Factorization-NCERT Solutions

Class VIII Math
NCERT Solution for Factorisation

EXERCISE: 14.1

1. Find the common factos of the given terms:

(i) 12x, 36

(ii) 2y, 22xy

(iii) 14pq, 28p²q²

(iv) 2x, 3x², 4

(v) 6abc, 24ab², 12a²b

(vi) 16x³, –4x², 32x

(vii) 10pq, 20qr, 30rp

(viii) 3x²y³, 10x³y², 6x²y²z

Sol. (i) The numerical coefficient in the given monomials are 12 and 36.

The highest common factor of 12 and 36 is 12.

But there is no common literal appearing in the given monomials 12x and 36.

The highest common factor = 12.

(ii) 2y = 2 × y

22xy = 2 × 11× x × y

The comoon factors are 2, y

And, 2 × y = 2y

(iii) The numerical coefficients of the given monomials are 14 and 28.

The highest common factor of 14 and 28 is 14.

The common literals appearing in the given monomials are p and q.

The smallest power of p and q in the two monomials = 1

The monomial of common literals with smallest powers = pq

∴The highest common factor = 14pq

(iv) 2x = 2 × x

3x² = 3 × x × x

4 = 2 × 2

The common factor is 1.

(v) The numerical coefficients of the given monomials are 6, 24 and 12.

The common literals appearing in the three monomials are a and b.

The smallest power of a in the three monomials = 1

The smallest power of b in the three monomials = 1

The monomial of common literals with smallest power = ab

Hence, the highest common factor = 6ab

(vi) 16x³ = 2 × 2 × 2 2 x × x × x

–4x² = –1 × 2 × 2 x × x

32x = 2 × 2 × 2 × 2 × 2 × x

The common factors are 2, 2, x.

And 2 × 2 × x = 4x

(vii) The numerical coefficients of the given monomials are 10, 20 and 30

The highest common factor of 10, 20 and 30 is 10

There is no common literal appearing in the three monomials.

(viii) 3x²y² = 3 × x × x × y × y × y

10x³y² = 2 × 5 × x × x × x × y × y

6x²y²z = 2 × 3 × x × x × y × y × z

The common factors are x, x, y, y

And

x × x × y ×y

2. Factorise the following expressions

(i) 7x – 42

(ii) 6p – 12q

(iii) 7a² + 14a

(iv) –16z + 20z³

(v) 20l2m + 30alm

(vi) 5x²y – 15xy²

(vii) 10a² – 15b² + 20c²

(viii) – 4a² + 4ab – 4ca

(x) ax²y + bxy² + cxyz

Sol. (i) Whe have 7x = 7 × x and 42 = 2 × 3 × 7

The two terms have 7 as a common factor

7x – 42 = (7 × x) – 2 × 3 × 7

= 7 × (x – 2 × 3) = 7(x – 6)

(ii) 6p = 2 × 3 × p

12q = 2 × 2 × 3 × q

The common factors are 2 and 3.

∴6p – 12q = (2 × 3 × p) – (2 × 2 × 3 × q)

= 2 × 3[p – (2 × q)]

= 6(p – 2q)

(iii) We have, 7a² = 7 × a × a

and , 14a = 2 × 7 × a

The two terms have 7 and a as common factors

7a² + 14a = (7 × a × a) + (2 × 7 × a)

= 7 × a × (a + 2) = 7a(a + 2)

(iv) 16z = 2 × 2 × 2 × 2 × z

20z³ = 2 × 2 × 5 × z × z × z

The common factors are 2, 2, and z.

∴ –16z + 20z³ = –(2 × 2 × 2 × 2 × z)

+ (2 × 2 × 5 × z × z × z

= (2 × 2 × z)[–(2 × 2) + (5 × z × z)]

= 4z(– 4 + 5z²)

(v) We have, 20l2m = 2 × 2 × 5 × l × l × m

and, 30alm = 3 × 2 × 5 × a × l × m

The two terms have 2, 5, l and m as common factors.

∴ 20 l2m+ 30alm = (2 × 2 × 5 × l × l × m)

+ (3 × 2 × 5 × a × l × m)

= 2 × 5 × l × m × (2 × l + 3 × a)

= 10lm(2l + 3a)

(vi) 5x²y = 5 × x × x × y

15xy² = 3 × 5 × x × y × y

The common factors are 5, x, and y.

∴5x²y – 15xy² = (5 × x × x × y)

– (3 × 5 × x × y × y)

= 5 × x × y[x – (3 × y)]

= 5xy(x – 3y)

(vii) We have, 10a² = 2 × 5 × a × a,

15b² = 3 × 5 × b × b

and 20c² = 2 × 2 × 5 × c × c

The three terms have 5 as a common factor

10a² – 15b² + 20c² = (2 × 5 × a × a)

– (3 × 5 × b × b) + (2 × 2 × 5 × c × c)

= 5 × (2 × a × a – 3 × b × b + 4 × c × c)

= 5(2a² – 3b² + 4c²)

(viii) We have, 4a² = 2 × 2 × a × a,

4ab = 2 × 2 × a × b

and, 4ca = 2 × 2 × c × a

The three terms have 2, 2 and a as common factors

∴–4a² + 4ab – 4ca = – (2 × 2 × a × a)

+ (2 × 2 × a × b) – (2 × 2 × c × a)

= 2 × 2 × a × (–a + b – c)

= 4a(–a + b – c)

(ix) x²yz = x × x × y × z

xy²z = x × y × y × z

xyz² = x × y × z × z

The common factors are x, y, and z.

∴ x²yz + xy²z + xyz² = (x × x × y × z) + (x × y × y × z) + (x × y × z × z)

= xyz(x + y + z)

(x) We have, ax²y = a × x × x × y

bxy² = b × x × y × y

and, cxyz = c × x × y × z

The three terms have x andy as common factors.

ax²y + bxy² + cxyz = (a × x × x × y)

+ (b × x × y × y) + (c × x × y × z)

= x × y × (a × x + b × y + c × z)

= xy(ax + by + cz)

3. Factorize:

(i) x² + xy + 8x + 8y

(ii) 15xy – 6x + 5y – 2

(iii) ax + bx – ay – by

(iv) 15pq + 15 + 9q + 25p

(v) z – 7 + 7xy – xyz

Sol. (i) (i) x² + xy + 8x + 8y = (x² + xy) + (8x + 8y) = x(x + y) + 8(x + y)

= (x + y)(x + 8)

[Taking (x + y) common]

(ii) 15xy – 6x + 5y – 2

= 3 × 5 × x × y – 3 × 2 × x + 5 × y – 2

= 3x(5y – 2) + 1(5y – 2)

= (5y – 2)(3x + 1)

(iii) ax + bx – ay – by = (ax + bx) – (ay + by)

[Grouping the terms]

= (a + b)x – (a + b)y

= (a + b)(x – y)

[Taking (a + b) common]

(iv) (iv) 15pq + 15 + 9q + 25p

= 15pq + 9q + 25p + 15

= 3 × 5 × p × q + 3 × 3 × q + 5 × 5 × p + 3 × 5

= 3q(5p + 3) + 5(5p + 3)

= (5p + 3)(3q + 5)

(v) z – 7 + 7xy – xyz

= z – 7 – xyz + 7xy

= 1(z – 7) – xy(z – 7)

= (z – 7)(1 – xy)

[Taking z – 7 common]

EXERCISE: 14.2

1. Factorize 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

(viii) a⁴ + 2a²b² + b⁴

Sol. (i) a² + 8a + 16

= a² + 2 × a × 4 + 42

= (a + 4)2

[Using: a² + 2ab + b² = (a + b)2]

= (a + 4)(a + 4)

(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

[Using: a² + 2ab + b² = (a + b)2]

= (5m + 3)(5m + 3)

(iv) 49y² + 84yz + 36z²

= (7y)2 + 2 × (7y) × (6z) + (6z)2

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

(v) 4x² – 8x + 4 = 4(x² – 2x + 1)

= 4(x² – 2 × x × 1 + 12)

= 4(x – 1)2

[Using: a² – 2ab + b² = (a – b)2]

= 4(x – 1)(x – 1)

(vi) 121b² – 88bc + 16c²

= (11b)2 – 2(11b)(4c) + (4c)2

= (11b – 4b)2

[∵(a – b)2 = a² – 2ab + b²]

(vii) (l + m)2 – 4lm

= l2 + 2lm + m2 – 4lm

= l2 – 2lm + m2

= (l – m)2 = (l – m)(l – m)

(viii) a⁴ + 2a²b² + b⁴

= (a²)2 + 2(a²)(b²) + (b²)2

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

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

Sol. (i) (i) 4p² – 9q²

= (2p²) – (3q)2

= (2p + 3q)(2p – 3q)

[Using: 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)

= 16x³[(x)2 – (3)2]

= 16x³(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)]

= (2l)(2m) = 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)

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

Sol. (i) (i) ax² + bx = 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 +bn + an2

= (am2

+ bm2) + (bn2 + an2)

= (a + b)m2 + (b + a)n2

= (a + b)(m2 + n2)

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

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

= (m + 1)(l + 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)

4. Factorise as far as you can:

(i) a⁴ – b⁴

(ii) p⁴ – 81

(iii) x⁴ – (y + z)4

(iv) x⁴ – (x – z)4

(v) a⁴ – 2a²b² + b⁴

Sol. (i) (i) a⁴ – b⁴

= (a²)2 – (b²)2

= (a² + b²)(a² – b²)

= (a² + b²)(a + b)(a – b)

(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

= [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)

= 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

= (a + b)(a + b)(a – b)(a – b)

5. Factorise the following expressions

(i) p² + 6p + 8

(ii) q²– 10q + 21

(iii) p² + 6p – 16

Sol. (i) p² + 6p + 8

= (p² + 6p + 9) – 1

[Using: 8 = 9 – 1]

= (p² + 2 × p × 3 + 32) – 1

= (p + 3)2 – 12

= (p + 3 + 1)(p + 3 – 1)

= (p + 4)(p + 2)

(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

= (p² + 6p + 9) – 9 – 16

= (p + 3)2 – 52

= (p + 3 + 5)(p + 3 – 5)

= (p + 8)(p – 2)

EXERCISE: 14.3

1. Carry out the following division

2. Divide the given polynomial by the given monomial.

(i) (5x² –6x) ÷ 3x

(ii) (3y⁸ – 4y⁶ + 5y⁴) ÷ y⁴

(iii) 8(x³y²z³ + x²y³z² +x²y²z³ ÷ 4x²y²z²

(iv) (x³ + 2x² + 3x) ÷ 2x

(v) (p³q⁶ – p⁶q³) ÷ p³q³

3. Work out of the following divisions

(i) (10x – 25 ÷ 5

(ii) (10x – 25) ÷ (2x – 5)

(iii) 10y(6y + 21) ÷ 5(2y + 7)

(iv) 9x²y²(3z – 24) ÷ 27xy(z – 8)

(v) 96abc(3a – 12)(5b – 30) ÷ 144(a – 4)(b – 6)

4. Divide as directed

(i) 5(2x + 1)(3x + 5) ÷ (2x + 1)

(ii) 26xy(x + 5)(y – 4) ÷ 13x(y – 4)

(iii) 52pqr(p + q)(q + r)(r + p)

÷ 104pq(q + r)(r + p)

(iv) 20(y + 4)(y² + 5y + 3) ÷ 5(y + 4)

(v) x(x + 1)(x + 2)(x + 3) ÷ x(x + 1)

5. Factorise the expressions and divide them as directed

(i) (y² + 7y + 10) ÷ (y + 5)

(ii) (m² – 14m – 32) ÷ (m + 2)

(iii) (5p² – 25p + 20) ÷ (p – 1)

(iv) 4yz(z² + 6z – 16) ÷ 2y(z + 8)

(v) 5pq(p² – q²) ÷ 2p(p + q)

(vi) 12xy(9x² – 16y²) ÷ 4xy(3x + 4y)

(vii) 39y³(50y² – 98) ÷ 26y²(5y + 7)

Sol. (i) (y² + 7y + 10)

= y² + 2y + 5y + 10

= y(y + 2) + 5(y + 2)

= (y + 2)(y + 5) ...(1)

(ii) m² – 14m – 32

= m² + 2m – 16m – 32

= m(m + 2)(m – 16)

= (m²

– 14m – 32) ÷ (m + 2)

(iii) (5p² –25p + 20)

= 5(p² – 5p + 4)

= 5(p² – p – 4p + 4)

= 5[p(p – 1) – 4(p – 1)]

= 5(p – 1)(p – 4) ...(1)

∴ (5p² – 25p �y 20) ÷ (p – 1)

= 5(p – 4)

(iv) 4yz(z² + 6z – 16)

= 4yz(z² + 8z – 2z – 16)

= 4yz[z(z + 8) – 2(z + 8)

= 4yz(z + 8)(z – 2) ...(1)

∴ 4yz(z² + 6z – 16) ÷ 2y(z + 8)

(v) 5pq(p² – q²)

= 5pq(p – q)(p + q) ...(1)

∴ 5pq(p² – q²) ÷ 2p(p + q)

(vi) 12xy(9x² – 16y²) = 12xy[(3x)² – (4y)²]

= 12xy(3x – 4y)(3x + 4y)

= 12xy(9x² – 16y²) ÷ 4xy(3x + 4y)

(vii) 39y³(50y² – 98)

= 3 × 13 × y × y × 2[(25y² – 49)]

= 3 × 13 × 2 × y × y × y × [(5y)² – 7)²]

= 3 × 13 × 2 × y × y × y × (5y – 7)(5y + 7)

26y²(5y + 7) = 2 × 13 × y × y ×(5y + 7)

39y³(50y² – 98) ÷ 26y²(5y + 7)

EXERCISE: 14.4

1. Find and correct the errors in the following mathematical statements.

(i) 4(x – 5) =4x – 5

(ii) x(3x + 2) = 3x² + 2

(iii) 2x + 3y = 5xy

(iv) x + 2x + 3x = 5x

(v) 5y + 2y + y – 7y = 0

(vi) 3x + 2x = 5x²

(vii) (2x)² + 4(2x) + 7 = 2x² + 8x + 7

(viii) (2x)² + 5x = 4x + 5x = 9x

(ix) (3x + 2)² = 3x² + 6x + 4

(x) Substituting x = – 3 in

(a) x² + 5x + 4 gives (–3)² + 5(–3) + 4

= 9 + 2 + 4 + 15

(b) x² – 5x + 4 gives (–3)² – 5(–3) + 4

= 9 – 15 + 4 = – 2

(xi) (y – 3)² = y² – 9

(xii) (z + 5)² = z² + 25

(xiii) (2a + 3b)(a – b) = 2a² – 3b²

(xiv) (a + 4)(a + 2) = a² + 8

(xv) (a – 4)(a – 2) = a² – 8

Sol. (i) LHS = 4(x – 5) = 4 × x – 4 × 5 = 4x – 20

∴ RHS should be 4x – 20.

Hence, 4(x – 5) = 4x – 20 is the correct statement.

(ii) x(3x + 2) = 3x² + 2

LHS = x(3x + 2)

= 3x² + 2x �j RHS

Hence, x(3x = 2) = 3x² + 2x is the correct statement.

(iii) The statement 2x + 3y = 5xy is incorrect.

Only like-terms can be grouped together.

Therefore, 2x + 3y = 2x + 3y is the correct statement.

(iv) LHS = x + 2x + 3x = 1x + 2x + 3x

= x(1 + 2 + 3) = 6x �j RHS

The correct statement is x + 2x + 3x = 6x

(v) By using the distributive property of multiplication over addition, we have

5y + 2y + y – 7y = (5 + 2 + 1 – 7)y

= (8 – 7)y = y

Thus, 5y + 2y + y – y – 7y �j 0, but

5y + 2y + y – 7y = y is the correct statement.

(vii) LHS = (2x)² + 4(2x) + 7

= 2x × 2x + 4 × 2x + 7

= 4x² + 8x + 7

Thus, (2x)² + 4(2x) + 7 �j 2x² + 8x + 7, but (2x)² + 4(2x) + 7 = 4x² + 8x + 7 is the correct

statement.

(viii) (2x)² + 5x = 4x + 5x = 9x

LHS = (2x)² + 5x = (2x)(2x) + 5x

= 4x² + 5x �j RHS

The correct statement is (2x)² + 5x

= 4x² + 5x

(ix) LHS = (3x + 2)² = (3x)² + 2 × 3x × 2 + 22

= 9x² + 12x + 4

Thus, (3x + 2)² �j 3x² + 6x + 4, but (3x + 2)²

= 9x² + 12x + 4 is the correct statement.

(x) (a) For x = –3,

x² + 5x + 4 = (–3)² + 5(–3) + 4

= 9 – 15 + 4 = 13 – 15 = – 2

(b) For x = –3,

x² – 5x + 4 = (–3)² + 5(–3) + 4

= 9 + 15 + 4 = 28

x² + 5x = (–3)² + 5(–3) = 9 – 15 = –6

(xi) LHS = (y – 3)² = y² – 2 × y × 3 + (–3)²

= y² – 6y + 9

Thus,

(y – 3)² �j y² – 9, but (y – 3)² = y² – 6y + 9 is the correct statement.

(xii) LHS = (z + 5)²

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

= z² + 10z + 25 �j RHS

The correct statement is (z + 5)²

= z² + 10z + 25

(xiii) LHS = (2a + 3b)(a – b)

= 2a(a – b) + 3b(a – b)

= 2a × a – 2a × b + 3b × a – 3b × b

= 2a² – 2ab + 3ab – 3b²

= 2a² + ab – 3b²

Thus, (2a + 3b)(a – b) �j 2a² – 3b², but (2a + 3b)(a – b) = 2a² + ab – 3b² is the correct

statement.

(xiv) LHS = (a + 4)(a + 2) = (a)² + (4 + 2)(a) +

4 × 2 = a² + 6a + 8 �j RHS

The correct statement is (a + 4)(a + 2)

= a² + 6a + 8

(xv) LHS = (a – 4)(a – 2)

= a² – (4 + 2)a + (–4 × –2)

= a² – 6a + 8

Thus, (a – 4)(a – 2) �j a² – 8, but (a – 4)(a – 2) = a² – 6a + 8 is the correct statement.