### Polynomials And Factorisation Class 9th Mathematics AP Board Solution

##### Question 1.Use suitable identities to find the following products(x + 5) (x + 2)Answer:using the identity (x + a) × (x + b) = x2 + (a + b)x + abhere a = 5 and b = 2⇒ (x + 5) (x + 2) = x2 + (5 + 2)x + 5 × 2Therefore (x + 5) (x + 2) = x2 + 7x + 10Question 2.Use suitable identities to find the following products(x - 5) (x - 5)Answer:(x - 5) (x - 5) = (x – 5)2Using identity (a – b)2 = a2 – 2ab + b2Here a = x and b = 5⇒ (x - 5) (x - 5) = x2 – 2 × x × 5 + 52⇒ (x - 5) (x - 5) = x2 – 10x + 25Therefore (x - 5) (x - 5) = x2 – 10x + 25Question 3.Use suitable identities to find the following products(3x + 2)(3x - 2)Answer:using the identity (a + b) × (a – b) = a2 – b2here a = 3x and b = 2⇒ (3x + 2)(3x - 2) = (3x)2 – 22Therefore (3x + 2)(3x - 2) = 9x2 – 4Question 4.Use suitable identities to find the following products Answer:using the identity (a + b) × (a – b) = a2 – b2here a = x2 and b = ⇒ = (x2)2 - Therefore = x4 - Question 5.Use suitable identities to find the following products(1 + x) (1 + x)Answer:(1 + x) (1 + x) = (1 + x)2Using identity (a + b)2 = a2 + 2ab + b2Here a = 1 and b = x⇒ (1 + x) (1 + x) = 12 + 2(1)(x) + x2Therefore (1 + x) (1 + x) = 1 + 2x + x2Question 6.Evaluate the following products without actual multiplication.101 × 99Answer:101 can be written as (100 + 1) and99 can be written as (100 - 1)⇒ 101 × 99 = (100 + 1) × (100 - 1)using the identity (a + b) × (a – b) = a2 – b2here a = 100 and b = 1⇒ 101 × 99 = 1002 – 12⇒ 101 × 99 = 10000 – 1⇒ 101 × 99 = 9999Question 7.Evaluate the following products without actual multiplication.999 × 999Answer:999 can be written as (1000 – 1)⇒ 999 × 999 = (1000 – 1) × (1000 – 1)⇒ 999 × 999 = (1000 – 1)2Using identity (a – b)2 = a2 – 2ab + b2Here a = 1000 and b = 1⇒ 999 × 999 = 10002 – 2(1000)(1) + 12⇒ 999 × 999 = 1000000 – 2000 + 1⇒ 999 × 999 = 998000 + 1⇒ 999 × 999 = 998001Question 8.Evaluate the following products without actual multiplication. Answer:⇒ 50 = and 49 = ⇒ 50 = and 49 = ⇒ 50 = and 49 = ⇒ = × ⇒ = Consider 101 × 99101 can be written as (100 + 1) and99 can be written as (100 - 1)⇒ 101 × 99 = (100 + 1) × (100 - 1)using the identity (a + b) × (a – b) = a2 – b2here a = 100 and b = 1⇒ 101 × 99 = 1002 – 12⇒ 101 × 99 = 10000 – 1⇒ 101 × 99 = 9999Therefore = Question 9.Evaluate the following products without actual multiplication.501 × 501Answer:501 can be written as (500 + 1)⇒ 501 × 501 = (500 + 1) × (500 + 1)⇒ 501 × 501 = (500 + 1)2⇒ 501 × 501 = (500 + 1) × (500 + 1)⇒ 501 × 501 = (500 + 1)2Using identity (a + b)2 = a2 + 2ab + b2Here a = 500 and b = 1⇒ 501 × 501 = 5002 + 2(500)(1) + 12⇒ 501 × 501 = 250000 + 1000 + 1⇒ 501 × 501 = 251001Question 10.Evaluate the following products without actual multiplication.30.5 × 29.5Answer:30.5 = and 29.5 = ⇒ 30.5 × 29.5 = × ⇒ 30.5 × 29.5 = × ⇒ 30.5 × 29.5 = …(i)Consider 61 × 5961 = (60 + 1)59 = (60 – 1)⇒ 61 × 59 = (60 + 1)(60 – 1)using the identity (a + b) × (a – b) = a2 – b2here a = 60 and b = 1⇒ 61 × 59 = 602 – 12⇒ 61 × 59 = 3600 – 1⇒ 61 × 59 = 3599From (i)⇒ 30.5 × 29.5 = Therefore 30.5 × 29.5 = 899.75Question 11.Factorise the following using appropriate identities.16x2 + 24xy + 9y2Answer:16x2 can be written as (4x)224xy can be written as 2(4x)(3y)9y2 can be written as (3y)2⇒ 16x2 + 24xy + 9y2 = (4x)2 + 2(4x)(3y) + (3y)2Using identity (a + b)2 = a2 + 2ab + b2Here a = 4x and b = 3y⇒ 16x2 + 24xy + 9y2 = (4x + 3y)2Therefore 16x2 + 24xy + 9y2 = (4x + 3y) (4x + 3y)Question 12.Factorise the following using appropriate identities.4y2 - 4y + 1Answer:4y2 can be written as (2y)24y can be written as 2(1)(2y)1 can be written as 12⇒ 4y2 - 4y + 1 = (2y)2 - 2(1)(2y) + 12Using identity (a - b)2 = a2 - 2ab + b2Here a = 2y and b = 1⇒ 4y2 - 4y + 1 = (2y - 1)2Therefore 4y2 - 4y + 1 = (2y - 1) (2y - 1)Question 13.Factorise the following using appropriate identities. Answer:4x2 can be written as (2x)2 can be written as ⇒ 4x2 - = (2x)2 - using the identity (a + b) × (a – b) = a2 – b2here a = 2x and b = Therefore 4x2 - = (2x + ) (2x - )Question 14.Factorise the following using appropriate identities.18a2 – 50Answer:Take out common factor 2⇒ 18a2 – 50 = 2 (9a2 - 25)Now9a2 can be written as (3a)225 can be written as 52⇒ 18a2 – 50 = 2 ((3a)2 – 52)using the identity (a + b) × (a – b) = a2 – b2here a = 3a and b = 5therefore 18a2 – 50 = 2 (3a + 5) (3a – 5)Question 15.Factorise the following using appropriate identities.x2 + 5x + 6Answer:Given is quadratic equation which can be factorised by splitting the middle term as shown⇒ x2 + 5x + 6 = x2 + 3x + 2x + 6= x (x + 3) + 2 (x + 3)= (x + 3) (x + 2)Therefore x2 + 5x + 6 = (x + 3) (x + 2)Question 16.Factorise the following using appropriate identities.3p2 - 24p + 36Answer:Take out common factor 3⇒ 3p2 - 24p + 36 = 3 (p2 – 8p + 12)Now splitting the middle term of quadratic p2 – 8p + 12 to factorise it⇒ 3p2 - 24p + 36 = 3 (p2 – 6p – 2p + 12)= 3 [p (p – 6) – 2 (p – 6)]= 3 (p – 2) (p – 6)Question 17.Expand each of the following, using suitable identities(x + 2y + 4z)2Answer:Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2caHere a = x, b = 2y and c = 4z⇒ (x + 2y + 4z)2 = x2 + (2y)2 + (4z)2 + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x)Therefore(x + 2y + 4z)2 = x2 + 4y2 + 16z2 + 4xy + 16yz + 8xzQuestion 18.Expand each of the following, using suitable identities(2a - 3b)3Answer:Using identity (x – y)3 = x3 - y3 – 3x2y + 3xy2Here x = 2a and y = 3b⇒ (2a - 3b)3 = (2a)3 – (3b)3 – 3(2a)2(3b) + 3(2a)(3b)2= 8a3 – 27b3 – 18a2b + 18ab2Therefore (2a - 3b)3 = 8a3 – 27b3 – 18a2b + 18ab2Question 19.Expand each of the following, using suitable identities(-2a + 5b - 3c)2Answer:(-2a + 5b - 3c)2 = [(-2a) + (5b) + (-3c)]2Using (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zxHere x = -2a, y = 5b and z = -3c⇒ (-2a + 5b - 3c)2 = (-2a)2 + (5b)2 + (-3c)2 + 2(-2a)(5b) + 2(5b)(-3c) + 2(-3c)(-2a)⇒ (-2a + 5b - 3c)2 = 4a2 + 25b2 + 9c2 + (-20ab) + (-30bc) + 12ac⇒ (-2a + 5b - 3c)2 = 4a2 + 25b2 + 9c2 - 20ab - 30bc + 12acTherefore(-2a + 5b - 3c)2 = 4a2 + 25b2 + 9c2 - 20ab - 30bc + 12acQuestion 20.Expand each of the following, using suitable identities Answer: = [ + + 1]2Using (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zxHere x = , y = and z = 1⇒ = + + 12 + 2 + 2 (1) + 2(1) ⇒ = + + 1 + + (-b) + ⇒ = + + 1 - – b + Therefore = + + 1 - – b + Question 21.Expand each of the following, using suitable identities(p + 1)3Answer:Using identity (x + y)3 = x3 + y3 + 3x2y + 3xy2Here x = p and y = 1⇒ (p + 1)3 = p3 + 13 + 3(p)2(1) + 3(p)(1)2= p3 + 13 + 3p2 + 3pTherefore (p + 1)3 = p3 + 13 + 3p2 + 3pQuestion 22.Expand each of the following, using suitable identities Answer:Using identity (a – b)3 = a3 - b3 – 3a2b + 3ab2Here a = x and b = y⇒ = x3 - – 3(x)2( ) + 3(x) ⇒ = x3 - – 2x2y + xy2Therefore = x3 - – 2x2y + xy2Question 23.Factorise25x2 + 16y2 + 4z2 - 40xy + 16yz - 20xzAnswer:25x2 + 16y2 + 4z2 - 40xy + 16yz - 20xz25x2 + 16y2 + 4z2 - 40xy + 16yz - 20xz = 25x2 + 16y2 + 4z2 + (-40xy) + 16yz + (-20xz)25x2 can be written as (-5x)216y2 can be written as (4y)24z2 can be written as (2z)2-40xy can be written as 2(-5x)(4y)16yz can be written as 2(4y)(2z)-20xz can be written as 2(-5x)(2z)⇒ 25x2 + 16y2 + 4z2 - 40xy + 16yz - 20xz = (-5x)2 + (4y)2 +(2z)2 + 2(-5x)(4y) + 2(4y)(2z) + 2(-5x)(2z) …(i)Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2caComparing (-5x)2 + (4y)2 + (2z)2 + 2(-5x)(4y) + 2(4y)(2z) + 2(-5x)(2z) with a2 + b2 + c2 + 2ab + 2bc + 2ca we geta = -5x, b = 4y and c = 2ztherefore(-5x)2 + (4y)2 + (2z)2 + 2(-5x)(4y) + 2(4y)(2z) + 2(-5x)(2z) = (- 5x + 4y + 2z)2From (i)25x2 + 16y2 + 4z2 - 40xy + 16yz - 20xz = (- 5x + 4y + 2z)2Question 24.Factorise9a2 + 4b2 + 16c2 + 12ab - 16bc - 24caAnswer:9a2 + 4b2 + 16c2 + 12ab - 16bc - 24ca9a2 + 4b2 + 16c2 + 12ab - 16bc - 24ca = 9a2 + 4b2 + 16c2 + 12ab + (-16bc) + (-24ca)9a2 can be written as (3a)24b2 can be written as (2b)216c2 can be written as (-4c)212ab can be written as 2(3a)(2b)-16bc can be written as 2(2b)(-4c)-24ca can be written as 2(-4c)(3a)⇒ 9a2 + 4b2 + 16c2 + 12ab - 16bc - 24ca = (3a)2 + (2b)2 + (-4c)2 + 2(3a)(2b) + 2(2b)(-4c) + 2(-4c)(3a) …(i)Using (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zxComparing (3a)2 + (2b)2 + (-4c)2 + 2(3a)(2b) + 2(2b)(-4c) + 2(-4c)(3a) with x2 + y2 + z2 + 2xy + 2yz + 2zx we getx = 3a, y = 2b and z = -4ctherefore(3a)2 + (2b)2 + (-4c)2 + 2(3a)(2b) + 2(2b)(-4c) + 2(-4c)(3a) = (3a + 2b + (-4c))2From (i)9a2 + 4b2 + 16c2 + 12ab - 16bc - 24ca = (3a + 2b – 4c)2

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