RBSE Solutions Class 8 Maths Chapter 10 Factorization Exercise 10.1

Question 1. Find the common factors of the given terms
(i) 12x, 36
(ii) 14pq, 28p²q²
(iii) 6abc, 24ab², 12a²b
(iv) 16x³, – 4x², 32x
(v) 10pq, 20qr, 30rp
(vi) 3x²y³, 10x²y², 6x²y²z
Answer:
(i) For 12x and 36:
First, we break down each term into its prime factors:
\( 12x = 2 \times 2 \times 3 \times x \)
\( 36 = 2 \times 2 \times 3 \times 3 \)
Now, we look for factors that appear in both expressions. The common factors are 2, 2, and 3.
\( \implies \) The common factor is \( 2 \times 2 \times 3 = 12 \).
(ii) For 14pq and 28p²q²:
Let's write down the prime factors for each term:
\( 14pq = 2 \times 7 \times p \times q \)
\( 28p^2q^2 = 2 \times 2 \times 7 \times p \times p \times q \times q \)
Next, we find the factors that are present in both terms. Here, the common factors are 2, 7, p, and q.
\( \implies \) The common factor is \( 2 \times 7 \times p \times q = 14pq \).
(iii) For 6abc, 24ab², and 12a²b:
We list the prime factors for each expression:
\( 6abc = 2 \times 3 \times a \times b \times c \)
\( 24ab^2 = 2 \times 2 \times 2 \times 3 \times a \times b \times b \)
\( 12a^2b = 2 \times 2 \times 3 \times a \times a \times b \)
Now, we identify the factors that are common to all three terms. These are 2, 3, a, and b.
\( \implies \) The common factor is \( 2 \times 3 \times a \times b = 6ab \).
(iv) For 16x³, – 4x², and 32x:
Let's find the prime factors for each term:
\( 16x^3 = 2 \times 2 \times 2 \times 2 \times x \times x \times x \)
\( -4x^2 = (-1) \times 2 \times 2 \times x \times x \)
\( 32x = 2 \times 2 \times 2 \times 2 \times 2 \times x \)
The common factors among these terms are 2, 2, and x.
\( \implies \) The common factor is \( 2 \times 2 \times x = 4x \).
(v) For 10pq, 20qr, 30rp:
The answer for this part is not provided in the source material.
(vi) For 3x²y³, 10x²y², and 6x²y²z:
We break down each term into its prime factors:
\( 3x^2y^3 = 3 \times x \times x \times y \times y \times y \)
\( 10x^2y^2 = 2 \times 5 \times x \times x \times y \times y \)
\( 6x^2y^2z = 2 \times 3 \times x \times x \times y \times y \times z \)
The factors common to all three expressions are x, x, y, and y.
\( \implies \) The common factor is \( x \times x \times y \times y = x^2y^2 \).
In simple words: To find the common factor, break each term into its basic parts (prime numbers and letters). Then, find all the parts that are exactly the same in every term and multiply them together. That gives you the biggest factor they all share.

🎯 Exam Tip: Always make sure to consider both numerical coefficients and variables when finding common factors. Break them down completely to avoid missing any common elements.

Question 2. Factorize the following expressions. (by common factors method)
(i) 6p - 12q
(ii) 7a² + 14a
(iii) 10a² - 15b² + 20c²
(iv) ax²y + bxy² + cxyz
(v) x²yz + xy²z + xyz²
(vi) – 16z + 20z³
Answer:
(i) For \( 6p - 12q \):
We can see that both 6p and 12q have a common factor of 6. We take this common factor out.
\( 6p - 12q = 6(p - 2q) \)
(ii) For \( 7a^2 + 14a \):
In both terms, 7a² and 14a, the common factors are 7 and a. So, we factor out 7a.
\( 7a^2 + 14a = 7a(a + 2) \)
(iii) For \( 10a^2 - 15b^2 + 20c^2 \):
Here, the numbers 10, 15, and 20 all share a common factor of 5. We factor out 5 from the expression.
\( 10a^2 - 15b^2 + 20c^2 = 5(2a^2 - 3b^2 + 4c^2) \)
(iv) For \( ax^2y + bxy^2 + cxyz \):
Each term in the expression has \( x \) and \( y \) as common factors. We take out \( xy \).
\( ax^2y + bxy^2 + cxyz = xy(ax + by + cz) \)
(v) For \( x^2yz + xy^2z + xyz^2 \):
The terms \( x^2yz \), \( xy^2z \), and \( xyz^2 \) all have \( x \), \( y \), and \( z \) as common factors. We factor out \( xyz \). This makes the expression simpler.
\( x^2yz + xy^2z + xyz^2 = xyz(x + y + z) \)
(vi) For \( -16z + 20z^3 \):
We look for common factors in \( -16z \) and \( 20z^3 \). Both numbers are divisible by 4, and both terms have \( z \). So, we factor out \( 4z \).
\( -16z + 20z^3 = 4z(-4 + 5z^2) \)
In simple words: To factorize means to rewrite an expression as a product of its factors. Find what number or letter is common to all parts of the expression and pull it out, writing the rest inside brackets.

🎯 Exam Tip: Always double-check your factorization by multiplying the factors back out. If you get the original expression, your factorization is correct.

Question 3. Factorize (by regrouping method).
(i) 2xy + 3 + 2y + 3x
(ii) z – 7 – 7xy + xyz
(iii) 6xy - 4y + 6 - 9x
(iv) 15pq + 15 + 9q + 25p
Answer:
(i) For \( 2xy + 3 + 2y + 3x \):
First, we arrange the terms to group common factors together:
\( 2xy + 2y + 3x + 3 \)
Now, we factor out common terms from each pair:
\( = 2y(x + 1) + 3(x + 1) \)
Since \( (x + 1) \) is common, we can factor it out:
\( = (x + 1)(2y + 3) \)
(ii) For \( z - 7 - 7xy + xyz \):
We regroup the terms to find common factors:
\( z - 7 + xyz - 7xy \)
Factor out common terms from each pair:
\( = 1(z - 7) + xy(z - 7) \)
Since \( (z - 7) \) is common, we factor it out:
\( = (z - 7)(1 + xy) \)
(iii) For \( 6xy - 4y + 6 - 9x \):
Regroup the terms to identify common factors:
\( 6xy - 4y - 9x + 6 \)
Factor out common terms from each pair. Note the sign change in the second pair:
\( = 2y(3x - 2) - 3(3x - 2) \)
Now, factor out the common term \( (3x - 2) \):
\( = (3x - 2)(2y - 3) \)
(iv) For \( 15pq + 15 + 9q + 25p \):
First, we rearrange the terms for easier grouping:
\( 15pq + 9q + 25p + 15 \)
Factor common terms from the first two and last two terms:
\( = 3q(5p + 3) + 5(5p + 3) \)
Now, factor out the common bracket \( (5p + 3) \):
\( = (5p + 3)(3q + 5) \)
In simple words: Regrouping means to rearrange the terms in a way that you can easily find common factors in pairs or small groups. After finding common factors for the groups, you often find another common factor that lets you simplify the whole expression.

🎯 Exam Tip: When using the regrouping method, always look for terms that share common factors, and remember that sometimes you might need to factor out a negative number to make the brackets match.