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Answer 1

Questions :

1. Find the common factors of the given terms :

(a) 9xy² , 27x²y

Common factors - 9xy .

(b) 12ab³c² , -24 a³b³c², - 15 a²b²c² .

Common factors > 3 a b²c² .

2. Factorise the following

(a) 45 xyz² - 5 xy²z

> 5xyz ( 9z - y )

(b) 4x²y³ - 12 x³y + 18 x⁴y³

> 2x²y ( 2y² - 6 x + 9 x²y²)

3. Factorise the following .

(a) . a³ - 2a²b + 3ab² - 6 b³

> a² ( a - 2b ) + 3b²( a - 2b )

> ( a - 2b )( a² + 3b² )

(b) 6pq + 6 - 9p - 4q

> 6pq - 4q + 6 - 9p

> 2q ( 3p - 2 ) - 3( 3p - 2)

> ( 3p - 2)( 2q - 3)

(c) 25 ( x - y )³ - 15( x - y )² .

> 5( x - y)² [ 5( x - y ) - 3 ]

>5( x - y )²( 5x - 5y - 3)

4. Factorise using Identities -

(a) 36x² - 49b².

> [ 6x ]² - [ 7b ]²

> [ 6x + 7b ][ 6x - 7b ]

(b) 36y² + 36y + 9

> 9 [ 4y² + 4y + 1 ]

> 9 [ 4y² + 2y + 2y + 1 ]

> 9[ 2y( 2y + 1) + ( 2y + 1) ]

> 9( 2y + 1)².

(c) 9x⁴ - 6x²y² + y⁴ .

> [ 3x² ]² - 2[ 3x² ][ y² ] + [ y² ]²

> [ 3x² - y² ]² .

(D) a² - b² - 4ac + 4c²

>[ a² - 4ac + 4c² ] - b²

> [ a - 2c ]² - b²

> [ a + b - 2c ][ a - b - 2c ]

(e) z⁴ - 256

> [ z² ]² - [ 16 ]²

> [ z² + 16 ][ z² - 16]

> [ z² + 16 ][ z + 4 ][ z - 4 ]

(f) . ( x + y )² - ( x - y )² .

> [ x + y + x - y ][ x + y - x + y ]

> [ 2x ][ 2y ]

> 4xy .

(g) . 9/25 a² - 16/64 b².

> 9/25 a² - 1/4 b²

> [ ⅗ a ]² - [ ½ b ]² .

> [ ⅗ a + ½ b ][ ⅗ a - ½ b ] .

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Additional Information -

(a + b)² = a² + 2ab + b²

(a + b)² = (a - b)² + 4ab

(a - b)² = a² - 2ab + b²

(a - b)² = (a + b)² - 4ab

a² + b² = (a + b)² - 2ab

a² + b² = (a - b)² + 2ab

2 (a² + b²) = (a + b)² + (a - b)²

4ab = (a + b)² - (a - b)²

ab = {(a + b)/2}² - {(a-b)/2}²

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

(a + b)³ = a³ + 3a²b + 3ab² b³

(a + b)³ = a³ + b³ + 3ab(a + b)

(a - b)³ = a³ - 3a²b + 3ab² - b³

a³ + b³ = (a + b)( a² - ab + b² )

a³ + b³ = (a + b)³ - 3ab( a + b)

a³ - b³ = (a - b)( a² + ab + b²)

a³ - b³ = (a - b)³ + 3ab ( a - b )

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Answer 2

Question (1) :- Find the given factors of the given terms .

• a) 9xy², 27x²y .
• b) 12ab³c² , (-24)a³b³c², (-15)a²b²c² .

Solution (a):-

writing both expressions in expand form we get,

→ 9xy² = 3 * 3 * x * y * y

→ 27x²y = 3 * 3 * 3 * x * x * y

therefore,

→ common factors = 3 * 3 * x * y = 9xy (Ans.)

Solution (b):-

writing given expressions in expand form we get,

→ 12ab³c² = 3 * 2 * 2 * a * b * b * b * c * c

→ (-24)a³b³c² = (-1) * 3 * 2 * 2 * 2 * a * a * a * b * b * b * c * c

→ (-15)a²b²c² = (-1) * 5 * 3 * a * a * b * b * c * c

therefore,

→ common factors = 3 * a * b * b * c * c = 3ab²c² (Ans.)

________________

Question (2) , (3):- Factorise the following :-

• a) 45xyz² - 5xy²z
• b) 4x²y³ - 12x³y + 18x⁴y³
• c) a³ - 2a²b + 3ab² - 6b³
• d) 6pq + 6 - 9p - 4q
• e) 25(x - y)³ - 15(x - y)²

Solution (a) :-

→ 45xyz² - 5xy²z

→ 5 * 9 * x * y * z * z - 5 * x * y * y * z

taking common , we get,

→ 5xyz(9z - y) (Ans.)

Solution (b) :-

→ 4x²y³ - 12x³y + 18x⁴y³

→ 2 * 2 * x * x * y * y * y - 2 * 2 * 3 * x * x * x * y + 2 * 3 * 3 * x * x * x * x * y * y * y

taking common , we get,

→ 2*x*x*y(2*y*y - 2*3*x + 3*3*x*x*y*y)

→ 2x²y(2y² - 6x + 9x²y²) (Ans.)

Solution (c) :-

→ a³ - 2a²b + 3ab² - 6b³

→ a²(a - 2b) + 3b²(a - 2b)

taking (a - 2b) common now,

→ (a - 2b)(a² + 3b²) (Ans.)

Solution (d) :-

→ 6pq + 6 - 9p - 4q

re - arranging the like terms,

→ 6pq - 4q + 6 - 9p

taking common now,

→ 2q(3p - 2) - 3(3p - 2)

taking (3p - 2) as common now,

→ (3p - 2)(2q - 3) (Ans.)

Solution (e) :-

→ 25(x - y)³ - 15(x - y)² .

taking 5 common first,

→ 5[5(x - y)³ - 3(x - y)²]

taking (x - y)² common now,

→ 5(x - y)²[ 5(x - y) - 3]

→ 5(x - y)²(5x - 5y - 3) (Ans.)

Question (4) :- Factorise using identities :-

• a) 36x² - 49b²
• b) 36y² + 36y + 9
• c) 9x⁴ - 6x²y² + y⁴
• d) a² - b² - 4ac + 4c²
• e) z⁴ - 256
• f) (x + y)² - (x - y)²
• g) (9/25)a² - (16/64)b²

Solution (a) :-

→ 36x² - 49b²

→ (6x)² - (7b)²

now we know that, (a)² - (b)² = (a - b)(a + b)

so,

→ (6x - 7b)(6x + 7b) (Ans.)

Solution (b) :-

→ 36y² + 36y + 9

taking 9 common first ,

→ 9(4y² + 4y + 1)

→ 9{(2y)² + 2 * 2y * 1 + (1)²}

now, we know that, (a² + 2ab + b²) = (a + b)² .

so,

→ 9(2y + 1)² (Ans.)

Solution (c) :-

→ 9x⁴ - 6x²y² + y⁴

→ (3x²)² - 2 * (3x²) * (y²) + (y²)²

now, we know that, (a² - 2ab + b²) = (a - b)² .

so,

→ (3x² - y²)² (Ans.)

Solution (d) :-

→ a² - b² - 4ac + 4c²

→ (a² - 4ac + 4c²) - b²

→ [(a)² - 2 * (2c) * a + (2c)²] - b²

now, we know that, (a² - 2ab + b²) = (a - b)² .

so,

→ (a - 2c)² - b²

→ (a - 2c)² - (b)²

now we know that, (a)² - (b)² = (a - b)(a + b)

so,

→ (a - 2c - b)(a - 2c + b)

→ (a - b - 2c)(a + b - 2c) (Ans.)

Solution (e) :-

→ z⁴ - 256

→ (z)⁴ - (4)⁴

→ {(z²)² - (4²)²}

now we know that, (a)² - (b)² = (a - b)(a + b)

so,

→ (z² - 4²)(z² + 4²)

again, using , (a)² - (b)² = (a - b)(a + b)

→ (z - 4)(z + 4)(z² + 16) (Ans.)

Solution (f) :-

→ (x + y)² - (x - y)²

using , (a)² - (b)² = (a - b)(a + b)

→ {(x + y) - (x - y)}{(x + y) + (x - y)}

→ (x + y - x + y)(x + y + x - y)

→ 2y * 2x

→ 4xy (Ans.)

Solution (g) :-

→ (9/25)a² - (16/64)b²

→ [{(3/5)a}² - {(1/2)b}²]

using , (a)² - (b)² = (a - b)(a + b)

→ {(3a/5) - (b/2)}{(3a/5) + (b/2)}

→ {(6a - 5b)/10}{(6a + 5b)/10}

→ (1/10)(6a - 5b)(6a + 5b) (Ans.)

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इस एक्स इक्वल टू रूट 3 प्लस वन डिवाइडेड बाय टू फाइंड द वैल्यू ऑफ एक्स क्यूब प्लस टू एक्स स्क्वायर माइनस 8 एक्स प्लस 7

https://brainly.in/question/20858452

if a^2+ab+b^2=25

b^2+bc+c^2=49

c^2+ca+a^2=64

Then, find the value of

(a+b+c)² - 100 = ...

https://brainly.in/question/16231132