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Math Notes For Class 8 Chapter 4 Ex 4.2 Solved & Quiz
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Math Notes For Class 8 Chapter 4 Ex 4.2 Solved & Quiz
Factorization is the mathematical process of expressing an algebraic expression as a product of two or more irreducible factors. In Chapter 4 (Factorization and Equations) of the Class 8 SNC textbook, factorization forms the cornerstone for simplifying rational algebraic expressions and solving quadratic equations.
Below is a quick reference summary table of all mandatory algebraic identities utilized throughout Exercise 4.2:
| Identity Type | Algebraic Formula | Application Key |
|---|---|---|
| Difference of Two Squares | $a^2 - b^2 = (a - b)(a + b)$ | Used when expressions contain two perfect square terms separated by a minus sign. |
| Perfect Square (Addition) | $a^2 + 2ab + b^2 = (a + b)^2$ | Applied to three-term polynomials with positive middle terms. |
| Perfect Square (Subtraction) | $a^2 - 2ab + b^2 = (a - b)^2$ | Applied to three-term polynomials with negative middle terms. |
| Trinomial Splitting | $x^2 + (p+q)x + pq = (x + p)(x + q)$ | Used for quadratic expressions of the form $x^2 + bx + c$. |
| Sum of Cubes | $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ | Used for two cubic terms added together. |
| Difference of Cubes | $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ | Used for two cubic terms subtracted from each other. |
Question 1 (Part i): Factorize $4x^2 - 25y^2$ using the difference of two squares identity.
Solution:
Step 1: Write each term as a perfect square:
$$4x^2 = (2x)^2$$
$$25y^2 = (5y)^2$$
Step 2: Apply the algebraic identity $a^2 - b^2 = (a - b)(a + b)$ where $a = 2x$ and $b = 5y$:
$$4x^2 - 25y^2 = (2x)^2 - (5y)^2 = (2x - 5y)(2x + 5y)$$
Answer: $$(2x - 5y)(2x + 5y)$$
Question 1 (Part ii): Factorize completely $16a^4 - 81b^4$.
Solution:
Step 1: Express as a difference of squares:
$$16a^4 - 81b^4 = (4a^2)^2 - (9b^2)^2$$
Step 2: Apply $a^2 - b^2 = (a - b)(a + b)$:
$$= (4a^2 - 9b^2)(4a^2 + 9b^2)$$
Step 3: Notice that $(4a^2 - 9b^2)$ is again a difference of two squares: $(2a)^2 - (3b)^2$. Factorize it further:
$$4a^2 - 9b^2 = (2a - 3b)(2a + 3b)$$
Step 4: Combine all terms together:
$$16a^4 - 81b^4 = (2a - 3b)(2a + 3b)(4a^2 + 9b^2)$$
Answer: $$(2a - 3b)(2a + 3b)(4a^2 + 9b^2)$$
Question 2 (Part i): Factorize the complete square expression $9x^2 + 24xy + 16y^2$.
Solution:
Step 1: Identify the squared terms:
$$a^2 = 9x^2 \implies a = 3x$$
$$b^2 = 16y^2 \implies b = 4y$$
Step 2: Check the middle term $2ab$:
$$2ab = 2(3x)(4y) = 24xy$$
Since the middle term matches $+24xy$, apply $(a + b)^2 = a^2 + 2ab + b^2$:
$$9x^2 + 24xy + 16y^2 = (3x + 4y)^2$$
Answer: $$(3x + 4y)^2$$
Question 2 (Part ii): Factorize $25a^2 - 70ab + 49b^2$.
Solution:
Step 1: Identify the squared terms:
$$a^2 = 25a^2 \implies a = 5a$$
$$b^2 = 49b^2 \implies b = 7b$$
Step 2: Check the middle term $-2ab$:
$$-2ab = -2(5a)(7b) = -70ab$$
Since the middle term matches $-70ab$, apply $(a - b)^2 = a^2 - 2ab + b^2$:
$$25a^2 - 70ab + 49b^2 = (5a - 7b)^2$$
Answer: $$(5a - 7b)^2$$
Question 3 (Part i): Factorize $x^2 + 7x + 12$ by splitting the middle term.
Solution:
Step 1: Find two integers $p$ and $q$ such that $p \times q = 12$ and $p + q = 7$.
The numbers are $3$ and $4$ because $3 \times 4 = 12$ and $3 + 4 = 7$.
Step 2: Rewrite the middle term $7x$ as $3x + 4x$:
$$x^2 + 3x + 4x + 12$$
Step 3: Group terms and factor out common factors:
$$x(x + 3) + 4(x + 3) = (x + 3)(x + 4)$$
Answer: $$(x + 3)(x + 4)$$
Question 3 (Part ii): Factorize $x^2 - 9x + 20$.
Solution:
Step 1: Find two integers $p$ and $q$ such that $p \times q = +20$ and $p + q = -9$.
The numbers are $-4$ and $-5$ because $(-4) \times (-5) = 20$ and $(-4) + (-5) = -9$.
Step 2: Rewrite the middle term $-9x$ as $-4x - 5x$:
$$x^2 - 4x - 5x + 20$$
Step 3: Group terms and factor out common factors:
$$x(x - 4) - 5(x - 4) = (x - 4)(x - 5)$$
Answer: $$(x - 4)(x - 5)$$
Question 4 (Part i): Factorize $2x^2 + 9x + 10$.
Solution:
Step 1: Multiply coefficient of $x^2$ by the constant term: $2 \times 10 = 20$.
Find two numbers that multiply to $20$ and add up to $9$. The numbers are $4$ and $5$.
Step 2: Rewrite the middle term $9x$ as $4x + 5x$:
$$2x^2 + 4x + 5x + 10$$
Step 3: Group the first two and last two terms:
$$2x(x + 2) + 5(x + 2)$$
Step 4: Factor out the common binomial $(x + 2)$:
$$(2x + 5)(x + 2)$$
Answer: $$(2x + 5)(x + 2)$$
Question 4 (Part ii): Factorize $3x^2 - 10x + 8$.
Solution:
Step 1: Product = $3 \times 8 = 24$. Sum = $-10$.
Find two numbers whose product is $24$ and sum is $-10$. The numbers are $-6$ and $-4$.
Step 2: Rewrite the middle term $-10x$ as $-6x - 4x$:
$$3x^2 - 6x - 4x + 8$$
Step 3: Group and factorize:
$$3x(x - 2) - 4(x - 2)$$
Step 4: Extract the common binomial factor $(x - 2)$:
$$(3x - 4)(x - 2)$$
Answer: $$(3x - 4)(x - 2)$$
Question 5 (Part i): Factorize by grouping terms: $ac + ad + bc + bd$.
Solution:
Step 1: Group terms into pairs:
$$(ac + ad) + (bc + bd)$$
Step 2: Factor out $a$ from the first pair and $b$ from the second pair:
$$a(c + d) + b(c + d)$$
Step 3: Factor out the common expression $(c + d)$:
$$(a + b)(c + d)$$
Answer: $$(a + b)(c + d)$$
Question 5 (Part ii): Factorize $x^3 + 3x^2 + 2x + 6$.
Solution:
Step 1: Group the four-term expression into two pairs:
$$(x^3 + 3x^2) + (2x + 6)$$
Step 2: Take out the greatest common factor from each group:
$$x^2(x + 3) + 2(x + 3)$$
Step 3: Factor out the binomial factor $(x + 3)$:
$$(x^2 + 2)(x + 3)$$
Answer: $$(x^2 + 2)(x + 3)$$
Question 6 (Part i): Factorize $8x^3 + 27y^3$ using the sum of cubes formula.
Solution:
Step 1: Rewrite each term as a perfect cube:
$$8x^3 = (2x)^3 \quad \text{and} \quad 27y^3 = (3y)^3$$
Step 2: Apply $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ where $a = 2x$ and $b = 3y$:
$$8x^3 + 27y^3 = (2x + 3y)\left((2x)^2 - (2x)(3y) + (3y)^2\right)$$
Step 3: Simplify the terms inside the second parentheses:
$$= (2x + 3y)(4x^2 - 6xy + 9y^2)$$
Answer: $$(2x + 3y)(4x^2 - 6xy + 9y^2)$$
Question 6 (Part ii): Factorize $64a^3 - 125b^3$ using the difference of cubes formula.
Solution:
Step 1: Express terms in cubic form:
$$64a^3 = (4a)^3 \quad \text{and} \quad 125b^3 = (5b)^3$$
Step 2