Factorisation

(i) Factor of 12 = 2× 2 × 3 ×

Factor of 36 = 2× 2 × 3× 3

Common factors are 2, 2, 3

Product of common factors = 2× 2 × 3 = 12

(ii)Factor of 2y = 2× y

Factor of 22xy = 2× 11 × x × y

Common factors are 2, y

Product of common factors = 2× y = 2y

(iii)Factor of 14pq = 2× 7 × p × q

Factor of 28p²q²= 2× 2 × 7 × p× p × q× q

Common factors are 2, 7, p, q

Product of common factors = 2× 7 × p× q = 14pq

(iv) Factor of 2x = 2 × x

Factor of 3x²= 3× x × x

Factor of 4 = 2 × 2

Common factor is 1

(v) _{Factor of 6abc = 2 × 3 × a × b × c}

Factor of 24ab² = 2× 2 × 2 × 3 × a × b × b

Factor of 12a²b = 2× 2 × 3 × a × a × b

Common factors are 2, 3, a, b

Product of common factors = 2× 3 × a × b = 6ab

(vi) Factor of 16x³ = 2 × 2 × 2 × 2 × x × x × x

Factor of -4x² = -2× 2 × x × x

Factor of 32x = 2× 2 × 2 × 2 × 2 × x

Common factors are 2, 2, x

Product of common factors = 2× 2 × x = 4x

(vii) Factor of 10pq = 2 × 5 × p × q

Factor of 20qr = 2× 2 × 5 × q × r

Factor of 30rp = 2× 3 × 5 × r × p

Common factors are 2, 5

Product of common factors = 2× 5 = 10

(viii) Factor of 3x²y³ = 3 × x × x × y × y × y

Factor of 10x³y² = 2× 5 × x × x × x × y × y

Factor of 6x²y²z = 2× 3 × x × x × y× y × z

Common factors are x × x, y × y

Product of common factors = x²y²

(i) Factor of 7x = 7 × x

Factor of 42 = 2× 3 × 7

Common factors is 7

7x-42 = (7× x)-( 2× 3 × 7) = 7(x-6)

(ii) Factor of 6p = 2 ×3 × p

Factor of -12q = -1× 2× 2 × 3 × q

Common factors is 2, 3 = 2× 3 = 6

6p-12q = (2 × 3 × p)-( 2× 2 × 3 × q) = 6(p-2q)

(iii) Factor of 7a² = 7 × a × a

Factor of 14a = 2 × 7 × a

Common factors is 7 a = 7 × a = 7a

7a²+14a = (7 × a × a)-( 2× 7 × a) = 7a(a-2)

(iv) Factor of -16z = - 2 × 2 × 2 × 2 × z

Factor of 20z³ = 2 × 2× 5 × z × z × z

Common factors is 2, 2, z = 2 × 2 × z = 4z

-16z+20z³ = -4z(z-5z²)

(v) Factor of 20l³m = 2 × 2 × 5 × l × l × l × m

Factor of 30alm = 2 × 3 × 5 × a × l × m

Common factors is 2, 5, l, m = 2 × 5 × l × m = 10lm

20l³m+30alm = 10lm(2l²-3a)

(vi) Factor of 5x²y = 5 × x × x × y

Factor of -15xy² = -3 × 5 × x × y × y

Common factors is 5, x, y = 5 × x × y = 5xy

5x²y-15xy²= 5xy(x-3y)

(vii) Factor of 10a² = 2 × 5 × a × a

Factor of -15b² = -3 × 5 × b × b

Factor of 20c² = 2 × 2 × 5 × c × c

Common factors is 5

10a²-15b²+20c² = 5(2a²-3b²+c²)

(viii) Factor of -4a² = -2 × 2 × a × a

Factor of 4ab = 2 × 2 × a × b

Factor of 4ca = 2 × 2 × c × a

Common factors is 2, 2, a

-4a²+4ab-4ca = 4a(-a+b-c)

(ix) Factor of x²yz = x × x × y × z

Factor of xy²z = x × y × y × z

Factor of xyz² = x × y × z × z

Common factors is x, y, z

X²yz + xy²z + xyz² = xyz(z+y+z)

(x) Factor of ax²y = a × x × x × y

Factor of bxy² = b × x × y × y

Factor of cxyz = c × x × y × z

Common factors is x, y

a x²y + bxy² + cxyz = xy(ax+by+cz)

(i) x² + xy + 8x + 8y

Taking x as common in first two terms and 8 as common in next two terms

x(x + y) +8 (x + y)

= (x + y)(x + 8) [ By taking (x+y) common from both terms]

(ii) 15xy - 6x + 5y - 2

Taking 3x as common in first two terms and 1 as common in next two terms

3x(5y - 2) + 1 (5y - 2)

= (3x + 1)(5y - 2)

(iii) ax + bx – ay - by

Taking a as common in ax and -ay and b as common in remaining two terms

a(x - y) + b (x - y)

= (a + b)(x - y)

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

Taking 3q as common in 15pq and 9q and 5 as common in remaining two terms

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

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

(v) z - 7 + 7xy - xyz

Taking xy as common in 7xy and -xyz and 1 as common in remaining two terms

1(z - 7) - xy (z - 7)

= (1 – x y) (z - 7)

(i) a² + 8a + 16

= a² + 2× a × 4 + 4²

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

Here a =a; b = 4

a² + 2× a × 4 + 4²

= (a + 4)²

^{= (a + 4)(a + 4)}

(ii) p² -10p + 25

= p² - 2× 5 × p + 5²

Using identity (a - b)² = a² - 2ab + b²

Here a =p; b = 5

p² - 2× 5 × p + 5²

= (p - 5)²

= (p-5)(p-5)

(iii) 25m² + 30m + 9

= (5m)² + 2 × 5 × 3 × m + 3²

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

Here a =5m; b = 3

(5m)² + 2 × 5 × 3 × m + 3²

= (5m + 3)²

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

(iv) 49y² + 84yz + 36z²

= (7y)² + 2 × 7 × 6 × y × z + (6z)²

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

Here a =7y; b = 6z

(7y)² + 2 × 7 × 6 × y × z + (6z)²

= (7y + 6z)²

= (7y + 6z)(7y + 6z)

(v) 4x² - 8x + 4

= (2x)² - 2 × 2 × 2× x + 2²

Using identity (a - b)² = a² - 2ab + b²

Here a =2x; b = 2

(2x)² - 2 × 2 × 2× x + 2²

= (2x - 2)²

= (2x - 2)(2x - 2)

(vi) 121b² - 88bc + 16c²

= (11b)² - 2 × 11b × 4c + (4c)²

Using identity (a - b)² = a² - 2ab + b²

Here a =11b; b = 4c

(11b)² - 2 × 11b × 4c + (4c)²

= (11b – 4c)²

= (11b – 4c)(11b – 4c)

(vii) (l + m)² - 4lm

Expand (l + m)² = l² + 2lm + m²

[using (a + b)² = a² + 2ab + b²]

l² + 2lm + m² - 4lm

⇒ l² - 2lm + m²

= l² - 2 × l × m + m²

Using identity (a - b)² = a² - 2ab + b²

Here a =l; b = m

l² - 2 × l × m + m²

= (l – m)²

= ( l - m)(l - m)

(viii) a⁴ + 2a²b² + b⁴

_{= (a}²)² + 2 × a² × b² + (b²)²

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

Here a = a²; b = b²

(a²)² + 2 × a² × b² + (b²)²

= (a² + b²)²

= (a² + b²)(a² + b²)

(i) 4p² - 9q²

4p² - 9q² = (2p)² - (3q)²

Using identity a² - b² = (a + b)(a - b)

Here a = 2p; b = 3q

(2p)² - (3q)² = (2p + 3q) (2p – 3q)

(ii) 63a² - 112b²

7(9a² - 16b²) = 7{(3a)² - (4b²)}

Using identity a² - b² = (a + b)(a - b)

Here a = 3a; b = 4b

7{(3a)² - (4b)²} = 7{(3a + 4b) (3a – 4b)}

(iii) 49x² - 36 = (7x)² - 6²

Using identity a² - b² = (a + b)(a - b)

Here a = 7x; b = 6

(7x)² - (6)² = (7x + 6) (7x – 6)

(iv) 16x⁵ - 144x³ = 16x³(x² - 9) ⇒ 16x³{x² - 3²}

Using identity a² - b² = (a + b)(a - b)

Here a = x; b = 3

16x³{(x)² - (3)²} = 16x³{(x + 3) (x – 3)}

(v) (l + m)² - (l - m)²

Using identity a² - b² = (a + b)(a - b)

Here a = (l + m); b = (l - m)

(l + m)² - (l - m)² = (l + m + l - m) (l + m – l + m)

⇒ (l + m)² - (l + m)² = 2l × 2m = 4lm

(vi) 9x²y² - 16 = (3xy)² - 4²

Using identity a² - b² = (a + b)(a - b)

Here a = 3xy; b = 4

(3xy)² - 4² = (3xy + 4) + (3xy - 4)

(vii) (x²- 2xy + y² ) = (x - y)² [using identity (a - b)² = a² - 2ab + b²]

(x - y)² - z² = (x - y + (z))(x - y - (z))

Using identity a² - b² = (a + b)(a - b)

Here a = (x - y)² ; b = z

(x - y)² - z² = (x - y + z) + (x - y - z)

(viii) 25a² - (4b² - 28bc + 49c²) ⇒ (5a)² - (2b – 7c )² [using identity (a - b)² = a² - 2ab + b²]

Using identity a² - b² = (a + b)(a - b)

Here a = 5a ; b = 4b – 7c

(5a)² - (2b – 7c )² = ((5a) + (2b – 7c)) + ((5a) – (2b - 7c))
(5a)² - (2b – 7c )² = (5a + 2b - 7c)(5a - 2b + 7c)

(i) ax² + bx

Here common factor is x

ax² + bx = x(ax + b)

(ii) 7p² + 21q²

Here common factor is 7

7p² + 21q² = 7(p² + 3q²)

(iii) 2x³ + 2xy² + 2xz²

Here common factor is 2x

2x³ + 2xy² + 2xz² = 2x(x² + y² + z²)

(iv) am² + bm² + bn² + an²

Here common factor is m² in first two terms and n² in the last two terms

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

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

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

On opening the bracket, we get

lm + l + m + 1

Here common factor is l in first two terms and 1 in the last two terms

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

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

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

In the brackets y + z is common,

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

(vii) 5y² - 20y – 8z + 2yz

Taking 5y common in first two pairs and 2z in the last two terms

5y(y - 4)-2z(y - 4) = (y - 4)(5y – 2z)

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

Taking 2a common in first two pairs and 1 in the last two terms

10ab + 4a + 5b + 2 = 2a(5b + 2) + 1(5b + 2)

10ab + 4a + 5b + 2 = (5b + 2) + (2a + 1)

(ix) 6xy – 4y + 6 – 9x

Taking 2y common in first two pairs and -3 in the last two terms

6xy – 4y + 6 – 9x = 2y(3x - 2) - 3(3x - 2)

6xy – 4y + 6 – 9x = (3x - 2) + (2y - 3)

(i) a⁴ - b⁴ = (a²)² - (b²)²

Using identity a² - b² = (a + b)(a - b)

Here a = a² ; b = b²

(a²)² - (b²)² = (a² + b²) (a² - b²)

Again Using identity a² - b² = (a + b)(a - b)

Here a = a ; b = b

a² - b² = (a + b)(a - b)

(a²)² - (b²)² = (a² + b²) (a + b)(a - b)

(ii) p⁴ - 81 = (p²)² - (3²)²

Using identity a² - b² = (a + b)(a - b)

Here a = p² ; b = 3²

(p²)² - (3²)² = (p² + 3²) (p² - 3²)

Again Using identity a² - b² = (a + b)(a - b)

Here a = p ; b = 3

p² - 3² = (p + 3)(p - 3)

(p²)² - (3²)² = (p² + 3²) (p + 3)(p - 3)

(iii) x⁴ - (y + z)⁴ = (x²)² - {(y + z)²}²

Using identity a² - b² = (a + b)(a - b)

Here a = x² ; b = (y + z)²

(x²)² - (y + z²)² = {x² + (y + z)²} {x² - (y + z)²}

Again Using identity a² - b² = (a + b)(a - b)

Here a = x ; b = y + z

x² - (y + z)² = {x + (y + z)}{(x – (y + z)}

(x²)² - (y + z²)² = {x² + (y + z)²} (x + y + z)(x – y - z)

(iv) x⁴ - (x - z)⁴ = (x²)² - {x - z)²}²

Using identity a² - b² = (a + b)(a - b)

Here a = x² ; b = (x - z)²

(x²)² - (x - z²)² = {x² + (x - z)²} {x² - (x - z)²}

Again Using identity a² - b² = (a + b)(a - b)

Here a = x ; b = x - z

x² - (x - z)² = {x + (x - z)}{(x – (x - z)}

(x²)² - (x - z²)² = {x² + (x - z)²} (x + x - z)(x – x + z)

(x²)² - (x - z²)² = {x² + (x - z)²} (2x - z)(z)

(x²)² - (x - z²)² = (2x² -2xz + z²) (2x - z)(z)

[using (a + b)² = a² + b² + 2ab]

(v) a⁴ - 2a²b² + b⁴ = (a²)² -2 × a× b + (b²)²

= (a²)² -2 × a× b + (b²)²

[using (a - b)² = a² -2ab + b²]

= (a² - b²)²

(i) P² + 6p + 8

Here middle term 6p can be written as 4p + 2p

P² + 6p + 8 = p² + 4p + 2p + 8

Taking p common in first two terms and 2 common in last two terms

P² + 6p + 8 = p² + 4p + 2p + 8 = p(p + 4) + 2(p + 4) = (p + 4)(p + 2)

(ii) q² - 10q + 21

Here middle term -10q can be written as -7q - 3q

q² - 10q + 21 = q² - 7q - 3q + 21

Taking q common in first two terms and 3 common in last two terms

q² - 10q + 21 = q² - 7q - 3q + 21 = q(q - 7) - 3(q - 7) = (q - 3)(q - 7)

(iii) P² + 6p - 16

Here middle term 6p can be written as 8p - 2p

P² + 6p - 16 = p² + 8p - 2p - 16

Taking p common in first two terms and 2 common in last two terms

P² + 6p + 8 = p² + 8p - 2p - 16 = p(p + 8) - 2(p + 8) = (p + 8)(p - 2)

(i) Factor of 28x⁴ = 2 × 2 × 7 × x × x × x × x

Factor of 56x = 2 × 2 × 2 7 × x

28x⁴ ÷ 56x =

(ii) Factor of -36y³ = 2 × 2 × 3 × 3 × y × y × y

Factor of 9y² = 3 × 3 × y × y

-36y³ ÷ 9y²₌

(iii) Factor of 66pq²r³ = 2 × 3 × 11 × p × q × q × r × r × r

Factor of 11qr² = 11 × q × r × r

66pq²r³ ÷ 11qr²₌

(iv) Factor of 34x³y³z³ =

Factor of 51xy²z³ =

₌

(v) Factor of 12a⁸b⁸ =

Factor of 6a⁶b⁴ =

₌

(i) (5x² -6x) ÷ 3x

Common factor of 5x² -6x is x

(ii)

Common factor of 3y⁸ - 4y⁶ + 5y⁴ is y⁴

(iii)

Common factor of 8(x³y²z² + x²y³z² + x²y²z³) is x²y²z²

(iv)

Common factor of x³ + 2x² + 3 is x

(v)

Common factor of p³q⁶ - p⁶q³ is p³q³

For carrying out divisions we will take common factors from numerator and denominator and then cancel out the common factors.

For Example:


In this example (a - b) as common in both numerator and denominator get cancelled out.


(i) (10x - 25)÷(5)

Common factor of 10x - 25 = 5

(ii) (10x - 25)÷(2x - 5)

Now, we can write the denominator as,


Now, clearly (2x - 5) is common in both numerator and denominator. Therefore, it get cancelled out.

(iii) 10y(6y + 21)÷5(2y + 7)

We can write the numerator as,



Now, Clearly 5(2y + 7) is common in both numerator and denominator and therefore it gets cancelled out.

Hence,

10y(6y + 21)÷5(2y + 7) = 6y

(iv) 9x² y² (3z - 24)÷27xy(z - 8)

Now, we can write the given division as,



From above we can see that, 27xy(z - 8) is common in both numerator and denominator and so it gets cancelled out.

Hence, 9x² y² (3z - 24)÷27xy(z - 8) = xy

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

(i)

Common factor of is 2x + 1

(ii)

Common factor of is y - 4

(iii) 52pqr(p + q)(q + r)(r + p) ÷ 104pq (q + r)(r + p)

(iv)

(v)

(i)

(ii)

(iii)

(iv)

(v)

[using identity a² - b² = (a + b) (a - b)]

(vi)

[using identity a² - b² = (a + b) (a - b)]

(vii)

[using identity a² - b² = (a + b) (a - b)]

Taking LHS

Hence RHS term is incorrect.

Therefore RHS will be 4x - 20

Therefore correct expression is:

Taking LHS

Hence RHS term is incorrect.

Therefore RHS will be

Therefore correct expression is:

2x + 3y = 5xy

Taking LHS

Hence RHS term is incorrect.

Therefore RHS will be

Therefore correct expression is:

x + 2x + 3x = 5x

Taking LHS

Hence RHS term is incorrect.

Therefore RHS will be

Therefore correct expression is:

Taking LHS

Hence RHS term is incorrect.

Therefore RHS will be

Therefore correct expression is:

Taking LHS

Hence RHS term is incorrect.

Therefore RHS will be

Therefore correct expression is:

Taking LHS

Hence RHS term is incorrect.

Therefore RHS will be

Therefore correct expression is:

(2 x)² + 5 x = 4 x² + 5 x = 9 x

Taking LHS

Hence RHS term is incorrect.

Therefore RHS will be

Therefore correct expression is:

Taking LHS

Hence RHS term is incorrect.

Therefore RHS will be

Therefore correct expression is:

(A) _{On substituting x = -3 in the given expression, we get}

= (-3)² + 5(-3) + 4 = 9 -15 + 4 = -2

Hence correct expression is:

= (-3)² + 5(-3) + 4 = 9 -15 + 4 = -2

(B)

On substituting x = -3 in the given expression, we get

= (-3)² - 5(-3) + 4 = 9 +15 + 4 = 28

Hence correct expression is:

= (-3)² - 5(-3) + 4 = 9 +15 + 4 = 28

(C) On substituting x = -3 in the given expression, we get

= (-3)² + 5(-3) = 9 -15 = -6

Hence correct expression is:

= (-3)² + 5(-3) = 9 -15 = -6

Taking LHS

Hence RHS term is incorrect.

Therefore RHS will be

Therefore correct expression is:

Taking LHS

Hence RHS term is incorrect.

Therefore RHS will be z² + 10z + 25

Therefore correct expression is:

(z + 5)² = z² + 10z + 25

Taking LHS

Hence RHS term is incorrect.

Therefore RHS will be

Therefore correct expression is:

Taking LHS

Hence RHS term is incorrect.

Therefore RHS will be

Therefore correct expression is:

Taking LHS

Hence RHS term is incorrect.

Therefore RHS will be

Therefore correct expression is:

Taking L.H.S

Hence R.H.S term is incorrect.

Therefore R.H.S will be 1

Therefore correct expression is:

Taking LHS

Hence RHS term is incorrect.

Therefore RHS will b

Therefore correct expression is:

Cross-multiplying, we get,

3x. 2 = 3x + 2

6x = 3x + 2

6x - 3x = 3x - 3x + 2

3x = 2

x = 2/3

Taking LHS

Hence RHS term is incorrect.

Therefore RHS will be

Therefore correct expression is:

Taking LHS

Hence RHS term is incorrect.

Therefore RHS will be

Therefore correct expression is: