Class-8 Maths- Factorisation Ex 14.2

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Question 1: Factorise the following expressions.

(i) a² + 8a + 16

(ii) p² − 10p + 25

(iii) 25m² + 30m + 9

(iv) 49y² + 84yz + 36z²

(v) 4x² − 8x + 4

(vi) 121b² − 88bc + 16c²

(vii) (l + m)² − 4lm (Hint: Expand (l + m)² first)

(viii) a⁴ + 2a²b² + b4

Answer:

(i) a² + 8a + 16 = (a)² + 2 × a × 4 + (4)²

= (a + 4)² [(x + y)² = x² + 2xy + y²]

(ii) p² − 10p + 25 = (p)² − 2 × p × 5 + (5)²

= (p − 5)² [(a − b)² = a² − 2ab + b²]

(iii) 25m² + 30m + 9 = (5m)² + 2 × 5m × 3 + (3)²

= (5m + 3)² [(a + b)² = a² + 2ab + b²]

(iv) 49y² + 84yz + 36z² = (7y)² + 2 × (7y) × (6z) + (6z)²

= (7y + 6z)² [(a + b)² = a² + 2ab + b²]

(v) 4x² − 8x + 4 = (2x)² − 2 (2x) (2) + (2)²

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

= [(2) (x − 1)]² = 4(x − 1)²

(vi) 121b² − 88bc + 16c² = (11b)² − 2 (11b) (4c) + (4c)²

= (11b − 4c)² [(a − b)² = a2 − 2ab + b²]

(vii) (l + m)² − 4lm = l² + 2lm + m² − 4lm

= l² − 2lm + m²

= (l − m)² [(a − b)² = a² − 2ab + b²]

(viii) a⁴ + 2a²b² + b⁴ = (a²)² + 2 (a²) (b²) + (b²)²

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

Question 2: Factorise

(i) 4p² − 9q²

(ii) 63a² − 112b²

(iii) 49x² − 36

(iv) 16x⁵ − 144x³

(v) (l + m)² − (l − m)²

(vi) 9x²y² − 16

(vii) (x² − 2xy + y²) − z²

(viii) 25a² − 4b² + 28bc − 49c²

Answer:

(i) 4p² − 9q² = (2p)² − (3q)²

= (2p + 3q) (2p − 3q) [a² − b² = (a − b) (a + b)]

(ii) 63a² − 112b² = 7(9a² − 16b²)

= 7[(3a)² − (4b)²]

= 7(3a + 4b) (3a − 4b) [a² − b² = (a − b) (a + b)]

(iii) 49x² − 36 = (7x)² − (6)²

= (7x − 6) (7x + 6) [a² − b² = (a − b) (a + b)]

(iv) 16x⁵ − 144x³ = 16x³(x² − 9)

= 16 x³ [(x)² − (3)²]

= 16 x³(x − 3) (x + 3) [a² − b² = (a − b) (a + b)]

(v) (l + m)² − (l − m)² = [(l + m) − (l − m)] [(l + m) + (l − m)]

[Using identity a² − b² = (a − b) (a + b)]

= (l + m − l + m) (l + m + l − m)

= 2m × 2l

= 4ml

= 4lm

(vi) 9x²y² − 16 = (3xy)² − (4)²

= (3xy − 4) (3xy + 4) [a² − b² = (a − b) (a + b)]

(vii) (x² − 2xy + y²) − z² = (x − y)² − (z)² [(a − b)² = a² − 2ab + b²]

= (x − y − z) (x − y + z) [a² − b² = (a − b) (a + b)]

(viii) 25a² − 4b² + 28bc − 49c² = 25a² − (4b² − 28bc + 49c²)

= (5a)² − [(2b)² − 2 × 2b × 7c + (7c)²]

= (5a)² − [(2b − 7c)²]

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

= [5a + (2b − 7c)] [5a − (2b − 7c)]

[Using identity a² − b² = (a − b) (a + b)]

= (5a + 2b − 7c) (5a − 2b + 7c)

Question 3:Factorise the expressions

(i) ax² + bx

(ii) 7p² + 21q²

(iii) 2x³ + 2xy² + 2xz²

(iv) am² + bm² + bn² + an²

(v) (lm + l) + m + 1

(vi) y(y + z) + 9(y + z)

(vii) 5y² − 20y − 8z + 2yz

(viii) 10ab + 4a + 5b + 2

(ix) 6xy − 4y + 6 − 9x

Answer:

(i) ax² + bx = a × x × x + b × x = x(ax + b)

(ii) 7p² + 21q² = 7 × p × p + 3 × 7 × q × q = 7(p² + 3q²)

(iii) 2x³ + 2xy² + 2xz² = 2x(x² + y² + z²)

(iv) am² + bm² + bn² + an² = am² + bm² + an² + bn²

= m²(a + b) + n²(a + b)

= (a + b) (m² + n²)

(v) (lm + l) + m + 1 = lm + m + l + 1

= m(l + 1) + 1(l + 1)

= (l + l) (m + 1)

(vi) y (y + z) + 9 (y + z) = (y + z) (y + 9)

(vii) 5y² − 20y − 8z + 2yz = 5y² − 20y + 2yz − 8z

= 5y(y − 4) + 2z(y − 4)

= (y − 4) (5y + 2z)

(viii) 10ab + 4a + 5b + 2 = 10ab + 5b + 4a + 2

= 5b(2a + 1) + 2(2a + 1)

= (2a + 1) (5b + 2)

(ix) 6xy − 4y + 6 − 9x = 6xy − 9x − 4y + 6

= 3x(2y − 3) − 2(2y − 3)

= (2y − 3) (3x − 2)

Question 4:-Factorise

(i) a⁴ − b⁴

(ii) p⁴ − 81

(iii) x⁴ − (y + z)4

(iv) x⁴ − (x − z)⁴

(v) a⁴ − 2a²b² + b⁴

Answer:

(i) a⁴ − b⁴ = (a²)² − (b²)²

= (a² − b²) (a² + b²)

= (a − b) (a + b) (a² + b²)

(ii) p⁴ − 81 = (p²)² − (9)²

= (p² − 9) (p² + 9)

= [(p)² − (3)²] (p² + 9)

= (p − 3) (p + 3) (p² + 9)

(iii) x⁴ − (y + z)⁴ = (x²)² − [(y +z)²]²

= [x² − (y + z)²] [x² + (y + z)²]

= [x − (y + z)][ x + (y + z)] [x2 + (y + z)2]

= (x − y − z) (x + y + z) [x² + (y + z)²]

(iv) x4 − (x − z)4 = (x²)2 − [(x − z)²]²

= [x² − (x − z)²] [x² + (x − z)²]

= [x − (x − z)] [x + (x − z)] [x² + (x − z)²]

= z(2x − z) [x² + x² − 2xz + z²]

= z(2x − z) (2x² − 2xz + z²)

(v) a⁴ − 2a²b² + b⁴ = (a²)² − 2 (a²) (b²) + (b²)²

= (a² − b²)²

= [(a − b) (a + b)]²

= (a − b)² (a + b)²

Question 5: Factorise the following expressions

(i) p² + 6p + 8

(ii) q² − 10q + 21

(iii) p² + 6p − 16

Answer:

(i) p² + 6p + 8

It can be observed that, 8 = 4 × 2 and 4 + 2 = 6

∴ p² + 6p + 8 = p² + 2p + 4p + 8

= p(p + 2) + 4(p + 2)

= (p + 2) (p + 4)

(ii) q² − 10q + 21

It can be observed that, 21 = (−7) × (−3) and (−7) + (−3) = − 10

∴ q² − 10q + 21 = q² − 7q − 3q + 21

= q(q − 7) − 3(q − 7)

= (q − 7) (q − 3)

(iii) p2 + 6p − 16

It can be observed that, 16 = (−2) × 8 and 8 + (−2) = 6

p² + 6p − 16 = p² + 8p − 2p − 16

= p(p + 8) − 2(p + 8)

= (p + 8) (p − 2)