DAV Class 7 Maths Chapter 6 Brain Teasers Solutions

The DAV Class 7 Maths Solutions and DAV Class 7 Maths Chapter 6 Brain Teasers Solutions of Algebraic Expressions offer comprehensive answers to textbook questions.

DAV Class 7 Maths Ch 6 Brain Teasers Solutions

Question 1.
A. Tick (S) the correct option:
(i) (3p² – 14pq + 2r) – (14pq + 3p² + 2r²) is a:
(a) monomial
(b) binomial
(c) trinomial
(d) none
Answer:
(3p² – 14pq + 2r) – (14pp + 3p² + 2r²)
= 3p² – 14pq + 2r – 14pq – 3p² – 2r²
= -2 r² – 28 pq + 2r
As this algebraic expression have three terms so it is a trinomial.
Hence, (c) is the correct answer.

(ii) H.C.F. of the terms of the expression (4p³q²r – 12pq²r² + 16p²q²r²) is:
(a) 4pq²r
(b) -4pq²r
(c) 16p³q²r²
(d) -16p³q²r²
Answer:
We know HCF shall be the product of the highest common coefficient and highest common powers of the variables respectively.
HCF of 4p³q²r, -12pq²r² and 16p²q²r² is 4pq²r.
Hence, (a) is the correct option.

(iii) For what value of’t’ the expressions (2x² – 5x + 10) amd (2x² – tx + 2t) are equal:
(a) 2
(b) 5
(c) 3
(d) 4
Answer:
The given expressions are (2x² – 5x + 10) and (2x² – tx + 2t).
Now, comparing the coefficients of x, we have -5 = -t
⇒ t = 5
Also, comparing the constant terms in the given expressions, we have
10 = 21
⇒ t = \(\frac{10}{2}\) = 5
Hence, (b) is the correct option.

(iv) Value of ‘p’ if the expression z² + 3z – p equals to 3 for z = 2 is:
(a) 6
(b) 5
(c) 7
(d) 4
Answer:
On putting the value of z = 2 in the expression z² + 3z – p, we have
(2)2 + 3(2) – p = 3
4 + 6 – p = 3
⇒ 10 – p = 3
⇒ p = 10 – 3
⇒ p = 7
Hence, (c) is the correct option.

(v) Factors of 2x³ + 5x – 6x² – 15 are:
(a) (2x² + 5) and (x + 3)
(b) (2x² – 5) and (x + 3)
(c) (2x² + 5) and (x – 3)
(d) (2x² – 5) and (x + 3)
Answer:
= (2x³ – 6x²) + (5x – 15) = 2x² (x – 3) + 5(x – 3)
= (x – 3)(2x² + 5)
= (2x² + 5) (x – 3)
Hence, (c) is the correct option.

B. Answer the following questions:
(i) In the given figure, ABCD is a rectangle with length (p²q – 5) and breadth (q² + 4p). Find the area of shaded triangle PAB.

Answer:

Draw PQ perpendicular to AB
Here, base (AB) = (p2q – 5) and altitude (PQ) = (q2 + 4p)
∴ Area of shaded triangle PAB = \(\frac{1}{2}\) × Base × Altitude
= \(\frac{1}{2}\) × AB × PQ
= \(\frac{1}{2}\) × (p²q – 5) × (q² + 4p)
= \(\frac{1}{2}\) (p²q – 5) (q² + 4P)
= \(\frac{1}{2}\)[p²q[q² + 4p) – 5(q² + 4p)]
= \(\frac{1}{2}\) (p²q³ + 4p³q – 5q² – 20p) sq. units

(ii) By how much does the expression 72z² – 45z + 4 exceed the expression 30z – 42z² – 17?
Answer:
The expression (72z² – 45z + 4) exceeds the expression (30z – 42z² -17) by (72z² – 45z + 4) – (30z – 42z² – 17).
Now, subtracting by horizontal method, we have (72z² – 45z + 4) – (30z – 42z² – 17)
= 72z² – 45z + 4 – 30z + 42z² + 17
= (72z² + 42z²) + (-45z – 30z) + (4 + 17)
= 114z² – 75z + 21
So, the expression (72z² – 45z + 4) exceeds the expression (30z – 42z² – 17) by (114z² – 75z + 21).

(iii) The perimeter of a triangle is (x²y + 10) units. One of the sides is of length (x²y – 4) units and another side is (3 – 2x²y) units. Find the length of the third side.
Answer:
Perimeter of a triangle = Sum of three sides of the triangle.
∴ (x²y + 10) = (x²y – 4) + (3 – 2x²y) + Length of the third side
⇒ (x²y + 10) = (-x²y -1) + length of the third side
⇒ Length of the third side = (x²y + 10) – (-x²y – 1)
= x²y + 10 + x²y + 1
= 2 x²y + 11 units
So, the length of the third side is given by the algebraic expression (2x²y + 11) units.

(iv) Find the HCF of the terms of the expression
2b³c + 4b⁴c² + 16b²c²
Answer:
HCF of 2b³c, 4b⁴c² and 16b²c² is 2b²c.
∴2b³c + 4b⁴c² + 16b²c²
= 2b²c (b + 2b²c + 8c)
So, factors of the expression 2b³c + 4b⁴c² + 16b²c² are 2b²c and (b + 2b²c + 8c).

(v) If area of a rectangle with length (a – b) and breadth (2a + b) is the same as the algebraic expression Kab + 2a² – b², find K.
Answer:
Length of rectangle = (a – b)
Breadth of rectangle = (2a + b)
∴ Area of rectangle = Length x Breadth
= (a – b)(2a + b)
= a(2a + b) – b(2a + b)
= 2a² + ab – 2ab – b²
= 2a² – ab – b²
Now, according to the question, we have
2 a² – a b – b²
= Kab + 2 a² – b²

Comparing the coefficients of term ‘ab’, we have
K = -1
So, the value of K is -1.

Questions 2.
Find the following products.
(i) (13ax – 4) (5ay + 1)
Answer:
(13ax – 4) (5ay + 1) = 13ax (5ay + 1) – 4 (5ay + 1)
= 13ax × 5ay + 1 × 13ax – 4 × 5ay – 4 × 1 .
= 65a²xy + 13 ax – 20ay – 4

(ii) (3x² + 5x – 7) (x + 5y)
Answer:
(3x² + 5x – 7) (x + 5y)
= 3x² (x + 5y) + 5x (x + 5y) – 7 (x + 5y)
= 3x³ + 15x²y + 5x² + 25xy – 7x – 35y

(iii) -6a² b (a⁴ + b⁴ – 3a² b²)
Answer:
– 6a²b (a⁴ + b⁴ – 3a²b²)
= – 6a²b × a4 – 6a²b × b⁴ – 6a²b × – 3a²b²
= – 6a⁶b – 6a²b⁵ + 18a⁴b³

(iv) (\(\frac{3}{5}\) x – \(\frac{2}{9}\)y)(15x – 9y)
Answer:
(\(\frac{3}{5}\) x – \(\frac{2}{9}\)y) (15x – 9y) = \(\frac{3}{5}\)x (15x – 9y) – \(\frac{2}{9}\)y (15x – 9y)
= \(\frac{3}{5}\)x × 15x – \(\frac{3}{5}\)x × 9y – \(\frac{2}{9}\)y × 15x – \(\frac{2}{9}\)y × – 9y
= 9x² – \(\frac{27}{5}\)xy – \(\frac{10}{3}\)y – xy + 2y²
= 9x² – \(\left(\frac{27}{5}+\frac{10}{3}\right)\)xy + 2y²
= 9x² – \(\frac{131}{15}\)xy + 2y²

(v) (x⁹) (- x¹⁰) (x¹¹) (-x¹²)
Answer:
(x⁹) (- x¹⁰) (x¹¹) (- x¹²)
= (-)( -) x^9+10+11+12
= x⁴²

(vi) 0.9 p³ q³ (10p – 20q)
Answer:
0.9p³q³ (10p – 20q)
= 0.9p³q³ × 10p – 0.9p³q³ × 20q
= 9p⁴q³ – 18p³q⁴

(vii) (7x² – 11x + 10) (x³ – x²)
Answer:
(7x² – 11x + 10) (x³ – x²)
= 7x²(x³ – x²) – 11x (x³ – x²) + 10(x³ – x²)
= 7x⁵ – 7x⁴– 11x⁴ + 11 x³ + 10x³ – 10x²
= 7x⁵ – 18x⁴ + 21x³ – 10x²

(viii) (\(\frac{10}{9}\)a⁵b⁶c⁶) (-\(\frac{3}{2}\)b²c) (\(\frac{6}{5}\)c³d⁴)(-a³d⁵)
Answer:
(\(\frac{10}{9}\)a⁵b⁶c⁶) (-\(\frac{3}{2}\)b²c) (\(\frac{6}{5}\)c³d⁴)(-a³d⁵)
= \(\) × -1 × a⁵ × a³ × b⁶ × b² × c⁶ × c × c³ × d⁴ × d⁵
= 2 a⁸b⁸c¹⁰d⁹

(ix) (0.7x³ – 0.5y³) (0.7x³ + 0.5y³)
Answer:
(0.7x³ – 0.5y³) (0.7x³ + 0.5y³)
= 0.7x³(0.7x³ + 0.5y³) – 0.5y³ (0.7x³ + 0.5y³)
= 0.7x³ x 0.7x³ + 0.7x³ x 0.5 y³ – 0.5y³ x 0.7x³ – 0.5y³ x 0.5 y³
= 0.49x⁶ + 0.35x³y³ – 0.35x³y³ – 0.25y⁶
= 0.49x⁶ – 0.25y⁶

(x) (64a² – 56ab + 49b²) (8a + 7b)
Answer:
(64a² – 56ab + 49b²) (8a + 7b)
= 64a²(8a + 7b) – 56ab (8a + 7b) + 49b² (8a + 7b)
= 64a² x 8a + 64a² x 7b – 56ab x 8a – 56ab x 7b + 49b² x 8a + 49b² x 7b
= 512a³ + 448a²b – 448a²b – 392ab² + 392ab² + 343b³
= 512a³ + 343b³

Question 3.
Find the variable part in the product of:
(i) 127x³y⁹, 255x⁶y⁵ and 313y³z⁴
(ii) \(\frac{61}{17}\)a²b⁴, \(\frac{97}{43}\)b³ , \(\frac{29}{41}\)a⁶b and \(\frac{111}{123}\)
Solution:
(i) Variable part in 127x³ y⁹ = x³ y⁹
Variable part in 255x⁶ y⁵ = x⁶ y⁵
Variable part in 313y² z⁴ = y² z⁴

(ii) Variable part in \(\frac{61}{17}\)a² b⁴ = a² b⁴
Variable part in \(\frac{97}{43}\)b³ = b³
Variable part in \(\frac{29}{41}\)a⁶ b = a⁶ b
Variable part in \(\frac{111}{123}\) = nil

Question 4.
Express (5x⁶)(12x²y) (\(\frac{3}{20}\)xy²) as a monomial and then evaluate at x = 1, y = 2.
Answer:
(5x⁶)(12x²y) (\(\frac{3}{20}\)xy²) = 5 × 12 × x⁶ × x²y × xy²
= 9x⁹y³
Put x = 1, y = 2
= 9(1 )⁹ (2)³
= 9 × 1 × 8 = 72

Question 5.
Express 1.5a²(10ab – 4b²) as a binomial and then evaluate at a = – 2, b = 3.
Answer:
1.5a²(10ab – 4b²) = 1.5a² × 10ab – 1.5a² × 4b²
= 15a³b – 6a²²
Put a = – 2, b = 3
= 15(-2)³(3) – 6 (- 2)²(3)²
= 15 × -8 × 3 – 6 × 4 × 9
= – 360 – 216
– 576

Question 6.
Simplify and then verify the result for the given values:
(i) (3x – 4y) (4x²y + 3 xy²); x = 2, y = -1
Answer:
(i) (3x – 4y) (4x²y + 3xy²); x = 2, y = – 1
= 3x(4x²y + 3xy²) – 4y (4x²y + 3xy²)
= 12x³y + 9x²y² – 16x²y² – 12xy³
= 12x³y – 7x²y² – 12xy³

Verification:
L.H.S = (3x – 4y) (4x²y + 3xy²)
Putting x = 2, y = -1
= [3(2) – 4(-1)] [4(2)² (-1) + 3(2)(-1)²]
= (6 + 4) (-16 + 6)
= (10)(-10) = -100

R.H.S = 12x³y – 7x²y² – 12xy³
Putting x = 2, y = -1
= 12(2)³(-1) – 7(2)²(-1)² – 12(2)(-2)³
= -96 – 28 + 24
= -100
Hence verified
L.H.S = R.H.S

(ii) (\(\frac{1}{4}\)a² + \(\frac{5}{9}\)b²)(a + b + ab); a = 2, b = 3
Answer:
(\(\frac{1}{4}\)a² + \(\frac{5}{9}\)b²)(a + b + ab); a = 2, b = 3
= \(\frac{1}{4}\)a²(a + b + ab) + \(\frac{5}{9}\)b²(a + b + ab)
= \(\frac{1}{4}\)a² × a + \(\frac{1}{4}\)a² × b + \(\frac{1}{4}\)a² × ab + \(\frac{5}{9}\)b² × a + \(\frac{5}{9}\)b² × b + \(\frac{5}{9}\)b² × ab
= \(\frac{1}{4}\)a³ + \(\frac{1}{4}\)a²b + \(\frac{1}{4}\)a³b + \(\frac{5}{9}\)ab² + \(\frac{5}{9}\)b³ + \(\frac{5}{9}\)ab³

Verification:
L.H.S = (\(\frac{1}{4}\)a² + \(\frac{5}{9}\)b²)(a + b + ab)
Putting a= 2, b = 3
= (\(\frac{1}{4}\)(2)² + \(\frac{5}{9}\)(3)²)(2 + 3 + 2 × 3)
= [\(\frac{1}{4}\) × 4 + \(\frac{5}{9}\) × 9] [5 + 6]
= (1 + 5)(11) = 77

R.H.S = \(\frac{1}{4}\)a³ + \(\frac{1}{4}\)a²b + \(\frac{1}{4}\)a³b + \(\frac{5}{9}\)ab² + \(\frac{5}{9}\)b³ + \(\frac{5}{9}\)ab³
= \(\frac{1}{4}\)(2)³ + \(\frac{1}{4}\)(2)²(3) + \(\frac{1}{4}\)(2)³(3) + \(\frac{5}{9}\)(2)(3)² + \(\frac{5}{9}\)(3)³ + \(\frac{5}{9}\)(2)(3)³
= \(\frac{1}{4}\) × 8 + \(\frac{1}{4}\) × 4 × 3 + \(\frac{1}{4}\) × 8 × 3 + \(\frac{5}{9}\) × 2 × 9 + \(\frac{5}{9}\) × 27 + \(\frac{5}{9}\) × 2 × 27
= 2 + 3 + 6 + 10 + 15 + 30 = 77
Hence verified L.H.S. = R.H.S

(iii) (x³y – y²)(x³y + y²);x = -1, y = -2
Answer:
(x³y – y²)(x³y + y²);x = -1, y = -2
= x³y(x³y + y²) – y² (x³y + y²)
= x³y × x³y + x³y × y² – y² × x³y – y²× y²
= x⁶y² + x³y³ – x³y³ – y⁴

Verification:
L.H.S = (x³y – y²)(x³y + y²)
= [(-1)³ (- 2) – (- 2)²] [(-1)³(- 2) + (- 2)²]
= [- 1 × -2 – 4][-1 × -2 + 4]
= (2 – 4) (2 + 4)
= – 2 × 6
= – 12
R.H.S = x⁶y² + x³y³ – x³y³ – y⁴
Putting x = -1, y = -2
= (-1)⁶(-2)² – (-2)⁴
= 1 × 4 – 16
= 4 – 16
= -12
Hence verified
L.H.S = R.H.S

(iv) (2p + 3q) (4p² + 12pq + 9q²); p = \(\frac{1}{2}\),q = \(\frac{1}{3}\)
= (2p + 3q) (4p² + 12pq + 9q²)
= 2p (4p² + 12pq + 9q²) + 3q (4p² + 12pq + 9q²)
= 8p³ + 24 p²q + 18 pq² + 12p²q + 36 pq² + 27q³
= 8p³ + 36 p²q + 54 pq² + 27 q³

Verification
L.H.S = (2p + 3q) (4p² + 12pq + 9q²)

Putting p = \(\frac{1}{2}\), q = \(\frac{1}{3}\)
= [2(\(\frac{1}{2}\)) + 3(\(\frac{1}{3}\))] [4(\(\frac{1}{2}\))² + 12(\(\frac{1}{2}\))(\(\frac{1}{3}\)) + 9(\(\frac{1}{3}\))²]
= (1 + 1) (4 × \(\frac{1}{4}\) + 12 × \(\frac{1}{6}\) + 9 × \(\frac{1}{9}\))
= 2(1 + 2 + 1)
= 2 × 4 = 8

R.H.S = 8p³ + 36p²q + 54 pq² + 27 q³
= 8(\(\frac{1}{2}\))3 – 36(\(\frac{1}{2}\))2(\(\frac{1}{3}\)) + 54(\(\frac{1}{2}\)) (\(\frac{1}{3}\))² + 27(\(\frac{1}{3}\))³
= 8 × \(\frac{1}{8}\) + 36 × \(\frac{1}{4}\) \(\frac{1}{3}\) + 54 × \(\frac{1}{2}\) × \(\frac{1}{9}\) + 27 × \(\frac{1}{27}\)
= 1 + 3 + 3 + 1
= 8
Hence verified L.H.S. = R.H.S.

(v) (m² + mn + n²)(m – n); m = 4, n = 3
Answer:
(m² + mn + n²) (m – n); m = 4, n = 3
= m²(m – n) + mn(m – n) + n² (m – n)
= m³ – m²n + m²n – mn² + mn² – n³
= nr-nr

Verification:
L.H.S. = (m² + mn + n²) (m – n)
Putting m = 4, n = 3
= [(4)² + 4 × 3 + (3)²](4 – 3)
= (16 + 12 + 9) (1) = 37
R.H.S. = m³ – n³
= (4)³ – (3)³
= 64 – 27 = 37
Hence verified L.H.S. = R.H.S.

Question 7.
Simplify:
(i) 3x² (3y² + 2) – x (x – 2xyz) + y (2x²y – 2y)
(ii) (2x + 7)(5x + 9)-(4x² + 1)(x – 3)
(iii) (y² – 7y + 4) (3y² – 2) + (y + 1) (y² + 2y)
Answer:
(i) 3x²(3y² + 2) – x(x – 2xy²) + y (2x²y – 2y)
= 3x² × 3y² + 3x² × 2 – x × x – x – 2xy² + y × 2x²y – 2y × y
= 9x²y² + 6x² – x² + 2x²y² + 2x²y² – 2 y²
= 13x²y² + 5x² – 2y²

(ii) (2x + 7) (5x + 9) – (4x² +1) (x – 3)
= [2x (5x + 9) + 7(5x + 9)] – [4x² (x – 3) + 1 (x – 3)]
= (10x² + 18x + 35x + 63) – (4x³ – 12x² + x – 3)
= 10x² + 53x + 63 – 4x³ + 12x² – x + 3
= – 4x³ + 22x² + 52x + 66

(iii) (y² – 7y + 4) (3y² – 2) + (y + 1) (y² + 2y)
= [y² (3y² – 2) – 7y(3y² – 2) + 4(3y² – 2)] + [y(y² + 2y) + 1 (y² + 2y)]
= 3y⁴ – 2y² – 21y³ + 14y + 12y² – 8 + y³ + 2y² + y² + 2y
= 3y⁴ – 20y³ + 13y² + 16y – 8

Question 8.
Find the HCF of the terms in the expression 3a²b² + 6ab²c² + 12a²b²c².
Answer:
3a²b² + 6ab²c² + 12a²b²c²
= 3 × a × a × b × b + 2 × 3 × a × b × b × c × c + 2 × 2 × 3 × a × a × b × b × c × c
H.C.F. = 3 × a × b × b
= 3 ab²

Question 9.
Factorise:
(a) ab² – bc² – ab + c²
Answer:
= (ab² – ab) – (bc² – c²)
= ab(b -1) – c²(b -1)
= (b – 1) (ab – c²)

(b) 4(p + q) (3a -b) + 6 (p + q) (2b – 3a)
Answer:
4(p + q)(3a – b) + 6(p + q)(2b – 3a)
= 2(p + q) [2(3a – b) + 3(2b – 3a)]
= 2(p + q) (6a -2b + 6b- 9a)
= 2(p + q) (4b – 3a)

(c) axy + bc xy – az – bcz
Answer:
axy + be xy – az – b c z
= xy (a + bc) – z(a + bc)
= xy (a + bc) – z(a + bc)
= (a + bc) (xy – z)

DAV Class 7 Maths Chapter 6 Enrichment Questions

Question 1.
Sum of first n natural numbers is given by the expression , \(\left(\frac{n^2+n}{2}\right)\)
Sume of first five natural numbers (i.e., 1+2 + 3 + 4 + 5) will be
\(\frac{5^2+5}{2}=\frac{30}{2}\) = 15

Sum of first seven natural numbers (i.e., 1+2 + 3 + 4 + 5 + 6 + 7) will be
\(\frac{7^2+7}{2}=\frac{56}{2}\) = 28
Now, write down the sum of first ten natural numbers.
Answer:
Sum of first n natural numbers is given by the expression \(\left(\frac{n^2+n}{2}\right)\).
Therefore, putting the value of n = 10 in the above expression and solving it, we will have the sum of first 10 natural numbers.
∴ Sum of first 10 natural numbers = \(\left(\frac{10^2+10}{2}\right)\)
= \(\left(\frac{100+10}{2}\right)\)
= \(\frac{110}{2}\)
= 55
So, the sum of first 10 natural numbers is 55.

Additional Questions

Question 1.
Find the product:
(i) (0.7x + 0.9y) (0.49x² – 0.63xy + 0.81y²)
Solution:
(i) (0,7x + 0.9y) (0.49x² – 0.63xy + 0.81y²)
= 0.7x (0.49x² – 0.63xy + 0.81y²) + 0.9y (0.49x² – 0.63xy + 0.81y²)

= 0.343x³ + 0.729y³

(ii) (\(\frac{1}{2}\)x – \(\frac{1}{2}\)y) (\(\frac{1}{2}\)x² + \(\frac{1}{2}\)y²)

Hence verified L.H.S. = R.H.S.

Question 2.
Simplify the following and verify the result for x = 1, y = – 1 and z = 1.
(i) (\(\frac{3}{2}\)x² – \(\frac{1}{5}\)y) (\(\frac{1}{2}\)x + \(\frac{3}{4}\)xy – \(\frac{1}{3}\)y)
Answer:
= \(\frac{3}{2}\)x²(\(\frac{1}{2}\)x + \(\frac{3}{4}\)xy – \(\frac{1}{3}\)y) – \(\frac{1}{5}\)y(\(\frac{1}{2}\)x + \(\frac{3}{4}\)xy – \(\frac{1}{3}\)y)
= \(\frac{3}{4}\)x³ + \(\frac{9}{8}\)x³y – \(\frac{1}{2}\)x²y – \(\frac{1}{10}\)xy – \(\frac{3}{20}\)xy² + \(\frac{1}{15}\)y²

Verification:
L.H.S = (\(\frac{3}{2}\)x² – \(\frac{1}{5}\)y) (\(\frac{1}{2}\)x + \(\frac{3}{2}\)xy – \(\frac{1}{3}\)y)
Put x = 1 and y = -1

Hence verified L.H.S = R.H.S

(ii) (x² + y² + z² – xy – yz – zx) (x + y + z)
= x²(x + y + z) + y²(x + y + z) + z² (x + y + z) – xy(x + y + z) – yz (x + y + z) – zx(x + y + z)

= x³ + y³ + z³ – 3xyz

Verification:
L.H.S = (x² + y² + z² – xy – yz – zx) (x + y + z)
= [(1)² + (-1)² + (1)² – (1) (- 1) – (-1) (1) – (1) (1)] [(1) + (-1) + (1)]
= [1 + 1 + 1 + 1 + 1 – 1] [1 – 1 + 1]
= 4 × 1 = 4

R.H.S = x³ + y³ + z³ – 3 xyz
Put x = 1, y = -1, z = 1
= (1)³ + (- 1)³ + (1)³ – 3(1) (- 1) (1)
= 1 – 1 + 1 + 3 = 4
Hence verified L.H.S. = R.H.S.

Question 3.
Simplify the following and verify the result also.
(i) (m³ + n³) (2m – 3n); m = -2, n = 0
Answer:
(m³ + n³) (2m – 3n)
= m³ (2m – 3n) + n³ (2m – 3n)
= m³ × 2m – m³ × 3n + n³ × 2m – n³ × 3n
= 2m³ – 3m³n + 2mn³ – 3n³

Verification:
L.H.S. = (m³ + n³) (2m – 3n)
Putting m = – 2, n = 0
= [(- 2)³ + (0)³] [2(-2) – 3 × (0)]
= -8 × -4
= 32
R.H.S. = 2m⁴ – 3m³n + 2mn³ – 3n³
Putting m = -2,n = 0
= 2(- 2)⁴ – 3 (- 2)³ (0) + 2(- 2) (0)³ – 3 (0)³
= 2 × 16 – 0 + 0 – 0
= 32
Hence verified L.H.S. = R.H.S.

(ii) (x²y – xy²) (2x + 1); x = 1, y = 2
Answer:
(x²y – xy²) (2x + 1)
= x²y (2x + 1) – xy²(2x + 1)
= x²y × 2x + x²y × 1 – xy² × 2x – xy² × 1
= 2x³y + x²y- 2x² y² – xy²

Verification:
L.H.S = (x²y – xy²) (2x +1)
Putting x = 1, y = 2
= [(1)² (2) – (1) (2)²] [2(1) + 1]
= (2 – 4) (2 + 1)
= -2 × 3 = -6

R.H.S = 2x³y + x²y – 2x²y² – xy²
= 2(1 )³ (2) + (1)² (2) – 2(1)² (2)² – (1) (2)²
= 2× 1 × 2 + 1 × 2 – 2 × 1 × 4 – 1 × 4
= 4 + 2 – 8 – 4
= 6 – 12
= -6
Hence verified L.H.S. – R.H.S.

(iii) (9b²c – 4c²) (2bc² + 5c); b = 0, c = 2
Answer:
(9b²c – 4c²) (2bc² + 5c)
= 9b²c (2bc² + 5c) – 4c² (2bc² + 5c)
= 9b²c x 2bc² + 9b²c x 5c – 4c² x 2bc² – 4c² x 5c
= 18b³c³ + 45b²c² – 8bc⁴ – 20c³

Verification:
L.H.S. = (9b²c – 4c²) (2bc² + 5c)
Putting b = 0 and c = 2
= [9(0)² (2) – 4(2)²] [2 × (0) (2)² + 5(2)]
= [0 – 16] [0 + 10]
= -16 × 10
= – 160

R.H.S. = 18b³c³ + 45b² – 8bc⁴ – 20c³
Putting b = 0,c = 2
= 18(0)³ (2)³ + 45(0)² (2)² – 8(0) (2)⁴ – 20(2)³
= 0 + 0 – 0 – 20 × 8
= -160
Hence verified L.H.S. = R.H.S.

Question 4.
Find the common factors of the following terms:
(i) 3x³y²,15 xy³, 60x²y²
Answer:
3x³y²,15xy³, 60x²y²
Factors of 3x³y² = 3 × x × x × x × y × y
Factors of 15xy³ = 3 × 5 × x × y × y × y
Factors of 60x²y² = 2 × 2 × 3 × 5 × x × x × y × y
Common factors = 3 × x × y × y = 3xy²

(ii) 4a³b, 12a²b⁴,16a⁴b²
Answer:
4a³b, 12a²b⁴,16a⁴b²
Factors of 4 a3b = 2 × 2 × a × a × a × b
Factors of 12 a2b4 = 2 × 2 × 3 × a × a × b × b × b × b
Factors of 16fl4fr2 = 2 × 2 × 2 × 2 × a × a × a × a × b × b
Common factors are 2 × 2 × a × a × b = 4 a2b

(iii) 8x²(x – y), 12x(x – y), 24x³y² (x – y)
Answer:
8x²(x – y), 12x(x – y), 24x³y² (x – y)
Factors of 8x²(x – y) = 2 × 2 × 2 × x × x × x – y)
Factors of 12x(x – y) =2 × 2 × 3 × x × (x – y)
Factors of 24x³y² (x – y) = 2 × 2 × 2 × 3 × x × x × x × y × y × (x – y)
Common factors are 2 × 2 × x × (x – y) = 4x(x – y)

(iv) 21x³y⁷z², 35x² y⁶z³, 42x⁵y³z⁴
Answer:
21x³y⁷z², 35x² y⁶z³, 42x⁵y³z⁴
Factors of 21x³y⁷z² = 3 × 7 × x × x × x × y × y × y × y × y × y × y × z × z
Factors of 35x² y⁶z³ = 5 × 7 × x × x × y × y × y × y × y × y × z × z × z
Factors of 42x⁵y³z⁴ = 2 × 3 × 7 × x × x × x × x × x × y × y × y × z × z × z × z
Common Factors are 7 × x × x × y × y × y × z × z = 7x²y³z²

Question 5.
Find H.C.F. of the terms and factorise the following expressions.
(i) 21x⁴y⁵ + 28x⁵y³ – 35x³y⁷
Answer:
21x⁴y⁵ + 28x⁵y³ – 35x³y⁷
Common factors = 7x³y³
H.C.F. of the terms = 7x³y³
∴ 21x⁴y⁵ + 28x⁵y³ – 35 x³y⁷ = 7x³y³(3xy³ + 4x² – 5y⁴)

(ii) 5p³q – 20pq³ + 3p²q²
Answer:
5p³q – 20pq³ + 3p²q²
Common factor = pq
∴ H.C.F. of the terms = pq
∴ 5p³q – 20pq³ + 3p²q² = pq(5p² + 4x² – 5y⁴)

Question 6.
Factorise the following expressions:
(i) (3x – 5y)² – 15x²y + 25xy²
Answer:
(3x – 5y)² – 15x²y + 25xy²
= (3x – 5y)² – 5xy (3x – 5y)
= (3x – 5y) (3x – 5y – 5xy)

(ii) 9x² – 16y² + 18x³y – 12xy³
Answer:
9x² – 16y² + 18x³y – 12xy³
= (3x)² – (4y)² + 6xy (3x – 4y)
= (3x – 4y) (3x + 4y) + 6xy (3x – 4y)
= (3x – 4y) (3x + 4y + 6xy)

Question 7.
Simplify the following and verify the result for the given values:
(i) (2a – 3b) (4a² + 6ab + 9b²); a = 1, b = 2
Answer:
(2a – 3b) (4a² + 6ab + 9b²) = 2a(4a² + 6ab + 9b²) – 3b(4a² + 6ab + 9b²)

= 8a³ – 27b³

Verification:
L.H.S. = (2a – 3b) (4a² + 6ab + 9b²)
Put a = 1,b = 2
= [2(1)-3 (2)] [4 (1)² + 6(1)(2) + 9 (2)²]
= (2 – 6) (4 + 12 + 36)
= – 4 × 52 = – 208
= 8a³ – 27b³

R.H.S
Put x = 1, b = 2
= 8(1)³ – 27 (2)³
= 8 – 27 × 8
= 8 – 216
= -208
Hence verified L.H.S = R.H.S

(ii) (\(\frac{1}{3}\)p² – \(\frac{2}{5}\)q²) (p + q – pq); p = -1, q = 1
Answer:
(\(\frac{1}{3}\)p² – \(\frac{2}{5}\)q²) (p + q – pq)
= \(\frac{1}{3}\)p²(p + q – pq) + \(\frac{2}{5}\)q²(p + q – pq)
= \(\frac{1}{3}\)p³ + \(\frac{1}{3}\)p²q – \(\frac{1}{3}\)p³q + \(\frac{2}{5}\)pq² + \(\frac{2}{5}\)q³ – \(\frac{2}{5}\)pq³

Verification:
L.H.S. = (\(\frac{1}{3}\)p² – \(\frac{2}{5}\)q²)
Put p = -1, q = 1
= [\(\frac{1}{3}\)(- 1)2 + \(\frac{2}{5}\)(1)2] [-1 + 1 – (- 1) (1)]
= \(\left(\frac{1}{3}+\frac{2}{5}\right)\)(-1 + 1 + 1)
= \(\left(\frac{5+6}{15}\right)\)(1)
= \(\frac{11}{15}\)

R.H.S.
= \(\frac{1}{3}\)p³ + \(\frac{1}{3}\)p²q – \(\frac{1}{3}\)p³q + \(\frac{2}{5}\)pq² + \(\frac{2}{5}\)q³ – \(\frac{2}{5}\)pq³

Put p = – 1 and q = 1

Hence verified L.H.S. = R.H.S.

Question 8.
Factorise:
(i) pxy + qr xy – pz – qrz
Answer:
pxy + qrxy – pz – qrz = xy (p + qr) – z(p + qr)
= (p + qr) (xy – z)

(ii) 8(x + y)(3p – q) + 12(x + y)(q – 3p)
Answer:
8(x + y) (3p – q) + 12(x + y)(q – 3p)
= 8(x + y) (3p – q) – 12(x + y) (3p – q)
= – 4(x + y) (3p – q)