Math Notes For Class 8 Chapter 4 Ex 4.1 Solved & Quiz

Question 1 (Part i): Factorize completely: $2ab + 4ac + 8ad$

Solution:
Identify the highest common factor (HCF) among the coefficients $2, 4, 8$, which is $2$.
Identify the common variable term present in all terms, which is $a$.
Extract $2a$ as a common factor:
$$2ab + 4ac + 8ad = 2a\left(\frac{2ab}{2a} + \frac{4ac}{2a} + \frac{8ad}{2a}\right)$$ $$= 2a(b + 2c + 4d)$$
Answer: $$2a(b + 2c + 4d)$$

Question 1 (Part ii): Factorize completely: $9x^2y - 12xy^2 + 15xyz$

Solution:
Determine the HCF of numerical coefficients $9, 12,$ and $15$, which is $3$.
Determine the common variable terms: $x$ and $y$.
Factor out $3xy$ from each term:
$$9x^2y - 12xy^2 + 15xyz = 3xy(3x - 4y + 5z)$$
Answer: $$3xy(3x - 4y + 5z)$$

Question 1 (Part iii): Factorize completely: $-6a^2b - 9ab^2 + 3ab$

Solution:
Determine the HCF of $-6, -9,$ and $3$, taking out a negative common factor $-3ab$ to simplify the remaining expression:
$$-6a^2b - 9ab^2 + 3ab = -3ab\left(\frac{-6a^2b}{-3ab} + \frac{-9ab^2}{-3ab} + \frac{3ab}{-3ab}\right)$$ $$= -3ab(2a + 3b - 1)$$
Answer: $$-3ab(2a + 3b - 1)$$

Question 2 (Part i): Factorize by grouping terms: $ax - ay + bx - by$

Solution:
Group the terms into two pairs:
$$(ax - ay) + (bx - by)$$ Factor out common elements from each group:
$$a(x - y) + b(x - y)$$ Take out the common binomial factor $(x - y)$:
$$= (x - y)(a + b)$$
Answer: $$(x - y)(a + b)$$

Question 2 (Part ii): Factorize by grouping terms: $2x^2 + 6x + xy + 3y$

Solution:
Group the expression into two logical pairs:
$$(2x^2 + 6x) + (xy + 3y)$$ Factor out $2x$ from the first pair and $y$ from the second pair:
$$2x(x + 3) + y(x + 3)$$ Extract the common binomial factor $(x + 3)$:
$$= (x + 3)(2x + y)$$
Answer: $$(x + 3)(2x + y)$$

Question 2 (Part iii): Factorize by grouping terms: $p^2 - pq - 5p + 5q$

Solution:
Group the terms:
$$(p^2 - pq) - (5p - 5q)$$ Factor out $p$ from the first group and $-5$ from the second group:
$$p(p - q) - 5(p - q)$$ Extract the common factor $(p - q)$:
$$= (p - q)(p - 5)$$
Answer: $$(p - q)(p - 5)$$

Question 3 (Part i): Factorize using the difference of two squares identity: $16x^2 - 49y^2$

Solution:
Write each term as a perfect square:
$$16x^2 = (4x)^2 \quad \text{and} \quad 49y^2 = (7y)^2$$ Apply identity $a^2 - b^2 = (a - b)(a + b)$ where $a = 4x$ and $b = 7y$:
$$(4x)^2 - (7y)^2 = (4x - 7y)(4x + 7y)$$
Answer: $$(4x - 7y)(4x + 7y)$$

Question 3 (Part ii): Factorize using algebraic identities: $25a^2b^2 - 81c^2$

Solution:
Rewrite both terms as perfect squares:
$$25a^2b^2 = (5ab)^2 \quad \text{and} \quad 81c^2 = (9c)^2$$ Applying $a^2 - b^2 = (a - b)(a + b)$ gives:
$$(5ab)^2 - (9c)^2 = (5ab - 9c)(5ab + 9c)$$
Answer: $$(5ab - 9c)(5ab + 9c)$$

Question 3 (Part iii): Factorize completely: $3x^3 - 75x$

Solution:
First, take out the overall common monomial factor $3x$:
$$3x^3 - 75x = 3x(x^2 - 25)$$ Next, recognize $x^2 - 25$ as a difference of two squares $x^2 - 5^2$:
$$x^2 - 25 = (x - 5)(x + 5)$$ Combine to obtain the complete factorization:
$$= 3x(x - 5)(x + 5)$$
Answer: $$3x(x - 5)(x + 5)$$

Question 4 (Part i): Factorize using complete square formulas: $x^2 + 10x + 25$

Solution:
Check if the expression fits $a^2 + 2ab + b^2$ form:
First term: $x^2 = (x)^2$
Last term: $25 = (5)^2$
Middle term check: $2(x)(5) = 10x$ (matches).
Applying $a^2 + 2ab + b^2 = (a + b)^2$ with $a = x$ and $b = 5$:
$$x^2 + 10x + 25 = (x + 5)^2$$
Answer: $$(x + 5)^2$$

Question 4 (Part ii): Factorize using complete square formulas: $4a^2 - 12ab + 9b^2$

Solution:
Check structure against $a^2 - 2ab + b^2$:
First term: $4a^2 = (2a)^2$
Last term: $9b^2 = (3b)^2$
Middle term check: $2(2a)(3b) = 12ab$ (matches with minus sign).
Using identity $a^2 - 2ab + b^2 = (a - b)^2$ with $a = 2a$ and $b = 3b$:
$$4a^2 - 12ab + 9b^2 = (2a - 3b)^2$$
Answer: $$(2a - 3b)^2$$

Question 4 (Part iii): Factorize using complete square formulas: $16y^2 + 56y + 49$

Solution:
First term: $16y^2 = (4y)^2$
Last term: $49 = (7)^2$
Middle term check: $2(4y)(7) = 56y$ (matches).
Applying $(a + b)^2 = a^2 + 2ab + b^2$:
$$16y^2 + 56y + 49 = (4y + 7)^2$$
Answer: $$(4y + 7)^2$$

Question 5 (Part i): Solve the quadratic equation by factorization: $x^2 - 9 = 0$

Solution:
Factorize the left side using the difference of two squares:
$$(x - 3)(x + 3) = 0$$ Set each factor to zero:
$$x - 3 = 0 \implies x = 3$$ $$x + 3 = 0 \implies x = -3$$
Answer: $$x = \pm 3$$

Question 5 (Part ii): Solve the equation: $2x^2 + 8x = 0$

Solution:
Factor out the common term $2x$:
$$2x(x + 4) = 0$$ Apply zero product property:
$$2x = 0 \implies x = 0$$ $$x + 4 = 0 \implies x = -4$$
Answer: $$x = 0, -4$$

Question 6 (Part i): The area of a rectangular garden is $(x^2 + 7x + 12)\text{ m}^2$. If its length is $(x + 4)\text{ m}$, find its width in terms of $x$.

Solution:
Formula: $\text{Area} = \text{Length} \times \text{Width}$
Therefore, $\text{Width} = \frac{\text{Area}}{\text{Length}} = \frac{x^2 + 7x + 12}{x + 4}$
Factorize the numerator $x^2 + 7x + 12$ by finding numbers that multiply to $12$ and sum to $7$ ($3$ and $4$):
$$x^2 + 7x + 12 = (x + 3)(x + 4)$$ Now substitute back:
$$\text{Width} = \frac{(x + 3)(x + 4)}{x + 4} = x + 3$$
Answer: $$(x + 3)\text{ m}$$

Question 6 (Part ii): Calculate the value of $x^2 - y^2$ if $x + y = 15$ and $x - y = 5$.

Solution:
Use the difference of two squares identity:
$$x^2 - y^2 = (x - y)(x + y)$$ Substitute the given values $x + y = 15$ and $x - y = 5$:
$$x^2 - y^2 = (5)(15) = 75$$
Answer: $$75$$