Algebraic Expressions and Identities

EXERCISE-9.1

P1. Identify the terms, their coefficients for each of the following expressions.

(i) 5xyz² − 3zy

(ii) 1 + x + x²

(iii) 4x²y² − 4x²y²z² + z²

(iv) 3 − pq + qr − rp

(v) $\frac{\mathbf{x}}{\mathbf{2}}\mathbf{+}\frac{\mathbf{y}}{\mathbf{2}}\mathbf{–}\mathit{x}\mathit{y}$

(vi) 0.3a − 0.6ab + 0.5b

Sol. The terms and the respective coefficients of the given expressions are as follows.

–	Terms	Coefficients
(i)	5xyz2 − 3zy	5 − 3
(ii)	1 x x2	1 1 1
(iii)	4x2y2 − 4x2y2z2 z2	4 − 4 1
(iv)	3 − pq qr − rp	3 −1 1 −1
(v)	$\begin{array}{c}\frac{x}{2}\\ \frac{y}{2}\end{array}$ − xy	$\begin{array}{l}\begin{array}{c}\frac{1}{2}\\ \frac{1}{2}\end{array}\end{array}$ − 1
(vi)	0.3a − 0.6ab 0.5b	0.3 − 0.6 0.5

P2. Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

x + y,  1000, x + x² + x³ + x⁴,   7 + y + 5x,   2y − 3y²,  2y − 3y² + 4y³,  5x − 4y + 3xy,  4z −15z²,  ab + bc + cd + da, pqr, p²q + pq², 2p + 2q

Sol. The given expressions are classified as

Monomials : 1000, pqr

Binomials : x + y, 2y − 3y², 4z − 15z², p²q + pq², 2p + 2q

Trinomials :  7 + y + 5x, 2y − 3y² + 4y³, 5x − 4y + 3xy

Polynomials that do not fit in any of these categories are

x + x² + x³ + x⁴, ab + bc + cd + da

P3. 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

Sol. The given expressions written in separate rows, with like terms one below the other and then the addition of these expressions are as follows.

(i)  $\begin{array}{r}\mathrm{ab}–\mathrm{bc}\\ + \mathrm{bc}–\mathrm{ca}\\ + –\mathrm{ab}+\mathrm{ca}\\ 0\end{array}$

Thus, the sum of the given expressions is 0.

(ii)  $$\begin{gathered} a – b + ab \\ + b – c + bc \\ + –a c + ac \\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ ab+bc+ac \end{gathered}$$

Thus, the sum of the given expressions is ab + bc + ac.

(iii)  $\begin{array}{r}2p^2{q}^2–3pq+4\\ + –3p^2{q}^2+7pq+5\\ –p^2{q}^2+4pq+9\end{array}$

Thus, the sum of the given expressions is −p²q² + 4pq + 9.

(iv) $$\begin{gathered} l^{2 }+ m^2 \\ + m^2 + n^2 \\ + l^2 + n^2 \\ + 2lm + 2mn + 2nl \\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ 2l^2 + 2m^2 + 2n^2 + 2lm + 2mn + 2nl \end{gathered}$$

Thus, the sum of the given expressions is 2(l² + m² + n² + lm + mn + nl).

P4. (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

Sol. The given expressions in separate rows, with like terms one below the other and then the subtraction of these expressions is as follows.

(a) $\begin{array}{l}12a–9ab+5b–3\\ 4a–7ab+3b+12\\ (–)(+) (–) (–)\\ \overline{ 8a–2ab+2b–15}\end{array}$

(b) $\begin{array}{l}5xy–2yz–2zx+10xyz\\ 3xy+5yz–7zx\\ (–) (–) (+) \\ \overline{2xy–7yz+5zx+10xyz}\end{array}$

(c) $$\begin{gathered} 18-3\mathrm{p}-11\mathrm{q}+5\mathrm{pq}-2{\mathrm{pq}}^2+5{\mathrm{p}}^2\mathrm{q} \\ -10-8\mathrm{p}+7\mathrm{q}-3\mathrm{pq}+5{\mathrm{pq}}^2+4{\mathrm{p}}^2\mathrm{q} \\ (+) (+) (-) (+)(-)(-) \\ \mathrm{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_} \\ 28+5\mathrm{p}-18\mathrm{q}+8\mathrm{pq}-7{\mathrm{pq}}^2+{\mathrm{p}}^2\mathrm{q} \end{gathered}$$

EXERCISE-9.2

P1. Find the product of the following pairs of monomials.

(i) 4, 7p

(ii) − 4p, 7p

(iii) − 4p, 7pq

(iv) 4p³, − 3p

(v) 4p, 0

Sol. The product will be as follows.

(i) 4 × 7p = 4 × 7 × p = 28p

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

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

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

(v) 4p × 0 = 4 × p × 0 = 0

P2. 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)

Sol. We know that,

Area of rectangle = Length × Breadth

Area of 1^st rectangle = p × q = pq

Area of 2^nd rectangle = 10m × 5n = 10 × 5 × m × n = 50 mn

Area of 3^rd rectangle = 20x² × 5y² = 20 × 5 × x² × y²  = 100 x²y²

Area of 4^th rectangle = 4x × 3x² = 4 × 3 × x × x² = 12x³

Area of 5^th rectangle = 3mn × 4np = 3 × 4 × m × n × n × p  = 12mn²p

P3. Complete the table of products.

$\frac{\text{ First monomial }\to }{\text{ Second monomial }\downarrow }$	2x	− 5y	3x2	− 4xy	7x2y	− 9x2y2
2x	4x2	…	…	…	…	…
− 5y	…	…	− 15x2y	…	…	…
3x2	…	…	…	…	…	…
− 4xy	…	…	…	…	…	…
7x2y	…	…	…	…	…	…
− 9x2y2	…	…	…	…	…	…

Sol. The table can be completed as follows.

$\frac{\text{ First monomial }\to }{\text{ Second monomial }\downarrow }$	2x	− 5y	3x2	− 4xy	7x2y	− 9x2y2
2x	4x2	− 10xy	6x3	− 8x2y	14x3y	− 18x3y2
− 5y	− 10xy	25 y2	− 15x2y	20xy2	− 35x2y2	45x2y3
3x2	6x3	− 15x2y	9x4	− 12x3y	21x4y	− 27x4y2
− 4xy	− 8x2y	20xy2	− 12x3y	16x2y2	− 28x3y2	36x3y3
7x2y	14x3y	− 35x2y2	21x4y	− 28x3y2	49x4y2	− 63x4y3
− 9x2y2	− 18x3y2	45 x2y3	− 27x4y2	36x3y3	− 63x4y3	81x4y4

P4. 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

Sol. We know that,

Volume = Length × Breadth × Height

(i) Volume = 5a × 3a² × 7a⁴ = 5 × 3 × 7 × a × a² × a⁴ = 105 a⁷

(ii) Volume = 2p × 4q × 8r = 2 × 4 × 8 × p × q × r = 64pqr

(iii) Volume = xy × 2x²y × 2xy² = 2 × 2 × xy ×x²y × xy² = 4x⁴y⁴

(iv) Volume = a × 2b × 3c = 2 × 3 × a × b × c = 6abc

P5. 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

 Sol. (i) xy × yz × zx = x²y²z²

(ii)  a × (− a²) × a³ = − a⁶

(iii)  2 × 4y × 8y² × 16y³ = 2 × 4 × 8 × 16 × y × y² × y³ = 1024 y⁶

(iv) a × 2b × 3c × 6abc = 2 × 3 × 6 × a × b × c × abc = 36a²b²c²

(v) m × (− mn) × mnp = − m³n²p

EXERCISE-9.3

P1. 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

Sol. (i) (4p) × (q + r) = (4p × q) + (4p × r) = 4pq + 4pr

(ii) (ab) × (a − b) = (ab × a) + [ab × (− b)] = a²b − ab²

(iii) (a + b) × (7a² b²) = (a × 7a²b²) + (b × 7a²b²) = 7a³b² + 7a²b³

(iv) (a² − 9) × (4a) = (a² × 4a) + (− 9) × (4a) = 4a³ − 36a

(v) (pq + qr + rp) × 0 = (pq × 0) + (qr × 0) + (rp × 0) = 0

P2. Complete the table

—	First expression	Second Expression	Product
(i)	a	b + c + d	–
(ii)	x + y − 5	5 xy	–
(iii)	p	6p2 − 7p + 5	–
(iv)	4p2q2	p2 − q2	–
(v)	a + b + c	abc	–

Sol. The table can be completed as follows.

–	First expression	Second Expression	Product
(i)	a	b + c + d	ab + ac + ad
(ii)	x + y − 5	5 xy	5x2y + 5xy2 − 25xy
(iii)	p	6p2 − 7p + 5	6p3 − 7p2 + 5p
(iv)	4p2q2	p2 − q2	4p4q2 − 4p2q4
(v)	a + b + c	abc	a2bc + ab2c + abc2

P3. Find the product.

(i)  (a²) × (2a²²) × (4a²⁶)

(ii) $\left(\frac{2}{3}xy\right)\mathbf{\times }\left(\frac{–9}{10}{x}^2{y}^2\right)$

(iii) $\left(\frac{–10}{3}p{q}^3\right)\mathbf{\times }\left(\frac{6}{5}{p}^{3}q\right)$

(iv) x × x² × x³ × x⁴

Sol. (i) (a²) × (2a²²) × (4a²⁶) = 2 × 4 ×a² × a²² × a²⁶ = 8a⁵⁰

(ii) $\left(\frac{2}{3}xy\right)\times \left(\frac{–9}{10}{x}^2{y}^2\right)$ $=\left(\frac{2}{3}\right)\times \left(\frac{–9}{10}\right)\times \mathrm{x}\times \mathrm{y}\times {\mathrm{x}}^2\times {\mathrm{y}}^2=\frac{–3}{5}{\mathrm{x}}^3{\mathrm{y}}^3$

(iii) $\left(\frac{–10}{3}{\mathrm{pq}}^3\right)\times \left(\frac{6}{5}{\mathrm{p}}^3\mathrm{q}\right)$ $=\left(\frac{–10}{3}\right)\times \left(\frac{6}{5}\right)\times {\mathrm{pq}}^3\times {\mathrm{p}}^3\mathrm{q}=–4{\mathrm{p}}^4{\mathrm{q}}^4$

(iv) x × x² × x³ × x⁴ = x¹⁰

P4. (a) Simplify 3x (4x −5) + 3 and find its values for

(i) x = 3,

(ii) x =$\frac{1}{2}$.

(b) a (a² + a + 1) + 5 and find its values for

(i) a = 0,

(ii) a = 1,

(iii) a = − 1.

Sol. (a) 3x (4x − 5) + 3 = 12x² − 15x + 3

(i) For x = 3, 12x² − 15x + 3

= 12 (3)² − 15(3) + 3

= 108 − 45 + 3

= 66

(ii) For $\mathrm{x}=\frac{1}{2},12{\mathrm{x}}^2–15\mathrm{x}+3=12{\left(\frac{1}{2}\right)}^2–15\left(\frac{1}{2}\right)+3=12\times \frac{1}{4}–\frac{15}{2}+3$

$=3–\frac{15}{2}+3=6–\frac{15}{2}$

$=\frac{12–15}{2}=\frac{–3}{2}$

(b) a (a² + a + 1) + 5 = a³ + a² + a + 5

(i)  For a = 0, a³ + a² + a + 5 = 0 + 0 + 0 + 5 = 5

(ii) For a = 1, a³ + a² + a + 5 = (1)³ + (1)² + 1 + 5 = 1 + 1 + 1 + 5 = 8

(iii) For a = −1, a³ + a² + a + 5 = (−1)³ + (−1)² + (−1) + 5 = − 1 + 1 − 1 + 5 = 4

P5. (a) Add : p (p − q), q (q _­­­− r) and r (r ­− p)

(b) Add : 2x (z − x − y) and 2y (z − y − x)

(c)  Subtract : 3l (l − 4m + 5n) from 4l (10n − 3m + 2l)

(d) Subtract : 3a (a + b + c) − 2b (a − b + c) from 4c (− a + b + c)

Sol. (a) First expression = p (p − q) = p² − pq

Second expression = q (q _­­­− r) = q² − qr

Third expression = r (r ­− p) = r² − pr

Adding the three expressions, we obtain

$$\begin{gathered} p^2–pq \\ + q^2–qr \\ + r^2–pq \\ \overline{p^2–pq+q^2–qr+r^2–pq} \end{gathered}$$

Therefore, the sum of the given expressions is p² + q² + r² − pq − qr − rp.

(b) First expression = 2x (z − x − y) = 2xz − 2x² − 2xy

Second expression = 2y (z − y − x) = 2yz − 2y² − 2yx

Adding the two expressions, we obtain

$$\begin{gathered} 2xz–2x^2–2xy \\ + –2yx+2yz–2y^2 \\ \overline{2xz–2x^2–4xy+2yz–2y^2} \end{gathered}$$

Therefore, the sum of the given expressions is − 2x² − 2y² − 4xy + 2yz + 2zx.

(c) 3l (l − 4m + 5n) = 3l² − 12lm + 15ln

4l (10n − 3m + 2l) = 40ln − 12lm + 8l²

Subtracting these expressions, we obtain

$$\begin{gathered} 40ln–12lm+8l^2 \\ 15ln–12lm+3l^2 \\ \left(–\right) \left(+\right) \left(+\right) \\ \overline{+25ln +5l^2} \end{gathered}$$

Therefore, the result is 5l² + 25ln.

(d) 3a (a + b + c) − 2b (a − b + c)

= 3a² +3ab + 3ac − 2ba + 2b² − 2bc

= 3a² + 2b² + ab + 3ac − 2bc

4c (− a + b + c) = − 4ac + 4bc + 4c²

Subtracting these expressions, we obtain

$$\begin{gathered} - 4ac + 4bc + 4c^2 \\ 3ac-2bc +3a^2+2b^2+ab \\ (-)(+) (-) (-) (-) (-) \\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ -7ac+6bc+4c^2-3a^2-2b^2-ab \end{gathered}$$

Therefore, the result is −3a² −2b² + 4c² − ab + 6bc − 7ac

EXERCISE-9.4

P1. Multiply the binomials.

(i) (2x + 5) and (4x − 3)

(ii) (y − 8) and (3y − 4)

(iii) (2.5l − 0.5m) and (2.5l + 0.5m)

(iv) (a + 3b) and (x + 5)

(v)  (2pq + 3q²) and (3pq − 2q²)

(vi) $(\frac{3}{4}{a}^2+3b^2)$ and $\mathbf{4}(a^2–\frac{2}{3}{b}^2)$

Sol. (i) (2x + 5) × (4x − 3)

= 2x × (4x − 3) + 5 × (4x − 3)

= 8x² − 6x + 20x − 15

= 8x² + 14x −15 (By adding like terms)

(ii) (y − 8) × (3y − 4)

= y × (3y − 4) − 8 × (3y − 4)

= 3y² − 4y − 24y + 32

= 3y² − 28y + 32 (By adding like terms)

(iii) (2.5l − 0.5m) × (2.5l + 0.5m)

= 2.5l × (2.5l + 0.5m) − 0.5m (2.5l + 0.5m)

= 6.25l² + 1.25lm − 1.25lm − 0.25m²

= 6.25l² − 0.25m²

(iv) (a + 3b) × (x + 5)

= a × (x + 5) + 3b × (x + 5)

= ax + 5a + 3bx + 15b

(v) (2pq + 3q²) × (3pq − 2q²)

= 2pq × (3pq − 2q²) + 3q² × (3pq − 2q²)

= 6p²q² − 4pq³ + 9pq³ − 6q⁴

= 6p²q² + 5pq³ − 6q⁴

(vi) $\left(\frac{3}{4}{a}^2+3b^2\right)\times \left[4\left(a^2–\frac{2}{3}{b}^2\right)\right]$$=\left(\frac{3}{4}{a}^2+3b^2\right)\times \left(4a^2–\frac{8}{3}{b}^2\right)$

$=\frac{3}{4}{a}^2\times \left(4a^2–\frac{8}{3}{b}^2\right)+3b^2\times \left(4a^2–\frac{8}{3}{b}^2\right)$ = 3a⁴ −2a²b² + 12b²a² − 8b⁴

= 3a⁴ + 10a²b² − 8b⁴

P2. Find the product.

(i) (5 − 2x) (3 + x)

(ii) (x + 7y) (7x − y)

(iii) (a² + b) (a + b²)

(iv) (p² − q²) (2p + q)

Sol. (i) (5 − 2x) (3 + x) = 5 (3 + x) − 2x (3 + x)

= 15 + 5x − 6x − 2x²

= 15 − x − 2x²

(ii) (x + 7y) (7x − y)

= x (7x − y) + 7y (7x − y)

= 7x² − xy + 49xy − 7y²

= 7x² + 48xy − 7y²

(iii) (a² + b) (a + b²)

= a² (a + b²) + b (a + b²)

= a³ + a²b² + ab + b³

(iv)  (p² − q²) (2p + q)

= p² (2p + q) − q² (2p + q)

= 2p³ + p²q − 2pq² − q³

P3. 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)

Sol. (i) (x² − 5) (x + 5) + 25

= x² (x + 5) − 5 (x + 5) + 25

= x³ + 5x² − 5x − 25 + 25

= x³ + 5x² − 5x

(ii) (a² + 5) (b³ + 3) + 5

= a² (b³ + 3) + 5 (b³ + 3) + 5

= a²b³ + 3a² + 5b³ + 15 + 5

= a²b³ + 3a² + 5b³ + 20

(iii) (t + s²) (t² − s)

= t (t² − s) + s² (t² − s)

= t³ − st + s²t² − s³

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

= a (c − d) + b (c − d) + a (c + d) − b (c + d) + 2 (ac + bd)

= ac − ad + bc − bd + ac + ad − bc − bd + 2ac + 2bd

= (ac + ac + 2ac) + (ad − ad) + (bc − bc) + (2bd − bd − bd)

= 4ac

(v) (x + y) (2x + y) + (x + 2y) (x − y)

= x (2x + y) + y (2x + y) + x (x − y) + 2y (x − y)

= 2x² + xy + 2xy + y² + x² − xy + 2xy − 2y²

= (2x² + x²) + (y² − 2y²) + (xy + 2xy − xy + 2xy)

= 3x² − y² + 4xy

(vi) (x + y) (x² − xy + y²)

= x (x² − xy + y²) + y (x² − xy + y²)

= x³ − x²y + xy² + x²y − xy² + y³

= x³ + y³ + (xy² − xy²) + (x²y − x²y)

= x³ + y³

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

= 1.5x (1.5x + 4y + 3) − 4y 

(1.5x + 4y + 3) − 4.5x + 12y

= 2.25 x² + 6xy + 4.5x − 6xy − 16y² − 12y − 4.5x + 12y

= 2.25 x² + (6xy − 6xy) + (4.5x − 4.5x) − 16y² + (12y − 12y) = 2.25x² − 16y²

(viii) (a + b + c) (a + b − c)

= a (a + b − c) + b (a + b − c) + c (a + b − c)

= a² + ab − ac + ab + b² − bc + ca + bc − c²

= a² + b² − c² + (ab + ab) + (bc − bc) + (ca − ca)

= a² + b² − c² + 2ab

EXERCISE-9.5

P1. Use a suitable identity to get each of the following products.

(i) (x + 3) (x + 3) (ii) (2y + 5) (2y + 5)

(iii) (2a ­− 7) (2a − 7)

(iv) $(3a–\frac{1}{2})(3a–\frac{1}{2})$

(v) (1.1m − 0.4) (1.1 m + 0.4)

(vi) (a² + b²) (− a² + b²)

(vii) (6x − 7) (6x + 7)

(viii) (− a + c) (− a + c)

(ix) $(\frac{x}{2}+\frac{3y}{4})(\frac{x}{2}+\frac{3y}{4})$

(x) (7a − 9b) (7a − 9b)

Sol. The products will be as follows.

 (i) (x + 3) (x + 3) = (x + 3)²

= (x)² + 2(x) (3) + (3)² [(a + b)² 

= a² + 2ab + b²]

= x² + 6x + 9

(ii) (2y + 5) (2y + 5) = (2y + 5)²

= (2y)² + 2(2y) (5) + (5)²

= 4y² + 20y + 25

(iii) (2a ­− 7) (2a − 7) = (2a − 7)²

= (2a)² − 2(2a) (7) + (7)² 

[(a − b)² = a² − 2ab + b²]

= 4a² − 28a + 49

(iv)  $\left(3a–\frac{1}{2}\right)\left(3a–\frac{1}{2}\right)={\left(3a–\frac{1}{2}\right)}^2$ 

$=(3\mathrm{a}{)}^2–2(3\mathrm{a})\left(\frac{1}{2}\right)+{\left(\frac{1}{2}\right)}^2$

[(a − b)² = a² − 2ab + b²]

$=9a^2–3a+\frac{1}{4}$

(v) (1.1m − 0.4) (1.1 m + 0.4)

= (1.1m)² − (0.4)²

[(a + b) (a − b) = a² − b²]

= 1.21m² − 0.16

(vi) (a² + b²) (− a² + b²) = (b² + a²) (b² − a²)

= (b²)² − (a²)² [(a + b) (a − b) = a² − b²]

= b⁴ − a⁴

(vii) (6x − 7) (6x + 7) = (6x)² − (7)² [(a + b) (a − b) = a² − b²] = 36x² −49

(viii) (− a + c) (− a + c) = (− a + c)²

= (− a)² + 2(− a) (c) + (c)² [(a + b)² 

= a² + 2ab + b²]

= a² − 2ac + c²

(ix) $\left(\frac{x}{2}+\frac{3y}{4}\right)\left(\frac{x}{2}+\frac{3y}{4}\right)$$={\left(\frac{x}{2}+\frac{3y}{4}\right)}^2$ 

$={\left(\frac{x}{2}\right)}^2+2\left(\frac{x}{2}\right)\left(\frac{3y}{4}\right)+{\left(\frac{3y}{4}\right)}^2$

[(a + b)² = a² + 2ab + b²]

$=\frac{x^2}{2}+\frac{3xy}{4}+\frac{9y^2}{16}$

(x) (7a − 9b) (7a − 9b) = (7a − 9b)²

= (7a)² − 2(7a)(9b) + (9b)² [(a − b)² 

= a² − 2ab + b²]

= 49a² − 126ab + 81b²

P2. Use the identity (x + a) (x + b) = x² + (a + b)x + ab to find the following products.

(i) (x + 3) (x + 7)

(ii) (4x +5) (4x + 1)

(iii) (4x − 5) (4x − 1)

(iv) (4x + 5) (4x − 1)

(v) (2x +5y) (2x + 3y)

(vi) (2a² +9) (2a² + 5)

(vii) (xyz − 4) (xyz − 2)

Sol. The products will be as follows.

(i) (x + 3) (x + 7) = x² + (3 + 7) x + (3) (7) = x² + 10x + 21

(ii) (4x + 5) (4x + 1) = (4x)² + (5 + 1) (4x) + (5) (1) = 16x² + 24x + 5

(iii) (4x  – 5) (4x  – 1) = (4x)²+[(– 5) + (–1)] (4x) + (– 5) (– 1) = 16x² − 24x + 5

(iv) (4x + 5) (4x  – 1) = (4x)² + [(5) + (–1)] (4x) + (5) (– 1) = 16x² + 16x − 5

(v) (2x +5y) (2x + 3y) = (2x)² + (5y + 3y) (2x) + (5y) (3y) = 4x² + 16xy + 15y²

(vi) (2a² +9) (2a² + 5) = (2a²)² + (9 + 5) (2a²) + (9) (5) = 4a⁴ + 28a² + 45

(vii) (xyz − 4) (xyz − 2) = (xyz)²+[(−4) + (−2)](xyz) + (−4)(− 2) = x²y²z² − 6xyz + 8

P3. Find the following squares by suing the identities.

(i) (b − 7)²

(ii) (xy + 3z)²

(iii) (6x² − 5y)²

(iv) ${(\frac{2}{3}m+\frac{3}{2}n)}^{\mathbf{2}}$      

(v) (0.4p − 0.5q)²

(vi) (2xy + 5y)²

Sol. (i) (b − 7)² = (b)² − 2(b) (7) + (7)²

[(a − b)² = a² − 2ab + b²] = b² − 14b + 49

(ii) (xy + 3z)² = (xy)² + 2(xy) (3z) + (3z)² [(a + b)² = a² + 2ab + b²]

= x²y² + 6xyz + 9z²

(iii) (6x² − 5y)² = (6x²)² − 2(6x²) (5y) + (5y)² [(a − b)² = a² − 2ab + b²]

= 36x⁴ − 60x²y + 25y²

(iv) $={\left(\frac{2}{3}m+\frac{3}{2}n\right)}^2$

$={\left(\frac{2}{3} \mathrm{m}\right)}^2+2\left(\frac{2}{3} \mathrm{m}\right)\left(\frac{3}{2}\mathrm{n}\right)+{\left(\frac{3}{2}\mathrm{n}\right)}^2$

 [(a + b)² = a² + 2ab + b²]

$=\frac{4}{9}{ \mathrm{m}}^2+2\mathrm{mn}+\frac{9}{4}{\mathrm{n}}^2$

(v) (0.4p − 0.5q)² = (0.4p)² − 2 (0.4p) (0.5q) + (0.5q)²

[(a − b)² = a² − 2ab + b²] = 0.16p² − 0.4pq + 0.25q²

(vi) (2xy + 5y)² = (2xy)² + 2(2xy) (5y) + (5y)²

[(a + b)² = a² + 2ab + b²] = 4x²y² + 20xy² + 25y²

P4. Simplify.

(i) (a² − b²)² (ii) (2x +5)² − (2x − 5)²

(iii) (7m − 8n)² + (7m + 8n)²

(iv) (4m + 5n)² + (5m + 4n)²

(v) (2.5p − 1.5q)² − (1.5p − 2.5q)²

(vi) (ab + bc)² − 2ab²c

(vii) (m² − n²m)² + 2m³n²

Sol. (i) (a² − b²)²  = (a²)² − 2(a²) (b²) + (b²)² [(a − b)² 

= [a² − 2ab + b² ] = a⁴ − 2a²b² + b⁴

(ii) (2x +5)² − (2x − 5)²  = (2x)² + 2(2x) (5) + (5)² − [(2x)² − 2(2x) (5) + (5)²]

[(a − b)² = a² − 2ab + b²]

[(a + b)² = a² + 2ab + b²]

= 4x² + 20x + 25 − [4x² − 20x + 25]

= 4x² + 20x + 25 − 4x² + 20x − 25 = 40x

(iii) (7m − 8n)² + (7m + 8n)²

= (7m)² − 2(7m) (8n) + (8n)² + (7m)² + 2(7m) (8n) + (8n)²

[(a − b)² = a² − 2ab + b² and (a + b)² = a² + 2ab + b²]

= 49m² − 112mn + 64n² + 49m² + 112mn + 64n²

= 98m² + 128n²

(iv) (4m + 5n)² + (5m + 4n)²

= (4m)² + 2(4m) (5n) + (5n)² + (5m)² + 2(5m) (4n) + (4n)²

= 16m² + 40mn + 25n² + 25m² + 40mn + 16n²

= 41m² + 80mn + 41n²

(v) (2.5p − 1.5q)² − (1.5p − 2.5q)²

= (2.5p)² − 2(2.5p) (1.5q) + (1.5q)² − [(1.5p)² − 2(1.5p)(2.5q) + (2.5q)²]

[(a − b)² = a² − 2ab + b² ]

= 6.25p² − 7.5pq + 2.25q² − [2.25p² − 7.5pq + 6.25q²]

= 6.25p² − 7.5pq + 2.25q² − 2.25p² + 7.5pq − 6.25q²] = 4p² − 4q²

(vi) (ab + bc)² − 2ab²c

= (ab)² + 2(ab)(bc) + (bc)² − 2ab²c [(a + b)² = a² + 2ab + b² ]

= a²b² + 2ab²c + b²c² − 2ab²c

= a²b² + b²c²

(vii) (m² − n²m)² + 2m³n²

= (m²)² − 2(m²) (n²m) + (n²m)² + 2m³n² [(a − b)² = a² − 2ab + b² ]

= m⁴ − 2m³n² + n⁴m² + 2m³n²

= m⁴ + n⁴m²

P5.  Show that

(i) (3x + 7)² − 84x = (3x − 7)²

(ii) (9p − 5q)² + 180pq = (9p + 5q)²

(iii) ${(\frac{4}{3}m–\frac{3}{4}n)}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathit{m}\mathit{n}\mathbf{=}\frac{\mathbf{16}}{\mathbf{9}}{\mathit{m}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{9}}{\mathbf{16}}{\mathit{n}}^{\mathbf{2}}$ 

(iv) (4pq + 3q)² − (4pq − 3q)² = 48pq²

(v) (a − b) (a + b) + (b − c) (b + c) + (c − a) (c + a) = 0

Sol. (i) L.H.S = (3x + 7)² − 84x

= (3x)² + 2(3x)(7) + (7)² − 84x

= 9x² + 42x + 49 − 84x

= 9x² − 42x + 49

R.H.S = (3x − 7)² 

= (3x)² − 2(3x)(7) +(7)²

= 9x² − 42x + 49

$\Rightarrow$L.H.S = R.H.S

(ii) L.H.S = (9p − 5q)² + 180pq

= (9p)² − 2(9p)(5q) + (5q)² − 180pq

= 81p² − 90pq + 25q² + 180pq

= 81p² + 90pq + 25q²

R.H.S = (9p + 5q)²

= (9p)² + 2(9p)(5q) + (5q)²

= 81p² + 90pq + 25q²

$\Rightarrow$ L.H.S = R.H.S

(iii) L.H.S = ${\left(\frac{4}{3}m–\frac{3}{4}n\right)}^2+2mn$

$={\left(\frac{4}{3} \mathrm{m}\right)}^2–2\left(\frac{4}{3} \mathrm{m}\right)\left(\frac{3}{4}\mathrm{n}\right)+{\left(\frac{3}{4}\mathrm{n}\right)}^2+2\mathrm{mn}$

$=\frac{16}{9}{ \mathrm{m}}^2–2\mathrm{mn}+\frac{9}{16}{\mathrm{n}}^2+2\mathrm{mn}$

$=\frac{16}{9}{ \mathrm{m}}^2+\frac{9}{16}{\mathrm{n}}^2=\mathrm{R}.\mathrm{H}.\mathrm{S}.$

(iv) L.H.S = (4pq + 3q)² − (4pq − 3q)²

= (4pq)² + 2(4pq)(3q) + (3q)² − [(4pq)² − 2(4pq) (3q) + (3q)²]

= 16p²q² + 24pq² + 9q² − [16p²q² − 24pq² + 9q²]

= 16p²q² + 24pq² + 9q² −16p²q² + 24pq² − 9q²

= 48pq² = R.H.S

(v) L.H.S = (a − b) (a + b) + (b − c) (b + c) + (c − a) (c + a)

= (a² − b²) + (b² − c²) + (c² − a²) = 0

= R.H.S.

P6. Using identities, evaluate.

(i) 71²                    (ii) 99²

(iii) 102²             (iv) 998²

(v) (5.2)²             (vi) 297 × 303

(vii) 78 × 82    (viii) 8.9²

(ix) 1.05 × 9.5

Sol. (i) 71² = (70 + 1)² = (70)² + 2(70) (1) + (1)²

[(a + b)² = a² + 2ab + b² ] = 4900 + 140 + 1 = 5041

(ii) 99² = (100 − 1)²  = (100)² − 2(100) (1) + (1)²

[(a − b)² = a² − 2ab + b² ] = 10000 − 200 + 1 = 9801

(iii) 102² = (100 + 2)² = (100)² + 2(100)(2) + (2)²

[(a + b)² = a² + 2ab + b² ] = 10000 + 400 + 4 = 10404

(iv) 998² = (1000 − 2)² = (1000)² − 2(1000)(2) + (2)²

[(a − b)² = a² − 2ab + b² ] = 1000000 − 4000 + 4 = 996004

(v) (5.2)² = (5.0 + 0.2)² = (5.0)² + 2(5.0) (0.2) + (0.2)²

[(a + b)² = a² + 2ab + b² ] = 25 + 2 + 0.04 = 27.04

(vi) 297 × 303 = (300 − 3) × (300 + 3) = (300)² − (3)² [(a + b) (a − b)

= [a² − b²] = 90000 − 9 = 89991

(vii) 78 × 82 = (80 − 2) (80 + 2)

= (80)² − (2)² [(a + b) (a − b) = a² − b²] = 6400 − 4 = 6396

(viii) 8.9² = (9.0 − 0.1)² = (9.0)² − 2(9.0) (0.1) + (0.1)²

[(a − b)² = a² − 2ab + b² ] = 81 − 1.8 + 0.01 = 79.21

(ix) 1.05 × 9.5 = 1.05 × 0.95 × 10 = (1 + 0.05) (1− 0.05) ×10

= [(1)² − (0.05)²] × 10 = [1 − 0.0025] × 10 [(a + b) (a − b)

= a² − b²] = 0.9975 × 10 = 9.975

P7. Using a² − b² = (a + b) (a − b), find

(i) 51² − 49²

(ii) (1.02)² − (0.98)²

(iii) 153² − 147²

(iv) 12.1² − 7.9²

Sol. (i) 51² − 49² = (51 + 49) (51 − 49) = (100) (2) = 200

(ii) (1.02)² − (0.98)²  = (1.02 + 0.98) (1.02 ­− 0.98)  = (2) (0.04) = 0.08

(iii) 153² − 147² = (153 + 147) (153 − 147) = (300) (6) = 1800

(iv) 12.1² − 7.9² = (12.1 + 7.9) (12.1 − 7.9) = (20.0) (4.2) = 84

P8. Using (x + a) (x + b) = x² + (a + b) x + ab, find

(i) 103 × 104

(ii) 5.1 × 5.2

(iii) 103 × 98

(iv) 9.7 × 9.8

Sol. (i) 103 × 104 = (100 + 3) (100 + 4)

= (100)² + (3 + 4) (100) + (3) (4)

= 10000 + 700 + 12 = 10712

(ii) 5.1 × 5.2 = (5 + 0.1) (5 + 0.2)

= (5)² + (0.1 + 0.2) (5) + (0.1) (0.2)

= 25 + 1.5 + 0.02 = 26.52

(iii) 103 × 98 = (100 + 3) (100 − 2)

= (100)² + [3 + (− 2)] (100) + (3) (− 2)

= 10000 + 100 − 6

= 10094

(iv) 9.7 × 9.8 = (10 − 0.3) (10 − 0.2)

= (10)² + [(− 0.3) + (− 0.2)] (10) + (− 0.3) (− 0.2)

= 100 + (− 0.5)10 + 0.06 = 100.06 − 5 = 95.06