NCERT Solutions for Class 8 Maths Chapter 12 Factorisation(Updated Pattern)

1. When we factorise an expression, we write it as a product of factors. These factors may be
   numbers, algebraic variables or algebraic expressions.
2. An irreducible factor is a factor which cannot be expressed further as a product of factors.
3. A systematic way of factorising an expression is the common factor method. It consists of
   three steps: (i) Write each term of the expression as a product of irreducible factors (ii) Look
   for and separate the common factors and (iii) Combine the remaining factors in each term in
   accordance with the distributive law.
4. Sometimes, all the terms in a given expression do not have a common factor; but the terms can
   be grouped in such a way that all the terms in each group have a common factor. When we do
   this, there emerges a common factor across all the groups leading to the required factorisation
   of the expression. This is the method of regrouping.
5. In factorisation by regrouping, we should remember that any regrouping (i.e., rearrangement)
   of the terms in the given expression may not lead to factorisation. We must observe the
   expression and come out with the desired regrouping by trial and error
6. In expressions which have factors of the type (x + a) (x + b), remember the numerical term gives ab. Its
   factors, a and b, should be so chosen that their sum, with signs taken care of, is the coefficient of x.
7. We know that in the case of numbers, division is the inverse of multiplication. This idea applies
   also to the division of algebraic expressions.
8. In the case of the division of a polynomial by a monomial, we may carry out the division either by
   dividing each term of the polynomial by the monomial or by the common factor method.
9. In the case of the division of a polynomial by a polynomial, we cannot proceed by dividing each term
   in the dividend polynomial by the divisor polynomial. Instead, we factorise both the polynomials
   and cancel their common factors.
10. In the case of divisions of algebraic expressions that we studied in this chapter, we have
    Dividend = Divisor × Quotient.
    In general, however, the relation is
    Dividend = Divisor × Quotient + Remainder
    Thus, we have considered in the present chapter only those divisions in which the remainder
    is zero.

NCERT Solutions for Class 8 Maths Chapter 11 Direct and Inverse Proportional (Updated Pattern)

——————————-Exercise 12.1——————————————

1. Find the common factors of the given terms.

(i) 12x, 36

(ii) 2y, 22xy

(iii) 14 pq, 28p²q²

(iv) 2x, 3x², 4

(v) 6 abc, 24ab², 12a²b

(vi) 16 x³, – 4x² , 32 x

(vii) 10 pq, 20qr, 30 rp

(viii) 3x²y³ , 10x³y² , 6x²y²z

Solution:

(i) Factors of 12x and 36

12x = 2×2×3×x

36 = 2×2×3×3

Common factors of 12x and 36 are 2, 2, 3

and , 2×2×3 = 12

(ii) Factors of 2y and 22xy

2y = 2×y

22xy = 2×11×x×y

Common factors of 2y and 22xy are 2, y

and ,2×y = 2y

(iii) Factors of 14pq and 28p²q²

14pq = 2x7xpxq

28p²q² = 2x2x7xpxpxqxq

Common factors of 14 pq and 28 p²q² are 2, 7 , p , q

and, 2x7xpxq = 14pq

(iv) Factors of 2x, 3x²and 4

2x = 2×x

3x²= 3×x×x

4 = 2×2

Common factors of 2x, 3x² and 4 is 1.

(v) Factors of 6abc, 24ab² and 12a²b

6abc = 2×3×a×b×c

24ab² = 2×2×2×3×a×b×b

12 a² b = 2×2×3×a×a×b

Common factors of 6 abc, 24ab² and 12a²b are 2, 3, a, b

and, 2×3×a×b = 6ab

(vi) Factors of 16x³ , -4x²and 32x

16 x³ = 2×2×2×2×x×x×x

– 4x² = -1×2×2×x×x

32x = 2×2×2×2×2×x

Common factors of 16 x³ , – 4x² and 32x are 2,2, x

and, 2×2×x = 4x

(vii) Factors of 10 pq, 20qr and 30rp

10 pq = 2×5×p×q

20qr = 2×2×5×q×r

30rp= 2×3×5×r×p

Common factors of 10 pq, 20qr and 30rp are 2, 5

and, 2×5 = 10

(viii) Factors of 3x²y³ , 10x³y² and 6x²y²z

3x²y³ = 3×x×x×y×y×y

10x³ y² = 2×5×x×x×x×y×y

6x²y²z = 3×2×x×x×y×y×z

Common factors of 3x²y³, 10x³y² and 6x²y²z are x², y²

and, x²×y² = x²y²

2.Factorise the following expressions.

(i) 7x–42

(ii) 6p–12q

(iii) 7a²+ 14a

(iv) -16z+20 z³

(v) 20l²m+30alm

(vi) 5x²y-15xy²

(vii) 10a²-15b²+20c²

(viii) -4a²+4ab–4 ca

(ix) x²yz+xy²z +xyz²

(x) ax²y+bxy²+cxyz

Solution:

(vii) 10a²-15b²+20c²

10a² = 2×5×a×a

– 15b² = -1×3×5×b×b

20c² = 2×2×5×c×c

Common factor of 10 a² , 15b² and 20c² is 5

10a²-15b²+20c² = 5(2a²-3b²+4c² )

(viii) – 4a²+4ab-4ca

– 4a² = -1×2×2×a×a

4ab = 2×2×a×b

– 4ca = -1×2×2×c×a

Common factor of – 4a² , 4ab , – 4ca are 2, 2, a i.e. 4a

So,

– 4a²+4 ab-4 ca = 4a(-a+b-c)

(ix) x²yz+xy²z+xyz²

x²yz = x×x×y×z

xy²z = x×y×y×z

xyz² = x×y×z×z

Common factor of x²yz , xy²z and xyz² are x, y, z i.e. xyz

Now, x²yz+xy²z+xyz² = xyz(x+y+z)

(x) ax²y+bxy²+cxyz

ax²y = a×x×x×y

bxy² = b×x×y×y

cxyz = c×x×y×z

Common factors of a x²y ,bxy² and cxyz are xy

Now, ax²y+bxy²+cxyz = xy(ax+by+cz)

3. Factorise.

(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

Solution:

——————————-Exercise 12.2——————————————

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

Solution:

(i) a²+8a+16

= a²+2×4×a+4²

= (a+4)²

Using the identity (x+y)² = x²+2xy+y²

(ii) p²–10p+25

= p²-2×5×p+5²

= (p-5)²

Using the identity (x-y)² = x²-2xy+y²

(iii) 25m²+30m+9

= (5m)²+2×5m×3+3²

= (5m+3)²

Using the identity (x+y)² = x²+2xy+y²

(iv) 49y²+84yz+36z²

=(7y)²+2×7y×6z+(6z)²

= (7y+6z)²

Using the identity (x+y)² = x²+2xy+y²

(v) 4x²–8x+4

= (2x)²-2×4x+2²

= (2x-2)²

Using the identity (x-y)² = x²-2xy+y²

(vi) 121b²-88bc+16c²

= (11b)²-2×11b×4c+(4c)²

= (11b-4c)²

Using the identity (x-y)² = x²-2xy+y²

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

Expand (l+m)² using the identity (x+y)² = x²+2xy+y²

(l+m)²-4lm = l²+m²+2lm-4lm

= l²+m²-2lm

= (l-m)²

Using the identity (x-y)² = x²-2xy+y²

(viii) a⁴+2a²b²+b⁴

= (a²)²+2×a^2×b²+(b²)²

= (a²+b²)²

Using the identity (x+y)² = x²+2xy+y²

GLOBAL FOOTBALL FEVER: UNVEILING THE TOP 10 SOCCER LEAGUES AROUND THE WORLD

2. Factorise.

(i) 4p²–9q²

(ii) 63a²–112b²

(iii) 49x²–36

(iv) 16x⁵–144x³ differ

(v) (l+m)²-(l-m) ²

(vi) 9x²y²–16

(vii) (x²–2xy+y²)–z²

(viii) 25a²–4b²+28bc–49c²

Solution:

(i) 4p²–9q²

= (2p)²-(3q)²

= (2p-3q)(2p+3q)

Using the identity x²-y² = (x+y)(x-y)

(ii) 63a²–112b²

= 7(9a² –16b²)

= 7((3a)²–(4b)²)

= 7(3a+4b)(3a-4b)

Using the identity x²-y² = (x+y)(x-y)

(iii) 49x²–36

= (7x)² -6²

= (7x+6)(7x–6)

Using the identity x²-y² = (x+y)(x-y)

(iv) 16x⁵–144x³

= 16x³(x²–9)

= 16x³(x²–9)

= 16x³(x–3)(x+3)

Using the identity x²-y² = (x+y)(x-y)

(v) (l+m) ²-(l-m) ²

= {(l+m)-(l–m)}{(l +m)+(l–m)}

Using the identity x²-y² = (x+y)(x-y)

= (l+m–l+m)(l+m+l–m)

= (2m)(2l)

= 4 ml

(vi) 9x²y²–16

= (3xy)²-4²

= (3xy–4)(3xy+4)

Using the identity x²-y² = (x+y)(x-y)

(vii) (x²–2xy+y²)–z²

= (x–y)²–z²

Using the identity (x-y)² = x²-2xy+y²

= {(x–y)–z}{(x–y)+z}

= (x–y–z)(x–y+z)

Using the identity x²-y² = (x+y)(x-y)

(viii) 25a²–4b²+28bc–49c²

= 25a²–(4b²-28bc+49c² )

= (5a)²-{(2b)²-2(2b)(7c)+(7c)²}

= (5a)²-(2b-7c)²

Using the identity x²-y² = (x+y)(x-y) , we have

= (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

Solution:

(i) ax²+bx = x(ax+b)

(ii) 7p²+21q² = 7(p²+3q²)

(iii) 2x³+2xy²+2xz² = 2x(x²+y²+z²)

(iv) am²+bm²+bn²+an² = m²(a+b)+n²(a+b) = (a+b)(m²+n²)

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

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

(vii) 5y²–20y–8z+2yz = 5y(y–4)+2z(y–4) = (y–4)(5y+2z)

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

(i) a⁴–b⁴

(ii) p⁴–81

(iii) x⁴–(y+z) ⁴

(iv) x⁴–(x–z) ⁴

(v) a⁴–2a²b²+b⁴

Solution:

(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)

= (p²-3²)(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)²-2a²b²+(b²)²

= (a²-b²)²

= ((a–b)(a+b))²

= (a – b)² (a + b)²

5. Factorise the following expressions.

(i) p²+6p+8

(ii) q²–10q+21

(iii) p²+6p–16

Solution:

(i) p²+6p+8

We observed that 8 = 4×2 and 4+2 = 6

p²+6p+8 can be written as p²+2p+4p+8

Taking Common terms, we get

p²+6p+8 = p²+2p+4p+8 = p(p+2)+4(p+2)

Again, p+2 is common in both the terms.

= (p+2)(p+4)

This implies that p²+6p+8 = (p+2)(p+4)

(ii) q²–10q+21

We observed that 21 = -7×-3 and -7+(-3) = -10

q²–10q+21 = q²–3q-7q+21

= q(q–3)–7(q–3)

= (q–7)(q–3)

This implies that q²–10q+21 = (q–7)(q–3)

(iii) p²+6p–16

We observed that -16 = -2×8 and 8+(-2) = 6

p²+6p–16 = p²–2p+8p–16

= p(p–2)+8(p–2)

= (p+8)(p–2)

So, p²+6p–16 = (p+8)(p–2)

——————————-Exercise 12.3——————————————

1. Carry out the following divisions.

(i) 28x⁴ ÷ 56x

(ii) –36y³ ÷ 9y²

(iii) 66pq²r³ ÷ 11qr²

(iv) 34x³y³z³ ÷ 51xy²z³

(v) 12a⁸b⁸ ÷ (– 6a⁶b⁴)

Solution:

(i)28x⁴ = 2×2×7×x×x×x×x

56x = 2×2×2×7×x

2. Divide the given polynomial by the given monomial.

(i)(5x²–6x) ÷ 3x

(ii)(3y⁸–4y⁶+5y⁴) ÷ y⁴

(iii) 8(x³y²z²+x²y³z²+x²y²z³)÷ 4x² y² z²

(iv)(x³+2x²+3x) ÷2x

(v) (p³q⁶–p⁶q³) ÷ p³q³

Solution:

3. Work out 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)

Solution:

(i) (10x–25) ÷ 5 = 5(2x-5)/5 = 2x-5

(ii) (10x–25) ÷ (2x–5) = 5(2x-5)/( 2x-5) = 5

(iii) 10y(6y+21) ÷ 5(2y+7) = 10y×3(2y+7)/5(2y+7) = 6y

(iv) 9x²y²(3z–24) ÷ 27xy(z–8) = 9x²y²×3(z-8)/27xy(z-8) = xy

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)

Solution:

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)

Solution:

(i) (y²+7y+10)÷(y+5)

First, solve the equation (y²+7y+10)

(y²+7y+10) = y²+2y+5y+10 = y(y+2)+5(y+2) = (y+2)(y+5)

Now, (y²+7y+10)÷(y+5) = (y+2)(y+5)/(y+5) = y+2

(ii) (m²–14m–32)÷ (m+2)

Solve for m²–14m–32, we have

m²–14m–32 = m²+2m-16m–32 = m(m+2)–16(m+2) = (m–16)(m+2)

Now, (m²–14m–32)÷(m+2) = (m–16)(m+2)/(m+2) = m-16

(iii) (5p²–25p+20)÷(p–1)

Step 1: Take 5 common from the equation, 5p²–25p+20, we get

5p²–25p+20 = 5(p²–5p+4)

Step 2: Factorise p²–5p+4

p²–5p+4 = p²–p-4p+4 = (p–1)(p–4)

Step 3: Solve original equation

(5p²–25p+20)÷(p–1) = 5(p–1)(p–4)/(p-1) = 5(p–4)

(iv) 4yz(z² + 6z–16)÷ 2y(z+8)

Factorising z²+6z–16,

z²+6z–16 = z²-2z+8z–16 = (z–2)(z+8)

Now, 4yz(z²+6z–16) ÷ 2y(z+8) = 4yz(z–2)(z+8)/2y(z+8) = 2z(z-2)

(v) 5pq(p²–q²) ÷ 2p(p+q)

p²–q² can be written as (p–q)(p+q) using the identity.

5pq(p²–q²) ÷ 2p(p+q) = 5pq(p–q)(p+q)/2p(p+q) = 5q(p–q)/2

(vi) 12xy(9x²–16y²) ÷ 4xy(3x+4y)

Factorising 9x²–16y² , we have

9x²–16y² = (3x)²–(4y)² = (3x+4y)(3x-4y) using the identity p²–q² = (p–q)(p+q)

Now, 12xy(9x²–16y²) ÷ 4xy(3x+4y) = 12xy(3x+4y)(3x-4y) /4xy(3x+4y) = 3(3x-4y)

(vii) 39y³(50y²–98) ÷ 26y²(5y+7)
st solve for 50y²–98, we have

50y²–98 = 2(25y²–49) = 2((5y)²–7²) = 2(5y–7)(5y+7)

Now, 39y³(50y²–98) ÷ 26y²(5y+7) =