9a^2-36b^2}=\frac{27(a^3-8b^3)}{9(a^2-4b^2)}\hfill \\ \hfill \hfill & \hfill =\frac{27(a-2b)(a^2+2ab+4b^2)}{9(a-2b)(a+2b)}\hfill \\ \hfill \hfill & \hfill =\frac{3(a^2+2ab+4b^2)}{a+2b},\hfill \end{array}$$

where $a\ne \pm 2b$.

(iii) $\frac{{𝐬}^{\mathrm{𝟑}}\mathbf{+}\mathrm{𝟏𝟐𝟓}{𝐭}^{\mathrm{𝟑}}}{{𝐬}^{\mathrm{𝟐}}\mathbf{-}\mathrm{𝟐}𝐬𝐭\mathbf{-}\mathrm{𝟑𝟓}{𝐭}^{\mathrm{𝟐}}}$

Answer:

$$s^3+125t^3=(s+5t)(s^2-5st+25t^2)$$

and

$$s^2-2st-35t^2=(s+5t)(s-7t).$$

Therefore,

$$\frac{s^2-5st+25t^2}{s-7t},$$

where $s\ne -5t,7t$.

5. Find possible expressions for the length and breadth of each of the following rectangles whose areas are given by the following expressions in square units.

(i) $\mathrm{𝟐𝟓}{𝐚}^{\mathrm{𝟐}}\mathbf{-}\mathrm{𝟑𝟎}𝐚𝐛\mathbf{+}\mathrm{𝟗}{𝐛}^{\mathrm{𝟐}}$

Answer:

$$25a^2-30ab+9b^2={(5a-3b)}^2.$$

Thus, possible dimensions are

$$\mathrm{\text{Length}}=5a-3b,\mathrm{\text{Breadth}}=5a-3b.$$

(ii) $\mathrm{𝟑𝟔}{𝐬}^{\mathrm{𝟐}}\mathbf{-}\mathrm{𝟒𝟗}{𝐭}^{\mathrm{𝟐}}$

Answer:

$$36s^2-49t^2=(6s-7t)(6s+7t).$$

Thus, possible dimensions are

$$\mathrm{\text{Length}}=6s+7t,\mathrm{\text{Breadth}}=6s-7t.$$

6. Find possible expressions for the length, breadth, and heights of each of the following cuboids whose volumes are given by the following expressions in cubic units.

(i) $\mathrm{𝟔}{𝐚}^{\mathrm{𝟐}}\mathbf{-}\mathrm{𝟐𝟒}{𝐛}^{\mathrm{𝟐}}$

Answer:

$$6a^2-24b^2=6(a^2-4b^2)=6(a-2b)(a+2b).$$

Possible dimensions are

$$6,a-2b,a+2b.$$

(ii) $\mathrm{𝟑}𝐩{𝐬}^{\mathrm{𝟐}}\mathbf{-}\mathrm{𝟏𝟓}𝐩𝐬\mathbf{+}\mathrm{𝟏𝟐}𝐩$

Answer:

$$\begin{array}{cc}\hfill \hfill & \hfill 3p{s}^2-15ps+12p=3p(s^2-5s+4)\hfill \\ \hfill \hfill & \hfill =3p(s-1)(s-4).\hfill \end{array}$$

Possible dimensions are

$$3p,s-1,s-4.$$

Printed Page 90

7. The village playground is shaped as a square of side 40 metres. A path of width $𝐬$ metres is created around the playground for people to walk. Find an expression for the area of the path in terms of $𝐬$.

Answer:

The outer square has side

$$40+2s.$$

Therefore,

$$\begin{array}{cc}\hfill \hfill & \hfill \mathrm{\text{Area of path}}={(40+2s)}^2-{40}^2\hfill \\ \hfill \hfill & \hfill =1600+160s+4s^2-1600\hfill \\ \hfill \hfill & \hfill =(4s^2+160s){\mathrm{\text{m}}}^2.\hfill \end{array}$$

Equivalently,

$$4s(s+40){\mathrm{\text{m}}}^2.$$

8. If a number plus its reciprocal equals $\frac{\mathrm{𝟏𝟎}}{\mathrm{𝟑}}$, find the number.

Answer:

Let the number be $x$, where $x\ne 0$.

$$x+\frac{1}{x}=\frac{10}{3}$$

$$3x^2-10x+3=0$$

$$(3x-1)(x-3)=0.$$

Therefore,

$$x=\frac{1}{3}\mathrm{\text{or}}x=3.$$

9. A rectangular pool has area $\mathrm{𝟐}{𝐱}^{\mathrm{𝟐}}\mathbf{+}\mathrm{𝟕}𝐱\mathbf{+}\mathrm{𝟑}$ square hastas. If its width is $\mathrm{𝟐}𝐱\mathbf{+}\mathrm{𝟏}$ hastas, find its length. Hasta was a unit used to measure length.

Answer:

$$2x^2+7x+3=(2x+1)(x+3).$$

Therefore, $\mathrm{\text{Length}}=\frac{(2x+1)(x+3)}{2x+1}=x+3\mathrm{\text{hastas}}$.

10. If both $𝐱\mathbf{-}\mathrm{𝟐}$ and $𝐱\mathbf{-}\frac{\mathrm{𝟏}}{\mathrm{𝟐}}$ are factors of $𝐩{𝐱}^{\mathrm{𝟐}}\mathbf{+}\mathrm{𝟓}𝐱\mathbf{+}𝐫$, show that $𝐩\mathbf{=}𝐫$.

Answer:

Since the two factors are $x-2$ and $x-\frac{1}{2}$,

$$p{x}^2+5x+r=p(x-2)(x-\frac{1}{2}).$$

Expanding,

$$\begin{array}{cc}\hfill \hfill & \hfill p(x-2)(x-\frac{1}{2})=p(x^2-\frac{5}{2}x+1)\hfill \\ \hfill \hfill & \hfill =p{x}^2-\frac{5p}{2}x+p.\hfill \end{array}$$

Comparing the coefficient of $x$,

$$-\frac{5p}{2}=5\Rightarrow p=-2.$$

Comparing the constant terms,

$$r=p=-2.$$

Hence,

$$p=r.$$

11. If $𝐚\mathbf{+}𝐛\mathbf{+}𝐜\mathbf{=}\mathrm{𝟓}$ and $𝐚𝐛\mathbf{+}𝐛𝐜\mathbf{+}𝐜𝐚\mathbf{=}\mathrm{𝟏𝟎}$, then prove that ${𝐚}^{\mathrm{𝟑}}\mathbf{+}{𝐛}^{\mathrm{𝟑}}\mathbf{+}{𝐜}^{\mathrm{𝟑}}\mathbf{-}\mathrm{𝟑}𝐚𝐛𝐜\mathbf{=}\mathbf{-}\mathrm{𝟐𝟓}$.

Answer:

Using

$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca),$$

and

$$\begin{array}{cc}\hfill \hfill & \hfill a^2+b^2+c^2-ab-bc-ca={(a+b+c)}^2-3(ab+bc+ca)\hfill \\ \hfill \hfill & \hfill =5^2-3\left(10\right)\hfill \\ \hfill \hfill & \hfill =-5,\hfill \end{array}$$

Thus, $a^3+b^3+c^3-3abc=5(-5)=-25$. Hence proved.

12. By factoring the expression, check that ${𝐧}^{\mathrm{𝟑}}\mathbf{-}𝐧$ is always divisible by 6 for all natural numbers $𝐧$. Give reasons.

Answer:

$$\begin{array}{cc}\hfill \hfill & \hfill n^3-n=n(n^2-1)\hfill \\ \hfill \hfill & \hfill =n(n-1)(n+1).\hfill \end{array}$$

These are three consecutive natural numbers.

• At least one of three consecutive numbers is divisible by $3$.
• At least one of them is even and therefore divisible by $2$.

Since $2$ and $3$ are coprime, their product is divisible by $6$. Therefore,

$$n^3-n\mathrm{\text{is always divisible by}}6.$$

13. Find the value of:

(i) ${𝐱}^{\mathrm{𝟑}}\mathbf{+}{𝐲}^{\mathrm{𝟑}}\mathbf{-}\mathrm{𝟏𝟐}𝐱𝐲\mathbf{+}\mathrm{𝟔𝟒}$, when $𝐱\mathbf{+}𝐲\mathbf{=}\mathbf{-}\mathrm{𝟒}$

Answer:

$$x^3+y^3-12xy+64=x^3+y^3+4^3-3xy\left(4\right).$$

Using the three-cube identity,

$$\begin{array}{cc}\hfill \hfill & \hfill x^3+y^3+4^3-3xy\left(4\right)=(x+y+4)\hfill \\ \hfill \hfill & \hfill \times (x^2+y^2+16-xy-4x-4y).\hfill \end{array}$$

Since $x+y=-4$,

$$x+y+4=0.$$

Therefore,

$$x^3+y^3-12xy+64=0.$$

(ii) ${𝐱}^{\mathrm{𝟑}}\mathbf{-}\mathrm{𝟖}{𝐲}^{\mathrm{𝟑}}\mathbf{-}\mathrm{𝟑𝟔}𝐱𝐲\mathbf{-}\mathrm{𝟐𝟏𝟔}$, when $𝐱\mathbf{=}\mathrm{𝟐}𝐲\mathbf{+}\mathrm{𝟔}$

Answer:

$$x^3-8y^3-36xy-216=x^3+{(-2y)}^3+{(-6)}^3-3x(-2y)(-6).$$

Using the three-cube identity, it contains the factor

$$x-2y-6.$$

Since $x=2y+6$, the factor $x-2y-6=0$. Therefore, $x^3-8y^3-36xy-216=0$.