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– pqr (p² + q² + r²) 

= – (pqr) × p² – (pqr) × q² – (pqr) × r² 

= – p³qr – pq³r – pqr³

5 Numeral expressions are 4, 100, – 17, 0, 2/3
5 algebric expressions are— 

2y², 3x² – 5, 13 – y + y², 4p²q – 3pq² + 5, xy + 4

Monomial Expressions—2y²

Binomial Expressions—3x² – 5, xy + 4

Trinomial Expressions—13 – y + y², 4p²q – 3pq² + 5

(i) 3 × 5x

= 3 × 5 × x

= 15x

(ii) – 5p,- 2q

= (- 5) × p × (- 2) × q

= (- 5) × (- 2) × p × q

= 10pq

(iii) 7l², – 3n²

= 7 × l² × (- 3) × n²

= 7 × (- 3) × l² × n²

= – 21l²n²

(iv) 6m, 3n

= 6 × m × 3 × n

= 6 × 3 × m × n

= 18 mn

(v) – 5x², – 2x

= (- 5) × x² × (- 2) × x

= (- 5) × (- 2) × x² × x

= 10x³

(b) (5x - 8)(5x + 1)

(5x –3)²  – (5x –3) – 20

Let (5x - 3) = a. Then, the expression becomes

a² - a - 20 = a² - 5a + 4a - 20

= a(a - 5) + 4(a - 5) = (a - 5)(a + 4)

= (5x - 3 - 5)(5x - 3 + 4)

= (5x - 8)(5x + 1)

3 × 4 × -5 × x × x × x

= -60 × x³

= -60x³

(a) (p – 19) (p + 2)

Factorised form of p² – 17p – 38 is = p² – 19p + 2p – 38

Take out the common factors,

= p (p – 19) + 2 (p – 19)

Again take out the common factor,

= (p – 19) (p + 2)

(px + qy) (ax – by) 

= px (ax – by) + qy (ax – by) 

= apx² – pbxy + aqxy – qby²

21x²y³ + 27x³y² 

= 3 × 7 × x × x × y × y × y + 3 × 3 × 3 × x × x × x × y × y 

= 3 × x × x × y × y (7y + 9x) (Using ab + ac = a (b + c)) 

= 3x²y² (7y + 9x)

(b) binomial

Let monomial = 2x, binomial = x + y

Then, product of a monomial and a binomial = (2x) × (x + y)

= 2x² + 2xy

(i) xy, x²y, xy, x

xy × x²y × xy × x
= x × y × x × x × y × x × y × x
= x × x × x × x × x × y × y × y
= x⁵ × y³
= x⁵y³

(ii) m, n, mn, m³n, mn³

m × n × mn × m³n × mn³

= m × n × m × n × m × m × m × n ×m × n × n × n
= m × m × m × m × m × m × n × n × n × n × n × n
= m⁶ × n⁶
= m⁶n⁶

(iii) kl, lm, km, klm

kl × lm × km × klm

= k × l × l × m × k × m × k × l × m
= k × k × k × l × l × l × m × m × m
= k³ × l³ × m³
= k³l³ × m³
= k³l³m³

B) (3k + 4l) (3k + 4l)

Correct option is (B) (3k + 4l) (3k + 4l)

\(9k^2+24\,kl+16\,l^2\) \(=(3k)^2+2\times3k\times4l+(4l)^2\)

\(=(3k+4l)^2\)

= \((3k+4l)(3k+4l)\).

i) (x + y) (2x – 5y + 3xy)

= x(2x – 5y + 3xy) + y(2x – 5y + 3xy)

= 2x² – 5xy + 3x²y + 2xy – 5y² + 3xy²

= 2x² – 5y² – 3xy + 3x²y + 3xy²

ii) (mn – kl + km) (kl – lm)

= kl(mn – kl + km) – lm(mn – kl + km)

= klmn – k²l² + k²lm – lm²n + kl²m – klm²

i) 2a – 9 and 3a + 4

= (2a – 9) (3a + 4) 

= 2a (3a + 4) – 9(3a + 4)

= 6a² + 8a – 27a – 36

= 6a² – 19a – 36

ii) x – 2y and 2x – y

= (x – 2y) (2x – y) 

= x(2x – y) – 2y(2x – y)

= 2x² – xy – 4xy + 2y²

= 2x² – 5xy + 2y²

iii) kl + lm and k – l

= (kl + lm) (k – l) 

= kl(k – l) + lm(k – l)

= k²l – l²k + klm – l²m

iv) m² – n² and m + n

= (m² – n²) (m + n) 

= m²(m + n) – n²(m + n)

= m³ + m²n – n²m – n³

(d) \(\big(a+\frac1a-1\big)^2\)

\(a^2 + \frac1{a^2}+3-2a-\frac2a\) 

= \(a^2 + \frac1{a^2}+1+2-2a-\frac2a\)

= \((a)^2+\big(\frac1a\big)^2+(-1)^2+2\times{a}\times\frac1a+2\times{a}\times(-1)+2\times\frac1a\times-1\)

= \(\big(a+\frac1a-1\big)^2\)

(c) (3 + 5a)(3 + 5a)(3 + 5a)

27 + 125a³ + 135a + 225a²

= (3)³ + (5a)³ + 3 x 3² x 5a + 3 x 3 x (5a)²

= (3 + 5a)³

4 × -3 × \(\frac{4}{5}\) × x² × x × x³

= \(\frac{-48}{5}\) × x⁶

= \(\frac{-48}{5}\)x⁶

(a) 1

Given exp. = \(\frac{(0.86)^3+(0.14)^3}{(0.86)^2-0.86\times0.14+(0.14)^2}\)

= \(\frac{(0.86+0.14)[(0.86)^2-0.86\times0.14+(0.14)^2]}{(0.86)^2-0.86\times0.14+(0.14)^2}\)                  [∴ (a3 + b3) = (a + b)(a² - ab + b²)]

= 0.86 + 0.14 = 1.

(b) (x + 1)(x² - 4x + 7)

x³ - 3x² + 3x + 7

= x³ - 3x² +3x - 1 + 8

= (x - 1)³ + (2)³

= (x - 1 + 2)[(x - 1)² - (x - 1) x 2 + 4)]

= (x + 1)(x² - 2x + 1 - 2x + 2 + 4)

= (x + 1)(x² - 4x + 7)

3.2 × 2.1 × x⁶ × x² × y³ × y²

= 6.72 × x⁸ × y⁵

= 6.72x⁸y⁵

Verify:

When x = 1 and y = 0.5

R.H.S = 6.72x³y⁵

= 6.72 × 1⁸ × 0.5⁵

= 0.21

L.H.S= 3.2 × 1⁶ × (-.5)³ × 2.1 × 1² × 0.5²= 0.21

Therefore,

L.H.S = R.H.S

5 × -1.5 × -12 × x⁶ × x² × x × y³ × y²

= 90 × x⁹ × y⁵

= 90x⁹y⁵

Verification:

x = 1 and y = 0.5

R.H.S = 90x⁹y⁵

= 90 (1)⁹ (05)⁵

= 2.8125

L.H.S = 2.8125

Therefore,

L.H.S = R.H.S

x⁶ × 2x × (-4x) × 5

= 2 × -4 × 5 × x⁶ × x × x

= -40 × x⁸

= -40 x⁸

-8 × -2 × x² × x × y⁶ × y

= 16 × x³ × y⁷

= 16x³y⁷

Verification is when,

x = 2.5 and y = 1

R.H.S = 16 (2.5)³ × (1)⁷

= 16 × 15.625

= 250

L.H.S = -8 × 2.5² × 1⁶ × -20 × 1 × 2.5

= 250

Therefore,

L.H.S = R.H.S

i) (7d – 9e) (7d – 9e)

= (7d – 9e)² is in the form of (a – b)².

= (7d)² – 2 × 7d × 9e + (9e)² [ ∵ (a – b)² = a² – 2ab + b²]

= 7d × 7d – 126de + 9e × 9e

= 49d² – 126de + 81e²

ii) (m² – n²) (m² + n²) is in the form of (a + b) (a – b).

∴ (a + b) (a – b) = a² – b²

∴ (m² + n²) (m² – n²) = (m²)² – (n²)² = m⁴ – n⁴

Since a² – b² = (a + b) (a – b), therefore 

38² – 22² 

= (38 – 22) (38 + 22)

= 16 × 60 

So,

(38² - 22²)/16

= (16 x 60)/16

= 60

(i) Given 7a²b from 3a²b

= 3a²b -7a²b

= (3 -7) a²b

= – 4a²b

(ii) Given 4xy from -3xy

= –3xy – 4xy

= –7xy

i) (3k + 4l) (3k 4l) = (3k + 4l)² is in the form of (a + b)².

= (3k)² + 2 × 3k × 4l+ (4l)² [ (a+ b)² = a² + 2ab + b²

= 3k × 3k + 24kl + 4l × 4l

= 9k² + 24kl + 16l²

ii) (ax² + by²) (ax² + by²) = (ax² + by²)² is in the form of (a + b)².

= (ax²)² + 2 × ax² × by² + (by²)²   [ ∵ (a + b)² = a² + 2ab + b² ]

= ax² × ax² + 2abx²y² + by² × by²

= a²x⁴ + 2ab x²y² + b²y⁴

i) (3t + 9s) (3t – 9s) = (3t)² – (9s)²  [ ∵ (a + b) (a – b) = a² – b² ]

= 3t × 3t – 9s × 9s

= 9t² – 81s²

ii) (kl – mn) (kl + mn) = (kl)² – (mn)²  [ ∵(a + b) (a – b) = a² – b² ]

= kl × kl – mn × mn

= k²l² – m²n²

i) 304² = (300 + 4)² is in the form of (a + b)².  [∵ (a + b)² = a² + 2ab + b² ]

a = 300, b = 4

(300 + 4)² = (300)² + 2 × 300 × 4 + (4)²

= 300 × 300+ 2400 + 4 × 4

= 90,000 + 2400 + 16

= 92,416

ii) 509² = (500 + 9)²

a  = 500, b = 9

= (500)² + 2 × 500 × 9 + (9)²     [ ∵ (a + b)² = a² + 2ab + b²]

= 500 × 500 + 9000 + 9 × 9

= 2,50,000 + 9000 + 81

= 2,59,081

(c) (a - 3)(a² + 3a + 9)(a + 1)(a² - a + 1)

a⁶ - 26a³ - 27

= a⁶ - 27 a³ + a³ - 27

 = a³ (a³ - 24) + 1(a³ - 27)

=(a³ - 27)(a³ + 1)

= (a - 3)(a² + 3a + 9)(a + 1)(a² - a + 1)

Given expression is 2x² – 4x + 5

When x = 0, then 

= 2(0)² – 4(0) + 5 

= 2(0) – 0 + 5 = 0 – 0 + 5 = 5 

When x = 0, then 2x² – 4x + 5 = 5

Correct option is B) – 3x² + 4x + 8 

(i) 4x² - 12x + 7

(2x)² – 2 (2x) (3) + 3² – 3² + 7 

= (2x – 3)² – 9 + 7

= (2x – 3)² – 2

Hence, 2 must be added to the expression in order to make a whole square

(ii) 4x² - 20x + 20

(2x)² – 2 (2x) (5) + 5² – 5² + 20 

= (2x – 5)² – 25 + 20

= (2x – 5)² – 5

Hence, 5 must be added to the expression in order to make it a whole square

(c) 4x

We know that, perimeter of the square = 4 × side

From the question it is given that, side length of the top of square table is x.

Then, perimeter = 4 × x

= 4x

i) (6x + 5) (6x + 6) is in the form of (ax + b) (ax + c).

(ax + b) (ax + c) = a²x² + ax(b + c) + bc

(6x + 5) (6x + 6) = (6)²x² + 6x (5 + 6) + 5 × 6

= 36x² + 6x × 11 + 30

= 36x² + 66x + 30

ii) (2b – a) (2b + c) is in the form of (ax – b) (ax + c).

(ax – b) (ax + c) = a²x² + ax(c – b) – cb

(2b – a) (2b + c) = (2)²(b)² + 2b (c – a) – ca

= 4b² + 2bc – 2ab – ca

\(\frac{1}{8}\) × \(\frac{1}{4}\) × 5 × x² × x⁴ × x × y⁴ × y² × y

= \(\frac{5}{32}\) × x⁶ × y⁶

= \(\frac{5}{32}\)x⁶y⁶

Verification:

When x = 1 and y = 2

R.H.S = \(\frac{5}{32}\) × 1⁶ × 2⁶

= \(\frac{5}{32}\) × 64

= 5 × 2

= 10

L.H.S = \(\frac{1}{8}\) × 1² × 2⁴ × \(\frac{1}{4}\) × 1⁴ × 2² × 1 × 2 × 5

 = 10

Therefore,

L.H.S = R.H.S

-8 × -20 × x² × x × y⁶ × y

= 160x³y⁷

Verify:

When,

x = 2.5 and y = 1

R.H.S = 160x³y⁷

= 160 × (2.5)³ × (1)⁷

= 2500

L.H.S = - 8 × 2.5² × 1 × -20 × 1 × 2.5

= 2500

Therefore,

L.H.S = R.H.S

-x × x³ × x × y³ × y × y

= -x⁵y⁵

Verify:

When x = 1 and y = 2

R.H.S = -x⁵y⁵

= (-1)⁵ × 2⁵

= -1 × 32

= -32

L.H.S = (-1) × 2³ × 2 × 1³ × 1 × 2

= -32

Therefore,

L.H.S = R.H.S

\(\frac{7}{9}\times \frac{15}{7}\times \frac{-3}{5}\) x a x a x a² x b² x b x c² x c

= -a⁴ x b³ x c³

= -a⁴b³c³

Let the added algebraic expression may be ‘A say

(1 + 2x – 3x²) + A = x² – x – 1

=A =(x² – x – 1) – (1 + 2x – 3x²)

= x² – x – 1 – 1 – 2x + 3x²

= (x² + 3x²) + ( – x – 2x) +( – 1 – 1)

∴ A = 4x² – 3x – 2

∴ The required added expression (A) 

= 4x² – 3x – 2

(c) (-2a + 3b + c)

4a² + 9b² + c² - 12ab + 6bc - 4ac

= (-2a)² + (3b)² + c² + 2 x (-2a) x 3b + 2 x 3b x c + 2 x (-2a) x c

= (-2a + 3b + c)²

∴ Reqd. sq. root = –2a + 3b + c

(a) 3(2x - 3y)(3y - 5z)(5z - 2x)

Since (2x - 3y) + (3y - 5z) + (5z - 2x) = 0

(2x - 3y)³ + (3y - 5z)³ + (5z - 2x)³ 

= 3(2x - 3y)(3y - 5z)(5z - 2x)

(i) Sum of 12ab, 9ab, ab

= 12ab + 9ab + ab 

= (12 + 9 + 1) ab 

= 22 ab.

(ii) Sum of 10x², – 3x², 5x²

= 10x² + (- 3x² ) + 5x² 

= 10x² – 3x² + 5x² 

= (10 – 3 + 5) x² 

= 12x²

(iii) Sum of – y², 5y², 8y², – 14y²

= – y² + 5y² + 8y² + (- 14y²) 

= – 1y² + 5y² + 8y² – 14y² 

= (- 1 + 5 + 8 – 14) y²

= – 2y²

(iv) Sum of 10mn, 6mn, – 2mn, – 7mn

= 10mn + 6mn + (- 2mn) + (- 7mn) 

= 10mn + 6mn – 2mn – 7mn 

= (10 + 6 – 2 – 7) mn 

= 7mn

Given that f he sum of 3 expressions is 8 + 13a + 7a² ……………..(1)

Two of them are 2a² + 3a + 2 and 3a² – 4a + 1.

∴ The sum of above two expressions (2a² + 3a + 2) + (3a² – 4a + 1)

= 2a² + 3a + 2 + 3a² – 4a + 1.

= (2a² + 3a²) + (3a – 4a) + (2 + 1)

= 5a² – a + 3

∴ The required 3rd expression (1) – (2)

(1) – (2) =(7a² + 13a + 8) – (5a² – a + 3) 

= 7a² + 13a + 8 – – 5a² – a – 3

=(7a² – 5a²) +(13a + a)+ (8 – 3)

= 2a² + 14a + 5

Given expression is 2x² – 4x + 5

When x = 1, then 

= 2(1)² – 4(1) + 5 

= 2 × 1 – 4 + 5 

= 2 – 4 + 5 = 3 

When x = 1, then 2x² – 4x + 5 = 3

(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

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

= (9p)² + 2(9p)(5q) + (5q)²

= 81p² + 90pq + 25q²

L.H.S = R.H.S

2a²b × (-5)ab²c × (-6)bc²

The coefficient of the product of two monomials is equal to the product of their coefficients.

The variable part in the product of two monomials is equal to the product of the variables in the given monomials.

Then,

= [(2) × (-5) × (-6)] × (a² × a) × (b × b² × b) × (c × c²)

= [60] × (a²⁺¹) × (b¹⁺²⁺¹) × (c¹⁺²) … [∵ a^m × aⁿ = a^m+n]

= [60] × (a³) × (b⁴) × (c³)

= [60]a³b⁴c³

\(\frac{-4}{7}\) × \(\frac{-2}{3}\) × \(\frac{-7}{6}\)× a² × a × b × b² × c × c²

= \(\frac{-4}{9}\)× a³ × b³ × c³

= \(\frac{-4}{9}\)a³b³c³

(d) (3a - b - 1)(9a² + b² + 1 + 3ab + 3a - b)

27a³ - b³ - 1 - 9ab

= (3a)³ + (-b)³ + (-1)³ - 3 x 3a x (-b) x (-1)

= [3a + (-b) + (-1)][(-3a)² + (-b)² + (-1)² - 3a x (-b) - (-b) x (-1) - 3a x (-1)]

 = (3a - b - 1)(9a² + b² + 1 + 3ab + b - 3a)

[∴ a³ + b³ + c³ - 3abc = (a + b + c) (a² + b² + c² - ab - bc - ca)]

2a³ (3a + 5b)

= 2a³ × 3a + 2a² × 5b

= 6 × a⁴ + 10a³b

\(\frac{-24}{25}\times \frac{-15}{16}\times x^{3}\times x\times z\times z^{2}\times y\)

= \(\frac{18}{20}\times x^{4}\times z^{3}\times y\)

= \(\frac{9}{10}x^4z^3y\)