Key Concept: Basic Concepts

d) 25

[Solution Description]

The exponent $5^2$ means 5 multiplied by itself. So, $5^2 = 5 \times 5 = 25$.

Key Concept: Factors of natural numbers, Factors of algebraic expressions

a) $6a^2 b^2 (2a - 3b)$

[Solution Description]

To find the prime factor form of $12a^3 b^2 - 18a^2 b^3$, we first factorize each term. The first term $12a^3 b^2$ can be written as $2 \times 2 \times 3 \times a \times a \times b \times b$. The second term $18a^2 b^3$ can be written as $2 \times 3 \times 3 \times a \times a \times b \times b \times b$. The common factors are $2 \times 3 \times a \times a \times b \times b$. Factoring out the common factors, we get $6a^2 b^2 (2a - 3b)$. This is the prime factor form of the expression.

Key Concept: Basic Concepts

c) 8

[Solution Description]

The square root of a number is a value that, when multiplied by itself, gives the original number. For 64, the square root is 8 because $8 \times 8 = 64$.

Key Concept: Factors of Algebraic Expressions, Prime Factors

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that the irreducible form of $10xy$ is $2 \times 5 \times x \times y$. Let's verify this:

The expression $10xy$ can be broken down into its prime factors as follows:

$10xy = 2 \times 5 \times x \times y$.

None of these factors can be further expressed as a product of other factors, making this the irreducible form. Thus, the assertion is true.

The reason explains what an irreducible form is in algebraic expressions. It correctly states that an irreducible form is a product of factors that cannot be further expressed as a product of other factors. This definition aligns with the assertion, as $2 \times 5 \times x \times y$ is indeed the irreducible form of $10xy$ because none of these factors can be broken down further.

Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Factors of Algebraic Expressions

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that the irreducible form of $10xy$ is $2 \times 5 \times x \times y$. This is true because $10xy$ can be factorized into its irreducible components as $2 \times 5 \times x \times y$, where 2, 5, x, and y are all irreducible factors. The reason states that an irreducible form of an algebraic expression cannot be further expressed as a product of factors. This is also true and correctly explains why the factors 2, 5, x, and y are considered irreducible in this context. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Factors of natural numbers

c) 7

[Solution Description]

To determine which number is not a factor of 120, we first list all the factors of 120. The factors of 120 are:

1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, and 120.

From the given options, we check which number is not in this list.

The number 7 is not a factor of 120, as it does not divide 120 evenly.

Key Concept: Importance

b) It provides context and outlines expectations

[Solution Description] An introduction is important because it sets the stage for the audience, providing context and outlining what to expect, making it easier for them to follow the main content.

Key Concept: Prime factors, Factors of natural numbers

a) 210

[Solution Description]

First, we find the prime factorization of 210. We start by dividing 210 by the smallest prime number, which is 2: $210 ÷ 2 = 105$. Next, we divide 105 by 3: $105 ÷ 3 = 35$. Then, we divide 35 by 5: $35 ÷ 5 = 7$. Finally, 7 is a prime number itself. Thus, the prime factorization of 210 is $2 × 3 × 5 × 7$. The product of these prime factors is $2 × 3 × 5 × 7 = 210$.

Key Concept: Factors of algebraic expressions, Irreducible form

a) $3xy(2x + 3y)$

[Solution Description]

To find the irreducible form of $6x^2y + 9xy^2$, we factor out the greatest common divisor (GCD) of the coefficients and the variables. The GCD of 6 and 9 is 3. The GCD of $x^2y$ and $xy^2$ is $xy$. Factoring out $3xy$ from both terms gives $3xy(2x + 3y)$. This is the irreducible form.

Key Concept: Factors of natural numbers, Factors of algebraic expressions

a) Both have the same prime factors.

[Solution Description]

The natural number N has prime factors $2^3$, $3^2$, and $5$. This means N can be written as $2 \times 2 \times 2 \times 3 \times 3 \times 5$.

For the algebraic expression $6a^2b$, factorise the coefficient 6 into its prime factors: $6 = 2 \times 3$. The variable part $a^2b$ can be written as $a \times a \times b$. Therefore, the irreducible factors of $6a^2b$ are $2 \times 3 \times a \times a \times b$.

Comparing both factorisations, we see that they share the common prime factors $2 \times 3$.

Key Concept: Factors of Natural Numbers and Algebraic Expressions

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

Key Concept: Factors of natural numbers

c) 18

[Solution Description]

To determine which option is not a factor of 120, we first find all the factors of 120.

Step 1: The prime factorization of 120 is $2^3 \times 3 \times 5$.

Step 2: The factors of 120 are all the numbers that can be formed by multiplying these prime factors in different combinations. They include 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, and 120.

From the options provided, 18 is not a factor of 120 because it cannot be formed by any combination of the prime factors of 120.

Key Concept: Prime Factors

d) $2^2 \times 3 \times 7$

[Solution Description]

To find the prime factor form of 84, we divide it by the smallest prime numbers until we get only prime factors.

84 divided by 2 gives 42.

42 divided by 2 gives 21.

21 divided by 3 gives 7.

7 is a prime number.

So, the prime factorization of 84 is $2 \times 2 \times 3 \times 7$, which can be written as $2^2 \times 3 \times 7$.

Key Concept: Factors of Natural Numbers

c) 8

[Solution Description]

To find the even factors of 60, we first write its prime factorization, which is $2^2 \times 3^1 \times 5^1$.

Even factors must include at least one 2.

For each prime factor, add 1 to its exponent and multiply them together but excluding the case where 2 is not included.

So, for 2, we use the exponent 2 directly, and for 3 and 5, we take $(1 + 1) \times (1 + 1)$.

Multiplying these, $2 \times 2 \times 2 = 8$.

Therefore, 60 has 8 even factors.

Key Concept: Factors of Natural Numbers, Prime Factors

a) 30

[Solution Description]

To determine which number has the least number of distinct prime factors, we will find the prime factors of each number:

- For 30: $30 = 2 \times 3 \times 5$ (3 distinct prime factors)

- For 42: $42 = 2 \times 3 \times 7$ (3 distinct prime factors)

- For 70: $70 = 2 \times 5 \times 7$ (3 distinct prime factors)

- For 90: $90 = 2 \times 3 \times 3 \times 5$ (3 distinct prime factors)

All numbers have 3 distinct prime factors, so none has fewer than the others.

Key Concept: Factors of natural numbers

c) $2^2 \times 3 \times 7$

[Solution Description]

To find the prime factor form of 84, we start by dividing it by the smallest prime number, which is 2. Dividing 84 by 2 gives 42. Next, divide 42 by 2 to get 21. Now, 21 is not divisible by 2, so we move on to the next prime number, which is 3. Dividing 21 by 3 gives 7. Finally, 7 is a prime number itself. Therefore, the prime factor form of 84 is $2 \times 2 \times 3 \times 7$ or $2^2 \times 3 \times 7$.

Key Concept: Finding Factors

c) 12

[Solution Description] To find a common factor of 24 and 36, we list the factors of each number. The factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24. The factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36. The common factors are 1, 2, 3, 4, 6, and 12.

Key Concept: Finding Factors

d) 18

[Solution Description] The largest factor of any number is the number itself. Therefore, the largest factor of 18 is 18.

Key Concept: Prime factorisation, Factors of natural numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

Let us verify the assertion and reason step by step. First, we factorise 60 into its prime factors. We know that 60 can be divided by 2, which gives 30. Dividing 30 by 2 again gives 15. Now, 15 is divisible by 3, giving 5. Finally, 5 is a prime number. Thus, the prime factorisation of 60 is $2^2 \times 3 \times 5$, which matches the assertion. The reason states that a number's prime factorisation includes all its prime factors raised to their respective powers, which is true as per the fundamental theorem of arithmetic. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Prime Factorisation

b) $2 × 2 × 3 × 7$

[Solution Description]

To find the prime factor form of 84, we start by dividing it by the smallest prime number, which is 2:

$84 ÷ 2 = 42$. Next, divide 42 by 2:

$42 ÷ 2 = 21$.

Now, 21 is not divisible by 2, so we move to the next prime number, which is 3:

$21 ÷ 3 = 7$.

Finally, 7 is a prime number itself. Thus, the prime factors of 84 are $2 × 2 × 3 × 7$, or $2^2 × 3 × 7$.

Key Concept: Prime Factorisation

c) $2 × 2 × 2 × 3 × 5$

[Solution Description]

To find the prime factor form of 120, we start by dividing it by the smallest prime number, which is 2:

$120 ÷ 2 = 60$. Next, divide 60 by 2:

$60 ÷ 2 = 30$.

Divide 30 by 2 again:

$30 ÷ 2 = 15$.

Now, 15 is not divisible by 2, so we move to the next prime number, which is 3:

$15 ÷ 3 = 5$.

Finally, 5 is a prime number itself. Thus, the prime factors of 120 are $2 × 2 × 2 × 3 × 5$, or $2^3 × 3 × 5$.

Key Concept: Factors of Algebraic Expressions

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The term $5xy$ is made up of the factors $5$, $x$, and $y$, which cannot be further broken down into other factors. These factors are called irreducible factors. Therefore, the Assertion is true. The Reason states that in algebraic expressions, terms are formed as products of irreducible factors, which is also true and correctly explains why $5xy$ can be expressed as $5 \times x \times y$.

Key Concept: Factors of Algebraic Expressions

c) $x - 2$

[Solution Description]

The expression $x^2 - 4$ can be factored using the difference of squares formula, which is $a^2 - b^2 = (a - b)(a + b)$. Applying this formula, we get:

$x^2 - 4 = (x - 2)(x + 2)$.

Therefore, the factors are $(x - 2)$ and $(x + 2)$.

Key Concept: Factorization

a) (2y + 3)(2y - 3)

[Solution Description]

The expression $4y^2 - 9$ is a difference of squares, which can be factorized using the formula $a^2 - b^2 = (a - b)(a + b)$. Here, $a = 2y$ and $b = 3$. Therefore, the factorized form is $(2y - 3)(2y + 3)$.

Key Concept: Regrouping Method

a) $(x + y)(a + b)$

[Solution Description]

To factorise $ax + ay + bx + by$, we use the regrouping method. We group the terms such that each group has a common factor. Here, we can group $ax + ay$ and $bx + by$. In the first group, $ax + ay$, the common factor is $a$, so we factor it out: $a(x + y)$. In the second group, $bx + by$, the common factor is $b$, so we factor it out: $b(x + y)$. Now, the expression becomes $a(x + y) + b(x + y)$. Next, we factor out the common factor $(x + y)$, which gives $(x + y)(a + b)$. Therefore, the factorised form of $ax + ay + bx + by$ is $(x + y)(a + b)$.

Key Concept: Breaking expressions into irreducible factors

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The expression $x^2 - 4$ is indeed a difference of squares, and it can be factored into $(x - 2)(x + 2)$. Both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Factors of Algebraic Expressions

b) $2^3 \times a \times b^2$

[Solution Description]

The term $8ab^2$ can be expressed as a product of its irreducible factors.

$8ab^2 = 8 \times a \times b^2$

The number 8 can be further broken down into its prime factors:

$8 = 2 \times 2 \times 2 = 2^3$

Therefore, the irreducible form of $8ab^2$ is $2^3 \times a \times b^2$.

Key Concept: Factorisation

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The expression $x^2 - 4$ can be written as $x^2 - 2^2$, which is a difference of squares. The difference of squares formula states that any expression of the form $a^2 - b^2$ can be factorised as $(a + b)(a - b)$. Therefore, applying this formula to $x^2 - 4$, we get $(x + 2)(x - 2)$. Thus, the Assertion is true, and the Reason correctly explains why the factorisation holds.

Key Concept: Examples of algebraic expressions in factorised form

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The expression $x^2 - 4$ is indeed a difference of squares, which can be factorised using the formula $a^2 - b^2 = (a + b)(a - b)$. Here, $a = x$ and $b = 2$, so the factorised form is $(x + 2)(x - 2)$. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Prime Factors

b) $2 \times 2 \times 3 \times 7$

[Solution Description]

To find the prime factors of 84, we start by dividing it by the smallest prime numbers. Step 1: Divide 84 by 2, which gives 42. Step 2: Divide 42 by 2, which gives 21. Step 3: Divide 21 by 3, which gives 7. Step 4: Divide 7 by 7, which gives 1. Therefore, the prime factors of 84 are $2 \times 2 \times 3 \times 7$.

Key Concept: Irreducible Factors

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that the expression $5xy$ is in its irreducible form. This is true because the factors $5$, $x$, and $y$ cannot be broken down further into a product of simpler factors. The reason correctly explains this by stating that the factors $5$, $x$, and $y$ of $5xy$ cannot be expressed as a product of factors. Hence, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Irreducible factors, Prime factors

d) Assertion is false, but Reason is true.

[Solution Description]

The assertion states that the expression $6xy$ can be written as $6 \times x \times y$ in its irreducible form.

According to the concept of irreducible factors in algebraic expressions, a factor is irreducible if it cannot be further expressed as a product of factors.

Here, 6 can be further factorized into $2 \times 3$, making it reducible. Therefore, the assertion is false.

The reason states that the factors 6, x, and y cannot be further expressed as a product of factors. However, 6 can be expressed as $2 \times 3$, so the reason is also false.

Hence, both the assertion and reason are incorrect.

Key Concept: Prime factors, Irreducible factors

b) 5

[Solution Description]

First, we find the prime factors of 180. We start by dividing 180 by 2, which gives 90. Then, we divide 90 by 2, which gives 45. Next, we divide 45 by 3, which gives 15. Then, we divide 15 by 3, which gives 5. Finally, we divide 5 by 5, which gives 1. So, the prime factors of 180 are $2 \times 2 \times 3 \times 3 \times 5$. The exponents of these prime factors are 2 for 2, 2 for 3, and 1 for 5. Adding these exponents together: 2 + 2 + 1 = 5.

Key Concept: Factorization

c) Assertion is true, but Reason is false.

[Solution Description]

To factorise the expression $x^2 + 5x + 6$, we look for two numbers that multiply to 6 and add to 5. These numbers are 2 and 3. Therefore, the expression can be written as $(x + 2)(x + 3)$. The assertion is correct. However, the reason states that factors are obtained by finding values of variables that make the expression zero, which is incorrect. Factors are obtained by breaking down the expression into simpler expressions whose product gives the original expression. Hence, the reason is false.

Key Concept: Factorisation

b) $(x + 2)(x + 3)$

[Solution Description]

To factorise $x^2 + 5x + 6$, we need to find two numbers that multiply to give $6$ (the constant term) and add up to give $5$ (the coefficient of $x$). The numbers $2$ and $3$ satisfy these conditions because $2 \times 3 = 6$ and $2 + 3 = 5$. Therefore, the expression can be factorised as:

$x^2 + 5x + 6 = (x + 2)(x + 3)$

So, the factorised form of the expression is $(x + 2)(x + 3)$.

Key Concept: Factorisation

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that the expression $2x + 4$ can be factorised as $2(x + 2)$. This is correct because we can take out the common factor of 2 from both terms, resulting in $2(x + 2)$. The reason states that factorising an expression involves writing it as a product of its factors, which is also true. Moreover, the reason correctly explains why the assertion is true because factorisation indeed involves breaking down an expression into its constituent factors. Therefore, both the assertion and the reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Factorisation

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

Both the Assertion and Reason are true. The expression $2x + 4$ can indeed be factorised as $2(x + 2)$ because $2$ is a common factor of both terms. The Reason is also true as factorisation involves breaking down an expression into a product of factors. Additionally, the Reason correctly explains the Assertion since the process described in the Reason directly leads to the factorisation given in the Assertion.

Key Concept: Factorisation, Quadratic Expressions

a) $(3x - 2)(2x + 5)$

[Solution Description]

To factorize $6x^2 + 11x - 10$, we look for two numbers that multiply to $6 \times (-10) = -60$ and add to $11$. These numbers are $15$ and $-4$. Rewrite the middle term using these numbers:

$6x^2 + 15x - 4x - 10$

Now, factor by grouping:

$3x(2x + 5) - 2(2x + 5)$

This gives us:

$(3x - 2)(2x + 5)$

Thus, the factorization of $6x^2 + 11x - 10$ is $(3x - 2)(2x + 5)$.

Key Concept: Factorisation

b) $(x - 2)(x^2 + 2x + 4)$

[Solution Description]

The expression $x^3 - 8$ is a difference of cubes, which can be factorised using the formula $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$. Here, $a = x$ and $b = 2$. Thus, the factorised form is $(x - 2)(x^2 + 2x + 4)$.

Key Concept: Common Factor Method, Regrouping

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

Let's consider the expression $x^3 + x^2 + x + 1$. We can group the terms as $(x^3 + x^2) + (x + 1)$. Now, factor out the common factor from each group: $x^2(x + 1) + 1(x + 1)$. Notice that $(x + 1)$ is a common factor in both groups. Using the distributive law, we can write this as $(x^2 + 1)(x + 1)$. Thus, the expression is successfully factorised by regrouping. The assertion is true because we were able to factorise the expression using regrouping. The reason is also true because regrouping allows us to form groups where each group has a common factor, which leads to successful factorisation of the above expression. Additionally, the reason correctly explains why the assertion is true.

Key Concept: Method of Common Factors

b) $3(x + 3)$

[Solution Description]

To factorise $3x + 9$, we first find the common factors in each term. The first term is $3x = 3 \times x$ and the second term is $9 = 3 \times 3$. The common factor is 3. Using the distributive law, we can write $3x + 9 = 3(x + 3)$. Thus, the factorised form is $3(x + 3)$.

Key Concept: Factorisation, Algebraic Expressions

b) $(2x - 3y)(2x + 3y)$

[Solution Description] The expression $4x^2 - 9y^2$ is a difference of squares, which can be factorised as $(2x)^2 - (3y)^2 = (2x - 3y)(2x + 3y)$.

Key Concept: Factorisation by Common Factors

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To factorise the expression $3x^2 + 6x$, we first identify the greatest common divisor (GCD) of the coefficients, which is 3, and the common variable factor, which is $x$. Taking out the GCD and the common variable factor, we get $3x(x + 2)$. The assertion is true. The reason is also true because factorisation by common factors indeed involves taking out the greatest common divisor from each term. Moreover, the reason correctly explains the assertion.

Key Concept: Factorisation by Common Factors, Distributive Law

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

First, let's break down the expression $6x^3 + 9x^2$ into its irreducible factors:

$6x^3 = 2 \times 3 \times x \times x \times x$

$9x^2 = 3 \times 3 \times x \times x$

Next, we identify the common factors in both terms. We observe that both terms have $3$, which is a numerical factor, and $x^2$, which is an algebraic factor. Therefore, the common factor is $3x^2$. Using the distributive law, we can rewrite the expression as:

$6x^3 + 9x^2 = 3x^2(2x + 3)$

Thus, the assertion (A) is true because the expression $6x^3 + 9x^2$ is correctly factorised as $3x^2(2x + 3)$. The reason (R) is also true because the common factor in the terms of the expression $6x^3 + 9x^2$ is indeed $3x^2$. Furthermore, the reason correctly explains the assertion.

Key Concept: Factorisation by Common Factors

c) $6(x + 2)$

[Solution Description]

To factorise $6x + 12$, we look for the greatest common factor (GCF) of the terms. The GCF of $6$ and $12$ is $6$. We then factor out $6$ from each term:

$6x + 12 = 6(x + 2)$.

Key Concept: Finding Common Factors

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To factorise the expression $6x + 12$, we first identify the common factors in each term. The term $6x$ can be written as $6 \times x$ and the term $12$ can be written as $6 \times 2$. Here, 6 is a common factor in both terms. Using the distributive law, we combine the terms as $6(x + 2)$. This shows that the assertion is true. The reason correctly identifies the common factor, which is indeed 6, and explains why the assertion is true.

Key Concept: Factorisation

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] To factorise the expression $3x + 6$, we first identify the common factor in both terms. The term $3x$ can be written as $3 \times x$, and the term $6$ can be written as $3 \times 2$. Both terms have a common factor of $3$. Using the distributive law, we can write $3x + 6$ as $3(x + 2)$. Therefore, the assertion is true. The reason given is also true, as the common factor in both terms is indeed $3$. Moreover, the reason correctly explains the assertion since the presence of the common factor $3$ allows us to factorise the expression as $3(x + 2)$.

Key Concept: Factorisation, Common Factors

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To verify the assertion and reason:

Step 1: Identify the terms in the expression $6x^2 + 12x$. The terms are $6x^2$ and $12x$.

Step 2: Find the irreducible factors of each term:

$6x^2 = 6 \times x \times x$

$12x = 12 \times x$

Step 3: Determine the greatest common factor (GCF) of $6x^2$ and $12x$. The GCF is the product of the common factors with the smallest exponents. Here, the common factors are $6$ and $x$, so GCF = $6x$.

Step 4: Factorise the expression by taking out the GCF $6x$:

$6x^2 + 12x = 6x(x + 2)$

Conclusion: The assertion is true because $6x(x + 2)$ is indeed the factorised form of $6x^2 + 12x$. The reason is also true as the GCF of $6x^2$ and $12x$ is indeed $6x$. Moreover, the reason correctly explains why the assertion is true since identifying the GCF is the key step in factorising the expression.

Key Concept: Factorisation using distributive property

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that the expression $2x + 4$ can be factorised as $2(x + 2)$. This is true because, when we factorise $2x + 4$, we find that 2 is a common factor of both terms. Writing each term as a product of irreducible factors, we get $2x = 2 \times x$ and $4 = 2 \times 2$. Now, using the distributive law, we can write $2x + 4$ as $(2 \times x) + (2 \times 2)$, which further simplifies to $2(x + 2)$. Thus, the factorised form of $2x + 4$ is indeed $2(x + 2)$. The reason given is that 2 is a common factor of both terms in the expression $2x + 4$, which is also true and directly explains why the expression can be factorised as $2(x + 2)$. Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.

Key Concept: Factorisation, Distributive Property

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To factorise the expression $6x^2 + 12x$, we first identify the common factors in both terms. The irreducible factor forms of the terms are:

$6x^2 = 2 \times 3 \times x \times x \quad \text{and} \quad 12x = 2 \times 2 \times 3 \times x$

The common factors are $2$, $3$, and $x$. Combining these, the greatest common factor is $6x$. Applying the distributive property, we can rewrite the expression as:

$6x^2 + 12x = 6x(x) + 6x(2) = 6x(x + 2)$

Thus, the assertion is true. The reason provided is also true and correctly explains why the assertion holds, as the distributive property allows us to factor out the common term $6x$ from the expression.

Key Concept: Factorisation by common factor

b) $10x(y + 2)$

[Solution Description]

To factorise $10xy + 20x$, we first identify the greatest common factor (GCF) of the two terms. The GCF of $10xy$ and $20x$ is $10x$. We can write each term as a product of the GCF and another factor:

$10xy = 10x \times y$

$20x = 10x \times 2$

Using the distributive property, we combine the terms:

$10xy + 20x = 10x(y + 2)$

Therefore, the factorised form of $10xy + 20x$ is $10x(y + 2)$.

Key Concept: Regrouping, Factorisation

c) $(2a + 3)(2b + 1)$

[Solution Description]

First, group the terms: $(4ab + 6b) + (2a + 3)$. Now, factor out the common factors from each group. From the first group $4ab + 6b$, the common factor is $2b$. So, $4ab + 6b = 2b(2a + 3)$. From the second group $2a + 3$, the common factor is $1$. So, $2a + 3 = 1(2a + 3)$. Now, combine both expressions: $2b(2a + 3) + 1(2a + 3)$ Notice that $(2a + 3)$ is a common factor. Thus, the expression can be written as $(2a + 3)(2b + 1)$.

Key Concept: Factorisation by Regrouping Terms

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion is true because the expression $2x + 4y + 3x + 6y$ can indeed be factorised by regrouping terms into $(2x + 3x) + (4y + 6y)$, which simplifies to $5x + 10y$. The reason is also true because regrouping terms does involve rearranging terms with common factors, such as grouping $2x$ and $3x$, and $4y$ and $6y$.

Key Concept: Factorisation by Regrouping Terms

a) $(x + 1)(4y + 5)$

[Solution Description]

First, group the terms: $(4xy + 4y) + (5x + 5)$.

Factor out the common factors from each group:

For $(4xy + 4y)$, the common factor is $4y$: $4y(x + 1)$.

For $(5x + 5)$, the common factor is $5$: $5(x + 1)$.

Now, the expression becomes $4y(x + 1) + 5(x + 1)$.

Factor out the common factor $(x + 1)$: $(x + 1)(4y + 5)$.

So, the factorised form of the expression is $(x + 1)(4y + 5)$.

Key Concept: Regrouping

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The expression $2xy + 3x + 2y + 3$ can be rearranged as $(2xy + 2y) + (3x + 3)$. This grouping allows us to identify common factors: $2y$ from the first group and $3$ from the second group. Therefore, the expression can be factorised as $(2y)(x + 1) + 3(x + 1)$, which further simplifies to $(x + 1)(2y + 3)$. Hence, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Identifying terms to regroup

b) $y + 1$

[Solution Description]

The expression $3xy + 3x + 4y + 4$ can be regrouped as follows:

$3xy + 3x + 4y + 4 = 3x(y + 1) + 4(y + 1) = (y + 1)(3x + 4)$

The common factor is $(y + 1)$.

Key Concept: Identifying terms to regroup, Factorisation

a) $(y - 2)(y + 2)(y^2 + 4)$

[Solution Description]

The expression $y^4 - 16$ can be rewritten as a difference of squares:

$y^4 - 16 = (y^2)^2 - 4^2 = (y^2 - 4)(y^2 + 4)$

Further factorize $y^2 - 4$:

$y^2 - 4 = (y - 2)(y + 2)$

Thus, the complete factorization is:

$(y - 2)(y + 2)(y^2 + 4)$

Key Concept: Factorisation using grouping

a) $(x + 1)(3y + 2)$

[Solution Description]

First, group the terms with common factors. The first two terms $3xy$ and $3y$ have a common factor of $3y$. Similarly, the last two terms $2x$ and $2$ have a common factor of $2$. Factor out the common terms:

$3xy + 3y + 2x + 2 = 3y(x + 1) + 2(x + 1)$

Now, observe that $(x + 1)$ is a common factor in both terms. Combine the terms by factoring out $(x + 1)$:

$3y(x + 1) + 2(x + 1) = (x + 1)(3y + 2)$

Therefore, the factorised form of the expression is $(x + 1)(3y + 2)$.

Key Concept: Factorisation by Regrouping

b)$q + 1$

[Solution Description]

First, group the terms with common factors: $(5pq + 5p)$ and $(2q + 2)$.

Factor out the common factors from each group:

$5pq + 5p = 5p(q + 1)$ and

$2q + 2 = 2(q + 1)$.

Now, the expression becomes $5p(q + 1) + 2(q + 1)$.

Here, $(q + 1)$ is the common factor.

Key Concept: Factorisation by grouping

a) $(q + 1)(7p + 8)$

[Solution Description]

First, group the terms with common factors:

$(7pq + 7p)$ and $(8q + 8)$.

Now, factorise each group:

$7pq + 7p = 7p(q + 1)$ and

$8q + 8 = 8(q + 1)$.

Next, combine the factorised groups:

$7p(q + 1) + 8(q + 1)$.

Notice that $(q + 1)$ is a common factor:

$(q + 1)(7p + 8)$.

Thus, the factorised form of the expression is $(q + 1)(7p + 8)$.

Key Concept: Factorisation Using Identities

a) $(2x - 3y)(2x + 3y)(4x^2 + 9y^2)$

[Solution Description]

The expression $16x^4 - 81y^4$ can be written as $(4x^2)^2 - (9y^2)^2$, which fits the form of Identity III, $(a^2 - b^2) = (a - b)(a + b)$. Applying this identity:

$16x^4 - 81y^4 = (4x^2)^2 - (9y^2)^2 = (4x^2 - 9y^2)(4x^2 + 9y^2)$

Now, the term $(4x^2 - 9y^2)$ can be further factorised using the same identity:

$4x^2 - 9y^2 = (2x)^2 - (3y)^2 = (2x - 3y)(2x + 3y)$

Therefore, the complete factorisation of $16x^4 - 81y^4$ is $(2x - 3y)(2x + 3y)(4x^2 + 9y^2)$.

Key Concept: Factorization Using Identities

c) $(x + 5)^2$

[Solution Description]

The given expression is $x^2 + 10x + 25$. We observe that it matches the form of Identity I: $(a + b)^2 = a^2 + 2ab + b^2$. Here, $a = x$ and $b = 5$ because $2ab = 2 \times x \times 5 = 10x$ and $b^2 = 5^2 = 25$. Therefore, the factorization is $(x + 5)^2$.

Key Concept: Factorisation Using Identities

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The given expression is $x^2 + 10x + 25$. Let us check if it fits the identity $(a + b)^2 = a^2 + 2ab + b^2$. Here, $a = x$ and $b = 5$, so $a^2 + 2ab + b^2 = x^2 + 2(x)(5) + 5^2 = x^2 + 10x + 25$, which matches the given expression. Therefore, the factorisation $(x + 5)^2$ is correct. The reason correctly explains that the expression fits the identity $(a + b)^2 = a^2 + 2ab + b^2$.

Key Concept: Standard Algebraic Identities

b) 29

[Solution Description]

We know the identity $x^2 + y^2 = (x + y)^2 - 2xy$. Given $x + y = 7$ and $xy = 10$, we substitute these values into the identity to find $x^2 + y^2$. First, calculate $(x + y)^2$: $(7)^2 = 49$. Next, calculate $2xy$: $2 \times 10 = 20$. Finally, subtract $2xy$ from $(x + y)^2$: $49 - 20 = 29$. Therefore, the value of $x^2 + y^2$ is 29.

Key Concept: Standard Algebraic Identities

b) $(4y - 3)^2$

[Solution Description]

The given expression is $16y^2 - 24y + 9$. We observe that it can be written in the form $a^2 - 2ab + b^2$, where $a = 4y$ and $b = 3$. Using Identity II, $(a - b)^2 = a^2 - 2ab + b^2$, we get:

$16y^2 - 24y + 9 = (4y)^2 - 2(4y)(3) + 3^2 = (4y - 3)^2$

Therefore, the factorised form of the expression is $(4y - 3)^2$.

Key Concept: Standard Algebraic Identities

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The given expression is $x^2 + 10x + 25$. Let us compare this with the identity $(a + b)^2 = a^2 + 2ab + b^2$. Here, $a = x$ and $b = 5$ because $a^2 = x^2$, $2ab = 10x$, and $b^2 = 25$. Therefore, $x^2 + 10x + 25 = (x + 5)^2$. The assertion is true as the expression is correctly factorised using the identity. The reason is also true and correctly explains why the assertion is valid.

Key Concept: Factorisation using Identity II

b) $(3y - 2)^2$

[Solution Description]

The expression is of the form $(a - b)^2 = a^2 - 2ab + b^2$. Here, $a = 3y$ and $b = 2$ because $a^2 - 2ab + b^2 = (3y)^2 - 2(3y)(2) + 2^2 = 9y^2 - 12y + 4$. Therefore, the factorisation is $(3y - 2)^2$.

Key Concept: Factoring using Algebraic Identities

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] The assertion states that $x^2 - 16$ can be factored as $(x - 4)(x + 4)$. This is true because the expression is a difference of squares, which can be factored using the identity $a^2 - b^2 = (a - b)(a + b)$ where $a = x$ and $b = 4$. The reason correctly explains the assertion by identifying the expression as a difference of squares and applying the correct identity. Therefore, both the assertion and the reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Factors of the form $(x + a

b) $(y - 4)(y - 5)$

[Solution Description]

We need to factorise $y^2 - 9y + 20$ which is of the form $y^2 + py + q$. We find two numbers $a$ and $b$ such that $a + b = -9$ and $ab = 20$. The numbers satisfying these conditions are $-4$ and $-5$, as $(-4) + (-5) = -9$ and $(-4) \times (-5) = 20$. Therefore, the factorised form of the expression is $(y - 4)(y - 5)$.

Key Concept: Factorisation of Quadratic Expressions

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] The given expression is $x^2 + 5x + 6$. To factorise this, we need to find two numbers $a$ and $b$ such that $a \times b = 6$ and $a + b = 5$. The numbers 2 and 3 satisfy these conditions because $2 \times 3 = 6$ and $2 + 3 = 5$. Therefore, the expression can be written as $(x + 2)(x + 3)$. Thus, the assertion is true. The reason given is the general method for factorising expressions of the form $x^2 + px + q$, which is correctly applied here. Hence, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Quadratic Expression, Factorization

d) $(z + 3)(z - 4)$

[Solution Description]

To factorise $z^2 - z - 12$, we need to find two numbers $a$ and $b$ such that $a + b = -1$ and $ab = -12$. The factors of -12 are 1, -1, 2, -2, 3, -3, 4, -4, 6, -6, 12, and -12. Testing the combinations, we find that 3 and -4 satisfy both conditions since $3 + (-4) = -1$ and $3 \times (-4) = -12$. Therefore, the factorised form of the expression is $(z + 3)(z - 4)$.

Key Concept: Quadratic Expression, Factorization

c) $(x - 3)(x - 5)$

[Solution Description]

To factorise $x^2 - 8x + 15$, we need to find two numbers $a$ and $b$ such that $a + b = -8$ and $ab = 15$. The factors of 15 are 1, 3, 5, and 15. Testing the combinations, we find that $-3$ and $-5$ satisfy both conditions since $-3 + (-5) = -8$ and $(-3) \times (-5) = 15$. Therefore, the factorised form of the expression is $(x - 3)(x - 5)$.

Key Concept: Splitting the middle term

b) $(x + 3)(x + 8)$

[Solution Description]

To factorise $x^2 + 11x + 24$, we need to find two numbers that multiply to 24 and add up to 11. The numbers 3 and 8 satisfy this condition because $3 \times 8 = 24$ and $3 + 8 = 11$. Therefore, the expression can be written as $(x + 3)(x + 8)$.

Key Concept: Splitting the middle term

b) $(x + 4)(x + 5)$

[Solution Description]

To factorise $x^2 + 9x + 20$, we need to find two numbers that multiply to 20 and add up to 9. The numbers 4 and 5 satisfy this condition because $4 \times 5 = 20$ and $4 + 5 = 9$. Therefore, the expression can be written as $(x + 4)(x + 5)$.

Key Concept: Factorisation, Identities, Division of Polynomials

a) $x^2 - x - 12$

[Solution Description]

To find the quotient when $p(x) = x^3 - 3x^2 - 10x + 24$ is divided by $d(x) = x - 2$, we use polynomial long division or synthetic division. Using synthetic division:

Step 1: Set up the synthetic division with root 2.

Step 2: Bring down the coefficient of $x^3$, which is 1.

Step 3: Multiply 1 by 2 and add to the next coefficient: $1 \times 2 = 2$, $-3 + 2 = -1$.

Step 4: Multiply -1 by 2 and add to the next coefficient: $-1 \times 2 = -2$, $-10 + (-2) = -12$.

Step 5: Multiply -12 by 2 and add to the next coefficient: $-12 \times 2 = -24$, $24 + (-24) = 0$.

The coefficients of the quotient are 1, -1, -12, so the quotient is $x^2 - x - 12$.

Key Concept: Factorisation

b) $(x + 2)(x + 3)$

[Solution Description]

To factorise the quadratic expression $x^2 + 5x + 6$, we look for two numbers that multiply to give 6 and add to give 5. The numbers 2 and 3 satisfy these conditions because $2 \times 3 = 6$ and $2 + 3 = 5$. Thus, the factorised form is $(x + 2)(x + 3)$.

Key Concept: Division of a polynomial by a monomial

a) $2y^2 + 3y + 4$

[Solution Description]

Divide each term of the polynomial by the monomial: $\frac{8y^3}{4y} + \frac{12y^2}{4y} + \frac{16y}{4y} = 2y^2 + 3y + 4$.

Key Concept: Division of Polynomial by Monomial, Factorization

a) $3a^2b^2 + 4ab - 2$

[Solution Description]

First, express each term of the polynomial in its irreducible factor form. The polynomial $15a^4b^3 + 20a^3b^2 - 10a^2b$ can be written as $(3 \times 5 \times a \times a \times a \times a \times b \times b \times b) + (2 \times 2 \times 5 \times a \times a \times a \times b \times b) - (2 \times 5 \times a \times a \times b)$. The divisor $5a^2b$ can be written as $5 \times a \times a \times b$. Now, divide each term of the polynomial by the monomial by canceling out the common factors. This results in $3a^2b^2 + 4ab - 2$.

Key Concept: Division of a Monomial by Another Monomial

a) $–4mnp$

[Solution Description]

To divide $–36m^4n^3p^2$ by $9m^3n^2p$, we first express both monomials in their irreducible factor forms:

$–36m^4n^3p^2 = –1 \times 2 \times 2 \times 3 \times 3 \times m \times m \times m \times m \times n \times n \times n \times p \times p$

$9m^3n^2p = 3 \times 3 \times m \times m \times m \times n \times n \times p$

Now, group the factors of $–36m^4n^3p^2$ to separate $9m^3n^2p$:

$–36m^4n^3p^2 = 3 \times 3 \times m \times m \times m \times n \times n \times p \times (–1 \times 2 \times 2 \times m \times n \times p)$

Therefore,

$–36m^4n^3p^2 \div 9m^3n^2p = –4mnp$

Key Concept: Division of a Monomial by Another Monomial

a) $-3x^2$

[Solution Description]

Break down both the monomials into their irreducible factors:

$-15x^4 = -1 \times 3 \times 5 \times x \times x \times x \times x$

$5x^2 = 5 \times x \times x$

Now divide $-15x^4$ by $5x^2$: $\frac{-15x^4}{5x^2} = \frac{-1 \times 3 \times 5 \times x \times x \times x \times x}{5 \times x \times x} = -3 \times x \times x = -3x^2$

Key Concept: Exponent laws for division, Polynomial division

a) $2x^3y^2 - 3x^2y + 4x$

[Solution Description] To simplify the expression $\frac{12x^5y^3 - 18x^4y^2 + 24x^3y}{6x^2y}$, we divide each term in the numerator by the denominator separately.

First term: $\frac{12x^5y^3}{6x^2y} = 2x^{5-2}y^{3-1} = 2x^3y^2$.

Second term: $\frac{-18x^4y^2}{6x^2y} = -3x^{4-2}y^{2-1} = -3x^2y$.

Third term: $\frac{24x^3y}{6x^2y} = 4x^{3-2}y^{1-1} = 4x$.

Therefore, the simplified expression is $2x^3y^2 - 3x^2y + 4x$.

Key Concept: Division of Monomials

a) $-3a^2b$

[Solution Description]

To simplify $\frac{-15a^4b^2}{5a^2b}$, first write both the numerator and denominator in their irreducible factor forms:

$-15a^4b^2 = -3 \times 5 \times a^4 \times b^2$

$5a^2b = 5 \times a^2 \times b$

Now, cancel out the common factors:

$\frac{-3 \times 5 \times a^4 \times b^2}{5 \times a^2 \times b} = -3 \times a^{4-2} \times b^{2-1} = -3a^2b$

Therefore, the simplified form is $-3a^2b$.

Key Concept: Simplification using cancellation method

a) $4m^2n - 3mn^2 + 2n^3$

[Solution Description]

To simplify $\frac{20m^3n^2 - 15m^2n^3 + 10mn^4}{5mn}$, divide each term in the numerator by the denominator $5mn$:

$\frac{20m^3n^2}{5mn} = 4m^2n$, $\frac{-15m^2n^3}{5mn} = -3mn^2$, and $\frac{10mn^4}{5mn} = 2n^3$. Adding these results together, we get $4m^2n - 3mn^2 + 2n^3$.

Key Concept: Simplification using cancellation method

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The given expression is $(6x^3 + 9x^2 + 12x) ÷ 3x$. To simplify this, we divide each term of the polynomial by $3x$.

$\frac{6x^3}{3x} = 2x^2$

$\frac{9x^2}{3x} = 3x$

$\frac{12x}{3x} = 4$

Therefore, the simplified form of the expression is $2x^2 + 3x + 4$. Thus, the Assertion is true. The Reason states that each term of the polynomial is divided by $3x$ using the cancellation method, which is indeed the correct method applied here to arrive at the simplified form. Hence, the Reason is also true and correctly explains the Assertion.

Key Concept: Division of a Polynomial by a Monomial, Multi-step Solutions

a) $3x^2y^2 + 4xy + 5$

[Solution Description]

To divide the polynomial $15x^4y^3 + 20x^3y^2 + 25x^2y$ by the monomial $5x^2y$, we can use the cancellation method. We divide each term of the polynomial by $5x^2y$ separately.

Step 1: Divide the first term $\frac{15x^4y^3}{5x^2y}$. Cancel out common factors: $\frac{15}{5} = 3$, $\frac{x^4}{x^2} = x^{4-2} = x^2$, and $\frac{y^3}{y} = y^{3-1} = y^2$. So, $\frac{15x^4y^3}{5x^2y} = 3x^2y^2$.

Step 2: Divide the second term $\frac{20x^3y^2}{5x^2y}$. Cancel out common factors: $\frac{20}{5} = 4$, $\frac{x^3}{x^2} = x^{3-2} = x$, and $\frac{y^2}{y} = y^{2-1} = y$. So, $\frac{20x^3y^2}{5x^2y} = 4xy$.

Step 3: Divide the third term $\frac{25x^2y}{5x^2y}$. Cancel out common factors: $\frac{25}{5} = 5$, $\frac{x^2}{x^2} = 1$, and $\frac{y}{y} = 1$. So, $\frac{25x^2y}{5x^2y} = 5$.

Therefore, the result of the division is $3x^2y^2 + 4xy + 5$.

Key Concept: Division of a Polynomial by a Monomial

b) $2m^2 + 3mn + 4n^2$

[Solution Description]

To divide the polynomial $8m^3n + 12m^2n^2 + 16mn^3$ by the monomial $4mn$, we divide each term of the polynomial by $4mn$.

$\frac{8m^3n}{4mn} = 2m^2$

$\frac{12m^2n^2}{4mn} = 3mn$

$\frac{16mn^3}{4mn} = 4n^2$

Adding these results together, we get $2m^2 + 3mn + 4n^2$.

Key Concept: Division of Algebraic Expressions

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that when dividing $6x^2 + 12x$ by $3x$, each term in the numerator is divided by the denominator separately. This is correct because it aligns with the rule that dividing a polynomial by a monomial involves dividing each term of the polynomial by the monomial. The reason states that dividing each term of a polynomial by a monomial follows the distributive property. This is also correct since division can be distributed over addition in this context. The reason correctly explains the assertion because it provides the underlying principle for the process described in the assertion.

Key Concept: Division of monomials

a) $3a^2bc^2$

[Solution Description]

To divide $18a^4b^3c^2$ by $6a^2b^2$, we separate the coefficients and variables. First, divide the coefficients: $18 ÷ 6 = 3$. Next, subtract the exponents of like variables. For $a$, it is $4 - 2 = 2$. For $b$, it is $3 - 2 = 1$. The variable $c$ remains unchanged as it only appears in the numerator. Therefore, the result is $3a^2bc^2$.

Key Concept: Common factor method

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] To solve the problem, we first factorise the numerator $y^2 + 7y + 10$. We look for two numbers that multiply to 10 and add to 7. These numbers are 5 and 2. Thus, the factorisation is $(y + 5)(y + 2)$. Now, when we divide $(y^2 + 7y + 10)$ by $(y + 5)$, we get $\frac{(y + 5)(y + 2)}{(y + 5)} = y + 2$. Therefore, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Factorisation, Division of Algebraic Expressions

b) $2(2y + 1)$

[Solution Description]

First, factorise the numerator and denominator. The numerator $8y^2 + 12y + 4$ can be written as $4(2y^2 + 3y + 1)$. Factorising further, $2y^2 + 3y + 1 = (2y + 1)(y + 1)$. So, the numerator becomes $4(2y + 1)(y + 1)$. The denominator $2y + 2$ can be written as $2(y + 1)$. Now, divide the numerator by the denominator: $\frac{4(2y + 1)(y + 1)}{ 2(y + 1) }$. Simplifying, we get $2(2y + 1)$.

Key Concept: Division of Polynomials

a) $5z^4$

[Solution Description]

First, factor the numerator $20z^5 - 40z^4$. The common factor is $20z^4$, so we can write $20z^5 - 40z^4 = 20z^4(z - 2)$. Rewrite the division as $\frac{20z^4(z - 2)}{4z - 8}$. Factor the denominator $4z - 8$ to get $4(z - 2)$. So, the expression becomes $\frac{20z^4(z - 2)}{4(z - 2)}$. Cancel the common factor $(z - 2)$ from the numerator and denominator to get $\frac{20z^4}{4}$. Simplify this by dividing $20z^4$ by $4$ to get $5z^4$.

Key Concept: Division of Polynomials

a) $3x$

[Solution Description]

First, we factorise the numerator $18x^2 + 36x$. Taking out the common factor $18x$, we get $18x^2 + 36x = 18x(x + 2)$. Now, we divide this by $6(x + 2)$. So, $\frac{18x(x + 2)}{6(x + 2)}$. The $(x + 2)$ terms cancel each other, leaving us with $\frac{18x}{6} = 3x$.

Key Concept: Division of Algebraic Expressions

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that when dividing a polynomial by a linear polynomial using long division, the degree of the remainder is always less than the degree of the divisor. This is true because, in polynomial long division, the remainder must indeed be of a lower degree than the divisor for the division process to terminate correctly. The reason explains why this is true by stating that the remainder must be of a lower degree than the divisor to ensure the division process terminates correctly. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Division of Algebraic Expressions

a) $3x + 4$

[Solution Description]

We use long division to divide $6x^2 + 11x + 4$ by $2x + 1$. First, we divide the leading term $6x^2$ by the leading term $2x$ to get $3x$. Then, we multiply $3x$ by $2x + 1$ to get $6x^2 + 3x$. Subtracting this from $6x^2 + 11x + 4$, we get $8x + 4$. Next, we divide $8x$ by $2x$ to get $4$. Multiplying $4$ by $2x + 1$, we get $8x + 4$. Subtracting this from $8x + 4$, we get $0$. Therefore, the result is $3x + 4$.

Key Concept: Division of Polynomial by Monomial, Factorisation

a) $2a^2b + 3ab^2 - 4b^3$

[Solution Description]

To divide the polynomial $18a^4b^3 + 27a^3b^4 - 36a^2b^5$ by the monomial $9a^2b^2$, we divide each term of the polynomial by the monomial separately.

First, divide $18a^4b^3$ by $9a^2b^2$: $\frac{18a^4b^3}{9a^2b^2} = 2a^{4-2}b^{3-2} = 2a^2b$.

Next, divide $27a^3b^4$ by $9a^2b^2$: $\frac{27a^3b^4}{9a^2b^2} = 3a^{3-2}b^{4-2} = 3ab^2$.

Finally, divide $-36a^2b^5$ by $9a^2b^2$: $\frac{-36a^2b^5}{9a^2b^2} = -4b^{5-2} = -4b^3$.

Adding all the results together, we get $2a^2b + 3ab^2 - 4b^3$.

Therefore, $(18a^4b^3 + 27a^3b^4 - 36a^2b^5) \div 9a^2b^2 = 2a^2b + 3ab^2 - 4b^3$.

Key Concept: Division of a Polynomial by a Monomial

a) $4a^2 + 5a + 6$

[Solution Description]

To solve $(12a^3 + 15a^2 + 18a) \div 3a$, we divide each term of the polynomial by the monomial:

$\frac{12a^3}{3a} = 4a^2$, $\frac{15a^2}{3a} = 5a$, and $\frac{18a}{3a} = 6$.

Therefore, the quotient is $4a^2 + 5a + 6$.

Key Concept: Factorisation and Division

a) $m - 16$

[Solution Description]

Factorise the numerator $m^2 - 14m - 32$. Find two numbers that multiply to -32 and add to -14. The numbers -16 and 2 satisfy this condition. So, the expression becomes:

$m^2 - 14m - 32 = (m - 16)(m + 2)$

Now, divide by $(m + 2)$:

$\frac{(m - 16)(m + 2)}{m + 2} = m - 16$

Key Concept: Division of Algebraic Expressions

b) $0.5x^3$

[Solution Description] To divide $28x^4$ by $56x$, follow these steps:

Step 1: Divide the coefficients, 28 and 56.

$28 ÷ 56 = 0.5$

Step 2: Subtract the exponents of $x$.

$x^4 ÷ x = x^{4-1} = x^3$

Step 3: Combine the results from Step 1 and Step 2.

$0.5x^3$

Therefore, the result of dividing $28x^4$ by $56x$ is $0.5x^3$.

Key Concept: Division of Polynomials

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To determine if the assertion and reason are correct, we perform the division of $x^2 - 4$ by $x - 2$. Using synthetic division, we find that the remainder is indeed zero. This confirms that $x^2 - 4$ is divisible by $x - 2$. Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.

Key Concept: Division of Polynomials

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

Let's verify the assertion and reason step by step.

Step 1: Factorise the polynomial $(7x^2 + 14x)$.

$7x^2 + 14x = 7x(x + 2)$

Step 2: Perform the division of $(7x^2 + 14x)$ by $(x + 2)$.

$\frac{7x^2 + 14x}{x + 2} = \frac{7x(x + 2)}{x + 2}$

Step 3: Cancel out the common factor $(x + 2)$ from the numerator and denominator.

$\frac{7x(x + 2)}{x + 2} = 7x$

The assertion is true as the division indeed results in $7x$. The reason is also true because $(7x^2 + 14x)$ can be factorised as $7x(x + 2)$, and this factorisation correctly explains the result of the division.