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(d) (y⁴ - 8)(y⁴ + 8)(y⁸ + 1)

y¹⁶ - 63y⁸ - 64

= y¹⁶ - 64y⁸ + y⁸ - 64

= y⁸ (y⁸ - 64) + 1(y⁸ - 64)

= (y⁸ - 64)(y⁸ + 1)

= (y⁴ - 8)(y⁴ + 8)(y⁸ + 1)

The process of writing a given algebraic expression as a product of two or more factors is called factorisation. 

For example, ax + bx + x² can be factorised as x (a + b + x )

\(16(2x - y)^2 - 24(4x^2 - y^2) + 9(2x + y)^2\)

= \((4(2x-y)^2-2\times4\times3x(2x-y)\times(2x+y)+(3(2x+y))^2\)

= \([4(2x-y)-3(2x+y)]^2\)        [∴ a² - 2ab + b² = (a - b)²]

= [ 8x - 4y - 6x - 3y]²

= (2x - 7y)(2x - 7y).

Correct option is   D) x² + 2 +\(\frac{1}{x^2}\)

(i) 2a + b, a – b 

= 2a + b – a + b

= (2a – a) + (b + b)

= a + 2b

ii) (x + 2y + z) – ( – x – y – 3z) 

= x + + z + x + y + 3z

=(x + x) + (2y + y) +(z + 3z)

= 2x + 3y + 4z

iii) (3a² – 8ab – 2b²) – (3a² – 4ab + 6b²) 

= 3a² – 8ab – 2b² – 3a² +4ab – 6b²

= (3a² – 3a²) + ( – 8ab + 4ab) + (2b² – 6b²)

= 0 – 4ab – 8b²

= – 4ab – 8b²

iv) (4pq – 6p² – 2q²) – (9p²) 

= 4pq – 6p² – 2q² – 9p²

= 4pq – 15p² – 2q²

v) (7 – 2x – 3x²) – (2x² – 5x – 3)

=7 – 2x – 3x² – 2x² + 5x + 3

= ( – 3x² – 2x²) + ( – 2x + 5x) + (7 + 3)

= – 5x² + 3x + 10

vi) (5x² – 3xy – 7y²) – (3x² – xy – 2y²) 

= 5x²– 3xy – 7y² – 3x² + xy + 2y²

= (5x² – 3x²) + ( – 3xy + xy) + ( – 7y² + 2y²)

= 2x² – 2xy – 5y²

viii) (6m³ + 4m² + 7m – 3) – (3m³ + 4)

= 6m³ + 4m² + 7m – 3 – 3m3 – 4

= (6m³ – 3m³) + 4m² + 7m + (- 3 – 4)

= 3m³ + 4m² + 7m – 7

\((x^2 + y^2 -z^2)^2 - 4x^2y^2\) = \((x^2 + y^2 -z^2)^2 - (2xy)^2\)

= (x² + y² + z² + 2xy)(x² + y² -z² -2xy)      [∴ a² - b² = (a-b)(a+b)]

= (x² + 2xy + y² - z²)(x² - 2xy + y² - z²)

= [(x + y)² -z²] [(x + y)² -z²]

= (x + y + z)(x + y - z)(x - y + z)(x - y - z).

(b) (x + 4√3)(x + √3)

x² + 5√3x + 12

= x² + 4√3x + √3(x + 4√3)

= x(x + 4√3) + √3(x + 4√3)

= (x + 4√3)(x + √3)

A) 49x² y² – 1 

Correct option is (A) 49x²y² – 1

(7xy – 1) (7xy + 1) \(=(7xy)^2-1^2\)

= \(49x^2y^2–1\)

By subtracting x – 2y + 3z from 3x + 5y – 7 we can get the required expression,

Required expression = (x – 2y + 3z) – (3x + 5y – 7)

= x – 2y + 3z – 3x – 5y + 7

Collecting positive and negative like terms together, we get

= x – 3x – 2y + 5y + 3z + 7

= –2x – 7y + 3z + 7

(b) (2x² + x + 1) (2x² – x + 1)

4x⁴ + 3x² + 1

= 4x⁴ + 4x² + 1 - x²

= (2x² + 1) - x²

Now, factorise using x² - y² =(x + y)(x - y)

Correct option is  D) Positive

Let ‘E’ be the required expression. Then, we have

x² – xy + y²– x + y + 3 – E = – x² + 3y² – 4xy + 1

Therefore, E = (x² – xy + y²– x + y + 3) – (- x² + 3y² – 4xy + 1)

= x² – xy + y²– x + y + 3 + x² – 3y² + 4xy – 1

Collecting positive and negative like terms together, we get

= x² + x²– xy + 4xy + y²– 3y² – x + y + 3 – 1

= 2x²+ 3xy- 2y²– x + y + 2

(c) (x - y)(x - 1)

x² - xy + y - x

= x(x - y) - 1(x - y)

= (x - y)(x - 1)

(d) (x² - 2x + 3)(x² + 3x + 3) 

x⁴ + 5x² + 9

= x⁴ + 6x² + 9 - x²         (Note the step)

= (x² + 3)² - x²

Now, factorise using x² - y² = (x + y)(x - y)

Correct option is  A) 84x⁶y⁶ 

Correct option is (A) 84x⁶y⁶

\(3x^2y\times4xy^2\times7x^3y^3\) \(=(3\times4\times7)(x^2\times x \times x^3)(y\times y^2\times y^3)\)

= \(84x^6y^6\)

(c) (a - 2d + b - 2c)(a - 2d - b + 2c)

a² - b² - 4c² + 4d² - 4(ad - bc)

= a² - b² - 4c² + 4d² - 4ad + 4bc

= (a² - 4ad + 4d²) - (b² - 4bc + 4c²)

= (a - 2d)² - (b - 2c)²

= (a - 2d + b - 2c)(a - 2d - b + 2c)

(b) (x⁴ + 5)(x⁴ - 6)

x⁸ - x⁴ - 30

= x⁸ - 6x⁴ + 5x⁴ - 30

= x⁴(x⁴ - 6) + 5(x⁴ - 6)

= (x⁴ + 5)(x⁴ - 6)

(b) 3m (m + 2)(m - 2)(m² + 4)

3m⁵ - 48m

= 3m (m⁴ - 16)

= 3m[(m²)² - 4²]

= 3m (m² - 4)(m² + 4)

= 3m (m + 2)(m - 2)(m² + 4)

(a) (x² + y² +3xy)(x² + y² - 3xy)

x⁴ - 7x²y² + y⁴

= x⁴ + 2x²y² + y⁴ - 9x²y²

= (x² + y²)² - (3xy)²

= (x² + y² +3xy)(x² + y² - 3xy)

(-1/27)a²b² × (-9/2)a³bc²

The coefficient of the product of two monomials is equal to the product of their coefficients.

The variable part in the product of two monomials is equal to the product of the variables in the given monomials.

Then,

= [(-1/27) × (-9/2)] × (a² × a³) × (b² × b) × c²

= [(-1×-9)/ (27×2)] × (a²⁺³) × (b²⁺¹) × c² … [∵ a^m × aⁿ = a^m+n]

= [(-1×-1)/ (3×2)] × (a⁵) × (b³) × c²

= [1/6]a⁵b³c²

(c) (x² + 2x +2)(x² - 2x + 2)

x⁴ + 4

= x⁴ + 4 + 4x² - 4x²            (Note the step)

= (x² + 2)² - (2x)²

= (x² + 2x +2)(x² - 2x + 2)

(-3/4)ab³ × (-2/3)a²b⁴

The coefficient of the product of two monomials is equal to the product of their coefficients.

The variable part in the product of two monomials is equal to the product of the variables in the given monomials.

Then,

= [(-3/4) × (-2/3)] × (a × a²) × (b³ × b⁴)

= [(-3×-2)/ (4×3)] × (a¹⁺²) × (b³⁺⁴) … [∵ a^m × aⁿ = a^m+n]

= [(-1×-1)/ (2×1)] × (a³) × (b⁷)

= [1/2]a³b⁷

(c) (1 + a - b)(a - a + b)

1 + 2ab – (a² + b²)

= 1 + 2ab - a² - b²

= 1 - (a² - 2ab + b²)

= 1 - (a - b)²

= (1 + a - b)(a - a + b)

-6x³ × 5x²

The coefficient of the product of two monomials is equal to the product of their coefficients.

The variable part in the product of two monomials is equal to the product of the variables in the given monomials.

Then,

= (-6 × 5) × (x³ × x²)

= (-30) × (x³⁺²) … [∵ a^m × aⁿ = a^m+n]

= (-30) × (x⁵)

= -30 a⁵

(b) (a + b - c) (a - b + c + 1)

a² - b² - c² + 2bc + a + b - c

= [a² - (b² + c² - 2bc)] + (a + b - c)

= [a² - (b - c)²] + (a + b + c)

= [(a + b - c)(a - b + c) + (a + b - c)]

= (a + b - c) (a - b + c + 1)

3a² × 8a⁴

The coefficient of the product of two monomials is equal to the product of their coefficients.

The variable part in the product of two monomials is equal to the product of the variables in the given monomials.

Then,

= (3 × 8) × (a² × a⁴)

= (24) × (a²⁺⁴) … [∵ a^m × aⁿ = a^m+n]

= (24) × (a⁶)

= 24 a⁶⁺

By using horizontal method, 

We have; 

= (9x² – x + 15) × (x² – 3) 

= x²(9x² – x +15) – 3(9x² – x + 15) 

= 9x⁴ – x³ + 15x² – 27x² + 3x – 45 

= 9x⁴ - x³ – 12x² + 3x – 45

(c) 16 (x + 9y)² 

49 (2x + 3y)²  – 70 ( 4x²  – 9y²) + 25 (2x – 3y)²

= (7(2x + 3y))² - 2 x 7(2x + 3y) x 5(2x - 3y) + (5(2x - 3y))²

= [7(2x + 3y) - 5(2x - 3y)]²

= [14x + 21y - 10x + 15y]²

= (4x + 36y)2 = (4(x + 9y))² = 16(x + 9y)²

Given (x³-y³) × (x² + y²)

To find the product of given expression we have to use horizontal method.

In that we have to multiply each term of one expression with each term of another

Expression so by multiplying we get,

(x³-y³) × (x² + y²)

⇒ x³ (x² + y²) -y³(x² + y²)

⇒ x⁵ + x³y² – x²y³ – y⁵

(c) (3a - √5) (3a - √5)

9a2 - 6√5a + 5

= (3a)² - 2 x 3a x √5 + (√5)²

= (3a - √5)² = (3a - √5) (3a - √5)

By using horizontal method,

We have;

= (x² – 5x + 8) × (x² + 2) 

= x²(x² – 5x + 8) + 2(x² – 5x + 8) 

= x⁴ – 5x³ + 8x² + 2x² – 10x + 16 

= x⁴ – 5x³ + 10x² – 10x +16

By using horizontal method, 

We have; 

= (3x² + 5x – 9) × (3x – 5) 

= 3x(3x² + 5x – 9) – 5(3x² + 5x – 9) 

= 9x³ + 15x² – 27x – 15x² - 25x + 45 

By arranging the expression in the form of descending powers of x, 

We get; 

= 9x³ + 15x² – 15x² – 27x – 25x + 45 

= 9x³ – 52x + 45

By using horizontal method, 

We have; 

= (x² – a²) × (x – a) 

= x²(x – a) – a²(x – a) 

= x³ – ax² – a²x + a³

(c) (a + bc) (ac + b) 

ab (c²  +1) + c (a²  + b²)

= abc² + ab + ca² + cb²

= abc² + ca² + ab + cb²

= ac(bc + a) + b(a + bc)

= (a + bc)(ac + b)

By using horizontal method, 

We have; 

= (x³ – 5x² + 3x + 1) × (x² – 3) 

= x²(x³ – 5x² + 3x + 1) – 3(x³ – 5x² + 3x +1) 

= x⁵ – 5x⁴ + 3x³ + x² – 3x³ + 15x² – 9x – 3 

By arranging the expression in the form of descending powers of x, 

We get; 

= x⁵ – 5x⁴ + 3x³ – 3x³ + x² + 15x² – 9x – 3 

= x⁵ – 5x⁴ +16x² – 9x – 3

By using horizontal method, 

We have; 

= (x² – 5x + 8) × (x² + 2x – 3) 

= x²(x² – 5x + 8) + 2x(x² – 5x + 8) – 3(x² – 5x + 8) 

= x⁴ – 5x³ + 8x² + 2x³ – 10x² + 16x – 3x² + 15x – 24 

By arranging the expression in the form of descending powers of x, 

We get; 

= x⁴ – 3x³ – 5x² + 31x – 24

Given (2x²-5y²) × (x²+3y²)

To find the product of given expression we have to use horizontal method.

In that we have to multiply each term of one expression with each term of another

Expression so by multiplying we get,

(2x²-5y²) × (x²+3y²)

⇒ 2x² (x²+3y²) – 5y²(x²+3y²)

⇒ 2x⁴ + x²y² – 15y⁴

Given that the formula for simple interest (I) = \(\frac {PTR}{100}\)

P = Rs.900, T = 2years, R = 5%

∴ The required simple interest (I) =  \(\frac {PTR}{100}\) = \(\frac {900\times2\times5}{100}\)

= 9 x 2 x 5 

= Rs. 90

Given expression is 3m – n 

If m = 2, n = – 1, then 

3m – n = 3(2) -(-1) = 6 + 1 = 7 

∴ If m = 2, n = – 1, then 3m – n = 7

By using horizontal method, 

We have; 

= (3x + 2y – 4) × (x – y + 2) 

= x(3x + 2y – 4) – y(3x + 2y – 4) + 2(3x + 2y – 4) 

= 3x² + 2xy – 4x – 3xy – 2y² + 4y + 6x + 4y – 8 

By arranging the expression in the form of descending powers of x, 

We get; 

= 3x² – 4x + 6x + 2xy – 3xy – 2y² + 4y + 4y – 8 

= 3x² + 2x – xy – 2y² + 8y – 8

(1.25)³  – 2.25 (1.25)²  + 3.75 (0.75)²  – (0.75)³

= (1.25)³ – 3 × 0.75 × (1.25)² + 3 × 1.25 × (0.75)² – (0.75)³

= (1.25 – 0.75)³ = (0.5)³ = \(\big(\frac12\big)^3=\frac18.\)        [∴ a³ - 3a²b + 3ab² - b³ = (a - b)³]   

By using horizontal method,

We have;

= (x² – y²) × (x + 2y) 

= x² (x + 2y) – y²(x + 2y) 

= x³ + 2x²y – xy² – 2y³

Given (3p²+q²) × (x²-y²)

To find the product of given expression we have to use horizontal method.

In that we have to multiply each term of one expression with each term of another

Expression so by multiplying we get,

(3p²+q²) × (x²-y²)

⇒ 3p² (x²-y²) + q² (x²-y²)

⇒ 6p4 – 7p²q² – 3q⁴

False

An equation is a statement of equality between two quantities or algebraic expressions. Most algebraic equations are TRUE when certain values are substituted for the variable (such as x) and are FALSE for all other values. The values that make equations TRUE are called "solutions". So, an equation is not necessarily true for all values of its variables.

Formula for area of a rectangle (A) = l × b

If l = 9cm, b = 6cm

then the area of the rectangle = l × b

= 9 × 6

= 54 cm²

Given expression is mn – 2 

If m = 2, n = – 1, then 

mn – 2 = 2 × (-1)-2 

= – 2 – 2 = – 4 

∴ If m = 2, n = – 1, then mn – 2 = – 4

Given that,

x + y = 4 and xy = 2

We take the equation: x + y = 4 and on squaring both sides, we get

(x + y)² = 4²

x² + y² + 2xy = 16

x² + y² + 2(2) = 16 (Because xy = 2 is given)

x² + y² + 4 = 16

x² + y² = 16 – 4

x² + y² = 12

Therefore, the value of x² + y² is 12

Given (x²-y²) × (x + 2y)

To find the product of given expression we have to use horizontal method.

In that we have to multiply each term of one expression with each term of another

Expression so by multiplying we get,

(x²-y²) × (x + 2y)

⇒ x²(x + 2y) – y²(x + 2y)

⇒ x³ + 2x²y – xy² – 2y³

The given algebraic expression 

= 5x² – 4 – 3x² + 6x + 8 + 5x – 13.

= (5x² – 3x²) (6x + 5x) + (-4 – 13)

= 2x² + 11x – 17

When x = – 2. then the value of 2x²+ 11x – 17 

= 2(- 2)² + 11(-2) – 17

= 2 × 4 – 22 – 17

= 8 – 39

= – 31

x² + \(\frac{1}{x\times x}\) = 18

Adding 2 on both sides, we get

x² + \(\frac{1}{x\times x}\) + 2 = 18 + 2

x² + \(\frac{1}{x\times x}\) + 2 × x × \(\frac{1}{x}\) = 20

(x + \(\frac{1}{x}\) )² = 20

x + \(\frac{1}{x}\) = \(2\sqrt{5}\)

Given that,

x² + \(\frac{1}{x\times x}\) = 18

Subtracting 2 from both sides, we get

x² +\(\frac{1}{x\times x}\) - 2 × x × \(\frac{1}{x}\) = 18 – 2

(x - \(\frac{1}{x}\))² = 16

x - \(\frac{1}{x}\) = 4