NCERT CLass 8 Maths

Class 8 NCERT Maths Ex 14.1 Question 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 = 2x2x7 x p x p x q x q

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, 3x²and 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 a²b = 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, – 4x²and 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

10x³y² = 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

Class 8 NCERT Maths Ex 14.1 Question 2

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:

(i) 7x – 42 = 7(x – 6)
(ii) 6p – 12q = 6(p – 2q)
(iii) 7a2 + 14a = 7a(a + 2)
(iv) -16z + 20z3 = 4z(-4 + 5z2)
(v) 20l2m + 30alm = 10lm(2l + 3a)
(vi) 5x2y – 15xy2 = 5xy(x – 3y)
(vii) 10a2 – 15b2 + 20c2 = 5(2a2 – 3b2 + 4c2)
(viii) -4a2 + 4ab – 4ca = 4a(-a + b – c)
(ix) x2yz + xy2z + xyz2 = xyz(x + y + z)
(x) ax2y + bxy2 + cxyz = xy(ax + by + cz)

Class 8 NCERT Maths Ex 14.1 Question 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:

(i) x2 + xy + 8x + 8y

Grouping the terms, we have
x2 + xy + 8x + 8y
= x(x + y) + 8(x + y)
= (x + y)(x + 8)
Hence, the required factors = (x + y)(x + 8)

(ii) 15xy – 6x + 5y – 2
Grouping the terms, we have
(15xy – 6x) + (5y – 2)
= 3x(5y – 2) + (5y – 2)
= (5y – 2)(3x + 1)

(iii) ax + bx – ay – by

Grouping the terms, we have
= (ax – ay) + (bx – by)
= a(x – y) + b(x – y)
= (x – y)(a + b)
Hence, the required factors = (x – y)(a + b)

(iv) 15pq + 15 + 9q + 25p
Grouping the terms, we have
= (15pq + 25p) + (9q + 15)
= 5p(3q + 5) + 3(3q + 5)
= (3q + 5) (5p + 3)
Hence, the required factors = (3q + 5) (5p + 3)

(v) z – 7 + 7xy – xyz
Grouping the terms, we have
= (-xyz + 7xy) + (z – 7)
= -xy(z – 7) + 1 (z – 7)
= (-xy + 1) (z – 1)
Hence the required factor = -(1 – xy) (z – 7)

Class 8 NCERT Maths Ex 14.2 Question 1

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

Using the identity (x+y)2 = x2+2xy+y2

= (a+4)²

(ii) p²–10p+25

= p²-2×5×p+5²

Using the identity (x-y)2 = x2-2xy+y2

= (p-5)²

(iii) 25m²+30m+9

= (5m)²+2×5m×3+3²

= (5m+3)²

(iv) 49y²+84yz+36z²

=(7y)²+2×7y×6z+(6z)²

= (7y+6z)²

(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)²using 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

Class 8 NCERT Maths Ex 14.2 Question 2

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(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)

Class 8 NCERT Maths Ex 14.2 Question 3

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)

Class 8 NCERT Maths Ex 14.2 Question 4

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)

Class 8 NCERT Maths Ex 14.3 Question 1

1. Carry out the following divisions:

i) 28x4 ÷ 56x

(ii) –36y3 ÷ 9y2

(iii) 66 pq2r3 ÷ 11qr2

(iv) 34x3y3z3 ÷ 51xy2z3

(v) 12a8b8 ÷ (-6a6b4)

Solution:

i) 28×4 ÷ 56x

28×4 = 2×2×7×x×x×x×x
56x = 2×2×2×7×x
28×4 ÷ 56x = $\frac{2\times 2\times 7\times x\times x\times x\times x}{2\times 2\times 2\times 7\times x}$

= $\frac{x^3}{2}$ = $\frac{1}{2}x^3$

(ii) –36y3 ÷ 9y2
-36y3 = – 2 × 2 × 3 × 3 × y × y × y
9y2 = 3 × 3 × y × y
-36y3 ÷ 9y2

= $\frac{- 2 \times 2 \times 3 \times 3 \times y \times y \times y}{3\times 3\times y\times y}$

= -4y

(iii) 66 pq2r3 ÷ 11qr2
66pq2r3 = 2 × 3 × 11 × p × q × q × r × r × r
11qr2 = 11 × q × r × r
66 pq2r3 ÷ 11qr2 = $\frac{2\times3\times 11 \times p\times q \times q \times r \times r \times r }{11\times q\times r\times r}$

= 6pqr

(iv) 34x3y3z3 ÷ 51xy2z3
34x3y3z3 = 2 × 17 × x × x × x × y × y × y × z × z × z
51xy2z3 = 3 × 17 × x × y × y × z × z × z

34x3y3z3 ÷ 51xy2z3 = $\frac{2 \times 17 \times x \times x \times x \times y \times y \times y \times z \times z \times z}{3\times 17\times x\times y\times y\times z\times z\times z}$

= $\frac{2x^2y}{3}$

(v) 12a8b8 ÷ (-6a6b4)
12a8b8 = 2 × 2 × 3 × a8 × b8
-6a6b4 = -2 × 3 × a6 × b4

12a8b8 ÷ (-6a6b4) = $\frac{2\times 2 \times 3 \times a^8 \times b^8}{-2\times 3\times a^6\times b^4}$
= -2a2b4

Class 8 NCERT Maths Ex 14.3 Question 2

2. Divide the given polynomial by the given monomial.
(i) (5×2–6x) ÷ 3x
(ii) (3y8–4y6+5y4) ÷ y4
(iii) 8(x3y2z2+x2y3z2+x2y2z3)÷ 4×2 y2 z2
(iv) (x3+2×2+3x) ÷2x
(v) (p3q6–p6q3) ÷ p3q3

Solution –

(i) (5×2 – 6x) ÷ 3x
(5×2 – 6x) = x(5x – 6)

(5×2 – 6x) ÷ 3x = $\frac{x(5x - 6)}{3x}$

= $\frac{(5x - 6)}{3}$

(ii) (3y8 – 4y6 + 5y4) ÷ y4
(3y8 – 4y6 + 5y4) = y4(3y4 – 4y2 + 5)

(3y8 – 4y6 + 5y4) ÷ y4 = $\frac{y^4(3y^4 - 4y^2 + 5)}{y^4}$
= 3y4 – 4y2 + 5

(iii) 8(x3y2z2 + x2y3z2 + x2y2z3) ÷ 4x2y2z2
8(x3y2z2 + x2y3z2 + x2y2z3) = 8x2y2z2(x + y + z)

8(x3y2z2 + x2y3z2 + x2y2z3) ÷ 4x2y2z2 = $\frac{8x^2y^2z^2(x + y + z)}{4x^2y^2z^2}$
= 2(x + y + z)

(iv) (x3 + 2×2 + 3x) ÷ 2x

(x3 + 2×2 + 3x) = x(x2 + 2x + 3)
(x3 + 2×2 + 3x) ÷ 2x = $\frac{x(x^2 + 2x + 3)}{2x}$

= $\frac{(x^2 + 2x + 3)}{2}$

(v) (p3q6 – p6q3) ÷ p3q3

(p3q6 – p6q3) = p3q3 (q3 – p3)

(p3q6 – p6q3) ÷ p3q3 = $\frac{p^3q^3 (q^3 - p^3)}{p^3q^3}$
= q3 – p3

Class 8 NCERT Maths Ex 14.3 Question 3

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
(10x – 25) = 5× 2 × x – 5 × 5
= 5(2x – 5)
(10x – 25) ÷ 5 = $\frac{5(2x - 5)}{5}$
= 2x – 5

(ii) (10x – 25) ÷ (2x – 5)
(10x – 25) = 5 × 2 × x – 5× 5
= 5(2x – 5)
(10x – 25) ÷ (2x – 5) = $\frac{5(2x - 5)}{(2x - 5)}$
= 5

(iii) 10y(6y + 21) ÷ 5(2y + 7)
10y(6y + 21) = 5 × 2 × y ×(2 × 3 × y + 3 × 7)
= 5 × 2 × y × 3(2 × y + 7)
= 30y(2y + 7)
10y(6y + 21) ÷ 5(2y + 7) = $\frac{30y(2y + 7)}{5(2y + 7)}$
= 6y

(iv) 9x2y2(3z – 24) ÷ 27xy(z – 8)
9x2y2(3z – 24) = 3 × 3 × x × x × y × y ×(3 × z – 2 × 2 × 2 × 3)
= 3 × 3 × x × x × y × y × 3( z – 2 × 2 × 2)
= 27x2y2(z – 8)
9x2y2(3z – 24) ÷ 27xy(z – 8) = $\frac{27x^2y^2(z - 8)}{27xy(z - 8)}$
= xy

(v) 96abc(3a – 12)(5b – 30) ÷ 144(a – 4)(b – 6)
96abc(3a – 12)(5b – 30) = 96abc ×(3 × a – 2 × 2 × 3) × (5 × b – 5 × 2 × 3)
= 96abc × 3(a – 2 × 2)× 5(b – 2 × 3)
= 1440abc(a – 4)(b – 6)
96abc(3a – 12)(5b – 30) ÷ 144(a – 4)(b – 6) = $\frac{1440abc(a - 4)(b - 6)}{144(a - 4)(b - 6)}$
= 10abc

Class 8 NCERT Maths Ex 14.3 Question 4

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 –

(i) 5(2x+1)(3x+5) ÷ (2x+1)

= $\frac{5(2x +1)(3x + 5) }{(2x +1)}$

= 5(3x + 5)

(ii) 26xy(x + 5)(y – 4) ÷ 13x(y – 4)

= $\frac{2 \times 13 \times xy(x + 5)(y - 4) }{13x(y - 4)}$

= 2y(x + 5)

(iii) 52pqr (p + q)(q + r)(r + p) ÷ 104pq(q + r)(r + p)
= $\frac{2\times 2 \times 13 \times p \times q \times r \times ( p + q) \times (q + r) \times (r + p)}{2 \times 2 \times 2 \times 13 \times p \times q \times (q + r) \times (r + p)}$

= $\frac{r(p + q)}{2}$

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

= $\frac{2 \times 2 \times 5 \times (y + 4) \times (y^2 + 5 y + 3)}{5 \times (y + 4)}$

= 4(y2 + 5 y + 3)

(v) x(x +1)(x + 2)(x + 3) ÷ x(x +1)
= (x + 2)(x + 3)

Class 8 NCERT Maths Ex 14.3 Question 5

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(9×2–16y2)÷4xy(3x+4y)
(vii) 39y3(50y2–98) ÷ 26y2(5y+7)

Solution –

(i) (y2+7y+10)÷(y+5)
(y2+7y+10) = y2+2y+5y+10
= y(y+2)+5(y+2)
= (y+2)(y+5)
(y2+7y+10)÷(y+5) = $\frac{(y+2)(y+5)}{(y+5)}$
= y+2

(ii) (m2–14m–32)÷ (m+2)
m2–14m–32
= m2+2m-16m–32
= m(m+2)–16(m+2)
= (m–16)(m+2)
(m2–14m–32)÷(m+2) = $\frac{(m - 16)(m+2)}{(m+2)}$
= m – 16

(iii) (5p2–25p+20)÷(p–1)
5p2–25p+20 = 5(p2 – 5p + 4)
= 5(p2 – p – 4p + 4)
= 5[p(p – 1) – 4(p – 1)]
= 5(p – 1)(p – 4)
(5p2–25p+20)÷(p–1) = $\frac{5(p-1)(p-4)}{(p-1)}$
= 5(p–4)

(iv) 4yz(z2 + 6z–16)÷ 2y(z+8)

4yz(z2 + 6z -16) = 4 yz (z2 – 2z + 8z -16)
= 4 yz [z(z – 2) + 8(z – 2)]
= 4 yz(z – 2)(z + 8)
4yz(z2+6z–16) ÷ 2y(z+8) = $\frac{4yz(z-2)(z+8)}{2y(z+8)}$
= 2z(z-2)

(v) 5pq(p2–q2) ÷ 2p(p+q)
5pq(p2 – q2) = 5pq(p–q)(p+q)
5pq(p2–q2) ÷ 2p(p+q) = $\frac{5pq(p-q)(p+q)}{2p(p+q)}$

= $\frac{5q(p-q)}{2}$

(vi) 12xy(9×2–16y2) ÷ 4xy(3x+4y)
12xy(9×2 -16y2) = 12xy[(3x)2 – (4y)2]
= 12xy(3x – 4y)(3x + 4y)
12xy(9×2–16y2) ÷ 4xy(3x+4y) = $\frac{12xy(3x+4y)(3x-4y)}{4xy(3x+4y)}$
= 3(3x – 4y)

(vii) 39y3(50y2–98) ÷ 26y2(5y+7)
39y3(50y2 – 98) = 3 × 13 × y × 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) = $\frac{3\times 13 \times 2 \times y \times y \times y(5y - 7)(5y + 7)}{2\times13\times y \times y \times (5y+7)}$
= 3y (5y – 7)

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