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## Karnataka Board Class 8 Maths Chapter 2 Algebraic Expressions Ex 2.4

Question 1.

Find the product:

i. (a + 3) ( a + 5)

ii. (3t + 1) (3t + 4)

iii. (a – 8) (a + 2)

iv. (a – 6) (a – 2)

Answer:

i. (a + 3) (a + 5) This is in the form.

(x + a)(x + b) = x² + (a + b)x + ab

Here x = a, a = 3, b = 5

(a + 3)(a + 5) = a² + (a + 5)a + 3.5

= a² + 8a + 15

(ii) (3t + 1)(3t + 4)This is in the form.

(x + a)(x + b) = x² +(a + b)x + ab

Here x =3t,a = 1, b = 4

(3t – 1)(3t + 4) = (3t)²+(1 + 4)3t + 1.4

= 9t² + 5(3t) + 4 = 9t² + 15t + 4

(iii) (a -8)(a + 2)

This is in the form (x + a)(x + b) = x² + (a + b)x + ab

Here x = a, a = -8, b = 2

(a – 8)(a + 2) = a² + (-8 + 2)a + (-8)(2)

= a² + (-6)a – 16

= a² – 6a – 16

(iv) (a – 6)(a – 2)

This is in the form (x + a)(x + b) = x² +(a + b)x + ab

Herex = a, a = -6,b = -2

(a – 6)(a – 2) = a² + (- 6 – 2) a + (-6)(-2)

=a² + (- 8)a + 12

= a² – 8a + 12

2. Evaluate using suitable identities

Question i.

53 × 55

Answer:

53 × 55 = (50 + 3) (50 + 5)

This is in the form (x + a)(x + b) = x² + (a + b)x + ab

Here x = 50, a = 3, b = 5

(50 + 3)(50 + 5) = 50² + (3 + 5) 50 + 3 × 5

= 2500 + (8)50 +15

= 2500 + 400 + 15

∴ 53 × 55 = 2915

Question ii.

102 × 106

Answer:

102 × 106 = (100 + 2) (100 + 6) This is in the form.

(x + a)(x + b) = x² +(a + b)x + ab

Here x = 100, a = 2, b = 6

(100 + 2)(100 + 6) = 100² +(2 + 6)100 + 2 × 6

= 10000 + (8)100 + 12

= 10000 + 800 + 12

∴102 × 106 = 10812

Question iii.

34 × 36

Answer:

34 × 36 = (30 + 4) (30 + 6) This is in the form.

(x + a)(x + b) = x²+ (a + b)x + ab

Here x =30, a = 4, b = 6

(30 + 4)(30 + 6) = 30² + (4 + 6)30 + 4 × 6

= 900 + 300 + 24

34 × 36 = 1224

Question iv.

103× 96

Answer: 103 × 96 = (100 + 3)(100-4)This is in the form.

(x + a)(x + b) = x² + (a + b)x + ab

Here x = 100,a = 3,b = -4 (100 + 3)(100-4) = 100² + (3 – 4)100 + 3 × -4

= 10000 + (- 1)100 – 12

= 10000 – 100 – 12

103 × 96 = 9888

Question 3.

Find the expression for the product (x+a) (x+b) (x+c) using the identity (x+a)(x+b) = x² + (a + b)x + ab

Answer:

(x + a) (x + b) (x + c)

Consider (x + a)(x + b)

= x² + (a + b)x + ab

= x² + ax + bx + ab (x + a)(x + b)(x + c)

= (x² + ax + bx + ab)(x + c)

= x(x² + ax + bx+ ab) + c

(x² + ax + bx + ab)

= x^{3} + ax² + bx² + abx + cx² + cax +cbx + abc

= x^{3} + ax² + bx² + cx² + abx + cax +cbx + abc [by rearranging]

= x^{3} + x²(a + b + c) + x(ab + ca +bc) + abc

∴ (x + a)(x + b)(x + c)

= x^{3} +x² (a + b + c) + x(ab + bc + ca) + abc

4. Using the identify (a+b)2 = a² + 2ab+b² simplify the following.

Question i.

(a + 6 )²

Answer:

Using the identity

(a + b)² = a² + 2ab + b²

(a + 6)² = a² +2.a.6 + 6²

= a² +12a + 36

Question ii.

(3x+2y)2

Answer:

Using the identity

(a + b)² = a² +2ab + b²

(3x + 2y)² =(3x)² + 2.3x.2y + (2y)²

= 9x² +12xy + 4y²

Question iii.

(2p + 3q)²

Answer:

Using the identity

(a + b)² =a² + 2ab + b²

(2p + 3q)² =(2p)² +2.2p.3q + (3q)²

= 4p² + 12pq + 9q²

Question iv.

(x² + 5)²

Answer:

Using the identity (a + b)² = a² +2ab + b²

(x² +5)² = (x²)² + 2.x².5 + 52 = x^{4} + 10x² + 25

5. Evaluate using the identity (a+b)² = a² + 2ab + b²

Question i.

(34)²

Answer:

34² = (30 + 4)²

(a + b)² = a² + 2ab+ b² Here a = 30, b = 4

(30 + 4)² = (30)² + 2 × 30 × 4 + (4)²

= 900 + 240 + 16

(34)² = 1156

Question ii.

(10.2)²

Answer:

(10.2)² = (10 + 0.2)²

(a + b)² = a² + 2ab + b² Here a = 10, b=0.2

[(10 + (0.2))]² = 10² +2.(10).(0.2) + (0.2)² ]

= 100 + 4 + 0.04

(10.2)² = 104.04

Question iii.

53²

Answer:

53² = (50 + 3)²

(a + b)² =a² +2ab + b²

Here a = 50, b = 3 (50 + 3)² =50² + 2 × 50 × 3 + (3)²

= 2500 + 300 + 9

53² = 2809

Question iv.

(41 )²

Answer:

41² = (40 + 1)²

(a + b)² = a² + 2ab + b²

Here a = 40,b = 1

(40 + 1)² = 40² + 2 × 40 × 1 + (1)²

= 1600 + 80 + 1

41² = 1681

6. Use the identify (a -b)² = a² – 2ab + b² to compute.

Question i.

(x – 6)²

Answer:

(a – b)² = a² – 2ab + b²

Herea = x, b=6

(x – 6)² = x² – (2)(x)(6) + 6²

= x² – 12x+ 36

Question ii.

(3x – 5y)²

Answer:

(a – b)² = a² – 2ab + b² Here a = 3x, b = 5y

(3x – 5y)² = (3x)² -(2)(3x)(5y) + (5y)² = 9x² – 30xy + 25y²

Question iii.

(5a – 4b)²

Answer:

(a – b)² = a² – 2ab + b²

Here a = 5a, b = 4b

(5a – 4b)² = (5a)² – (2)(5a)(4b) + (4b)²

= 25a² – 40ab + 16b²

Question iv.

(p² – q²)²

Answer:

(a – b)² = a² – 2ab + b² Here a = p² , b = q2

(p² – q²)² = (p²)² – (2)(p²)(q²) + (q²)²

= p^{4} – 2p²q² + q^{4}

7. Evaluate using the identify (a – b)² = a² – 2ab + b².

Question i.

(49)²

Answer:

49² = (50 – 1)²

(a – b)² = a² – 2ab + b² Here a = 50, b = 1

(50 – 1)² = 50² – (2)(50)(1) + 1² = 2500 – 100 + 1

(49)² = 2401

Question ii.

(9.8)²

Answer:

(9.8)² = (10 – 0.2)²

(a – b)² =a² – 2ab + b² Here a = 10, b=0.2

(10 – 0.2)² = 10² – (2)(10)(0.2) + (0.2)²

= 100 – 4 + 0.04

(9.8)² =96.04

Question iii.

59²

Answer:

59² = (60 – 1)²

(a – b)² = a² – 2ab + b² Here a = 60, b = 1

(60 – 1)² = 60² – (2)(60)(1) +12 = 3600 – 120 + 1

59² =3481

Question iv.

(198)²

Answer:

(198)² = (200 – 2)²

(a – b)² = a² – 2ab + b² Here a = 200, b = 2

(200 – 2)² = 200² – (2)(200)(2) + 2²

= 40000 – 800 + 4

(198)² = 39204

8. Use the identify (a + b) (a – b) = a² – b² to find the product.

Question i.

(x + 6) (x – 6)

Answer:

(a + b) (a – b) = a2 – b2

Here a = x,b = 6

(x + 6)(x – 6) = x² – 6² = x² – 36

Question ii.

(3x + 5) (3x – 5)

Answer:

(a + b) (a – b) = a² – b²

Here a = 3x, b = 5

(3x + 5)(3x – 5) = (3x)² – (5)²

= 9x² -25

Question iii.

(2a + 4b) (2a – 4b)

Answer:

(a + b) (a – b) = a² – b²

Here a = 2a,b = 4b

(2a + 4b)(2a-4b) = (2a)² – (4b)²

= 4a² – 16b²

Question iv.

\(\left(\frac{2 x}{3}+1\right)\left(\frac{2 x}{3}-1\right)\)

Answer:

(a + b) (a – b) = a² – b²

Here a = \(\frac{2 x}{3}\) ,b = 1

\(\left(\frac{2 x}{3}+1\right)\left(\frac{2 x}{3}-1\right)\) = \(\left(\frac{2 x}{3}\right)^{2}-1^{2}\)

= \(\frac{4 x^{2}}{9}-1\)

9. Evaluate these using identity :

Question i.

55 × 45

Answer:

55 × 45 = (50 + 5) (50 – 5)

Identity (a + b)(a – b) = a² – b²

Here a = 50,b=5

(50 + 5)(50 – 5) = (50)² – 5²

= 2500 – 25

55 × 45 = 2475

Question ii.

33 × 27

Answer:

33 × 27 = (30 + 3) (30 – 3)

identity (a + b)(a – b) = a² – b²

Here a = 30,b = 3

(30 + 3)(30 – 3) = 30² – 3² = 900 – 9

33 × 27 = 891

Question iii.

8.5 × 9.5

Answer:

8.5 × 9.5 = (9 – 0.5) (9 + 0.5)

Here a = 9, b = 0.5

(9 – 0.5)(9 + 0.5) = (9²) – (0.5)² = 81 – 0.25

= 80.75

Question iv.

102 × 98

Answer:

102 × 98 = (100+ 2) (100 – 2)

Identity (a + b)(a – b) = a^{2} – b^{2}

Here a = 100, b= 2

(100 + 2)(100 – 2) = 100² – 2² =10000 – 4

102 × 98 = 9996

10. Find the product.

Question i.

(x – 3)(x + 3) (x2 + 9)

Answer:

(x – 3)(x + 3) (x2 + 9)

Identity(a + b)(a-b) = a² – b²

= (x² – 3²)(x² + 9)

= (x² – 9)(x² + 9)

= (x²)² – 9²

= x^{4} – 81

Question ii.

(2a + 3)(2a – 3) (4a² + 9)

Answer:

(2a + 3)(2a – 3) (4a² + 9)

Identity (a + b)(a-b) = a² – b²

= [((2a)² – 3²)(4a²+9)²

= (4a² – 9)(4a² + 9)

= (4a²)² – 9² = 16a^{4} – 81

Question iii.

(P + 2) (P – 2) (P² + 4)

Answer:

(P + 2) (P – 2) (P² + 4)

Identity (a + b)(a – b) = a² – b²

= (P² – 2²)(P²+ 4)

= (P² – 4)(P² + 4)

= (P²)² – 4² = P^{4} – 16

Question iv.

\(\left(\frac{1}{2} m-\frac{1}{3}\right)\left(\frac{1}{2} m+\frac{1}{3}\right)\left(\frac{1}{4} m^{2}+\frac{1}{9}\right)\)

Answer:

Question v.

(2x – y)(2x + y) (4x² + y²)

Answer:

(2x – y)(2x + y) (4x² + y²)

Identity (a + b)(a – b) = a² – b²

[(2x)² – y²](4x² + y²)

= (4x² – y²)(4x² + y²)

= I6x^{4} – y^{4}

Question vi.

(2x – 3y)(2x + 3y) (4x² + 9y²)

Answer:

(2x – 3y)(2x + 3y) (4x² + 9y²)

Identity (a + b)(a-b) = a² -b²

= [(2x)² – (3y)²](4x²+ 9y²)

= (4x² – 9y²)(4x² + 9y²)

= (4x²)² – (9y²)²

= 16x^{4} – 81y^{4}