GSEB Class 8 Maths Solutions Chapter 14 Factorization InText Questions

Try These (Page 219)

Question 1. Factorise.
1. \( 12x + 36 \)
Answer:
We have \( 12x = 2 \times 2 \times 3 \times x \).
And \( 36 = 2 \times 2 \times 3 \times 3 \).
So, \( 12x + 36 = (2 \times 2 \times 3 \times x) + (2 \times 2 \times 3 \times 3) \).
We can extract the common factor \( (2 \times 2 \times 3) \) from both terms.
\( = (2 \times 2 \times 3)(x + 3) \)
\( = 12(x + 3) \)
In simple words: First, find the greatest common factor shared by both numbers in the expression. Then, write the expression as this common factor multiplied by the remaining parts inside parentheses.

Exam Tip: Always double-check your factorization by multiplying out the result to ensure it precisely matches the original expression.

Question 1. Factorise.
2. \( 22y - 33z \)
Answer:
We have \( 22y = 2 \times 11 \times y \).
And \( 33z = 3 \times 11 \times z \).
Therefore, \( 22y - 33z = (2 \times 11 \times y) - (3 \times 11 \times z) \).
We can take out the common factor \( 11 \).
\( = 11(2y - 3z) \)
In simple words: Identify the shared numerical factor between the two terms. Then, pull this common factor outside the parentheses, leaving the remaining parts inside.

Exam Tip: Look for both numerical and variable common factors. If the signs are different, ensure the common factor is extracted correctly.

Question 1. Factorise.
3. \( 14pq + 35pqr \)
Answer:
We have \( 14pq = 2 \times 7 \times p \times q \).
And \( 35pqr = 5 \times 7 \times p \times q \times r \).
Therefore, \( 14pq + 35pqr = (2 \times 7 \times p \times q) + (5 \times 7 \times p \times q \times r) \).
We can extract the common factor \( (7 \times p \times q) \).
\( = 7pq(2 + 5r) \)
In simple words: Find any numbers and letters that both parts of the expression share. Take these common elements outside, then write what's left inside the brackets.

Exam Tip: When finding common factors for terms with multiple variables, make sure to include all variables that are common to every term.

Try These (Page 225)

Question 1. Divide:
(i) \( 24xy^2z^3 \) by \( 6yz^2 \)
Answer:
We have the expression \( \frac{24xy^2z^3}{6yz^2} \).
Let's expand the terms into their prime factors and variables:
\( \frac{2 \times 2 \times 2 \times 3 \times x \times y \times y \times z \times z \times z}{2 \times 3 \times y \times z \times z} \)
Cancelling the common factors from the numerator and denominator, we obtain:
\( 2 \times 2 \times x \times y \times z \)
\( = 4xyz \)
In simple words: To divide, write the first term over the second as a fraction. Break down both parts into their prime numbers and individual variables. Then, cross out anything that appears on both the top and bottom. Multiply what's left to get the final answer.

Exam Tip: When dividing terms with exponents, remember to subtract the powers of identical bases. For example, \( y^2 / y = y^{2-1} = y \).

Question 1. Divide:
(ii) \( 63a^2b^4c^6 \) by \( 7a^2b^2c^3 \)
Answer:
We have the expression \( \frac{63a^2b^4c^6}{7a^2b^2c^3} \).
We can divide the numerical coefficients and variable terms separately.
\( \frac{63}{7} \times \frac{a^2}{a^2} \times \frac{b^4}{b^2} \times \frac{c^6}{c^3} \)
Simplifying each fraction using exponent rules:
\( = 9 \times a^{2-2} \times b^{4-2} \times c^{6-3} \)
\( = 9 \times a^0 \times b^2 \times c^3 \)
Since \( a^0 = 1 \), the expression becomes:
\( = 9 \times 1 \times b^2 \times c^3 \)
\( = 9b^2c^3 \)
In simple words: To divide algebraic terms, divide the numbers first. Then, for each letter, subtract the power of the bottom letter from the power of the top letter. If the powers are the same, the letter disappears.

Exam Tip: Any variable raised to the power of zero is equal to 1. Remember to apply this rule correctly when simplifying expressions.