OP Malhotra Class 9 Maths Solutions Chapter 4 Factorisation Exercise 4 (B)

Factorise the Following:

Question 1. \( bx + 2b + cx + 2c \)
Answer: First, we group terms that have common factors. Group \( bx \) with \( cx \) and \( 2b \) with \( 2c \).
\( bx + 2b + cx + 2c \)
\( = bx + cx + 2b + 2c \)
Next, factor out the common term from each group. From \( bx + cx \), factor out \( x \). From \( 2b + 2c \), factor out \( 2 \).
\( = x(b + c) + 2 (b + c) \)
Now, we see that \( (b + c) \) is a common factor in both parts. So, we factor out \( (b + c) \).
\( = (b + c) (x + 2) \)
In simple words: We rearrange the terms, take out common factors from pairs of terms, and then take out the common bracket to get the final factorized form.

🎯 Exam Tip: Always look for common factors by grouping terms. Rearranging terms might be necessary to find suitable pairs.

Question 2. \( y^3 + 2y^2 + 3y + 6 \)
Answer: We begin by grouping the terms in pairs to find common factors. Group \( y^3 \) with \( 2y^2 \) and \( 3y \) with \( 6 \).
\( y^3 + 2y^2 + 3y + 6 \)
From the first pair, \( y^3 + 2y^2 \), we can factor out \( y^2 \).
\( = y^2(y + 2) + 3y + 6 \)
From the second pair, \( 3y + 6 \), we can factor out \( 3 \).
\( = y^2(y + 2) + 3(y + 2) \)
Now, we observe that \( (y + 2) \) is a common factor in both terms. We factor it out.
\( = (y + 2) (y^2 + 3) \)
In simple words: We split the expression into two groups. We take out what's common in each group, and if the leftover parts match, we pull that matching part out too.

🎯 Exam Tip: Remember to check if the terms in the brackets are identical after the first step of factoring. If they are, you're on the right track.

Question 3. \( xa + 3b + xb + 3a \)
Answer: First, we rearrange the terms to group common factors together. Let's group terms with \( x \) and terms with \( 3 \).
\( xa + 3b + xb + 3a \)
\( = xa + xb + 3a + 3b \)
Now, we factor out \( x \) from the first two terms and \( 3 \) from the last two terms.
\( = x (a + b) + 3 (a + b) \)
We can see that \( (a + b) \) is a common factor for both parts. So, we factor it out.
\( = (a + b) (x + 3) \)
In simple words: We put similar parts next to each other, take out what's common from each pair, and then combine the parts that are now the same.

🎯 Exam Tip: When rearranging, try to find combinations that will result in identical bracketed expressions for further factoring.

Question 4. \( 8xy + 5zy - 8xt - 5zt \)
Answer: We start by rearranging the terms to put similar groups together for easy factoring. Group the terms with \( 8x \) and the terms with \( 5z \).
\( 8xy + 5zy - 8xt - 5zt \)
\( = 8xy - 8xt + 5zy - 5zt \)
Next, factor out \( 8x \) from the first two terms and \( 5z \) from the last two terms.
\( = 8x (y - t) + 5z (y - t) \)
Now, \( (y - t) \) is a common factor in both terms, so we factor it out.
\( = (y - t) (8x + 5z) \)
In simple words: We move the terms around to make pairs that share something. Then, we take out the shared part from each pair. Finally, we take out the part that is now common to both results.

🎯 Exam Tip: Pay close attention to negative signs when factoring. If you factor out a negative number, the signs inside the bracket will flip.

Question 5. \( 8kl + 12ml - 12mn - 8kn \)
Answer: To factor this expression, we first rearrange the terms to group those with common factors.
\( 8kl + 12ml - 12mn - 8kn \)
Group \( 8kl \) with \( -8kn \) and \( 12ml \) with \( -12mn \).
\( = 8kl - 8kn + 12ml - 12mn \)
Now, factor out \( 8k \) from the first pair and \( 12m \) from the second pair.
\( = 8k (l - n) + 12m (l - n) \)
We can see that \( (l - n) \) is a common factor to both terms. So, we factor it out.
\( = (l - n) (8k + 12m) \)
In simple words: We group the terms so that each pair has something in common. We pull out that common part from each pair. If what's left in the brackets is the same, we pull that out too.

🎯 Exam Tip: Always double-check your factoring by multiplying the final factors to ensure you get the original expression.

Question 6. \( 32 (x + y)^2 - 2x - 2y \)
Answer: We begin by looking for common factors in the expression. The last two terms \( -2x - 2y \) have a common factor of \( -2 \).
\( 32 (x + y)^2 - 2x - 2y \)
Factor \( -2 \) from \( -2x - 2y \).
\( = 32 (x + y)^2 - 2(x + y) \)
Now we see that \( (x + y) \) is a common factor in both terms. We can factor out \( 2(x + y) \).
\( = 2(x + y) [16(x + y) - 1] \)
Finally, distribute the \( 16 \) inside the bracket.
\( = 2 (x + y) (16x + 16y - 1) \)
In simple words: First, we find a common number and letter pair that can be taken out from some terms. Then, we see if the remaining parts also share something. We keep pulling out common factors until nothing else can be factored.

🎯 Exam Tip: Recognize that \( (x + y)^2 \) is the same as \( (x + y)(x + y) \), which helps in identifying common binomial factors.

Question 7. \( x^3 - x^2 + ax + x-a-1 \)
Answer: We start by rearranging the terms in a way that allows for grouping and factoring.
\( x^3 - x^2 + ax + x-a-1 \)
Rearrange the terms to group those sharing \( x-1 \).
\( = x^3 - x^2 + ax - a + x - 1 \)
Now, factor out common terms from each group: \( x^2 \) from \( x^3 - x^2 \), \( a \) from \( ax - a \), and \( 1 \) from \( x - 1 \).
\( = x^2 (x - 1) + a (x - 1) + 1 (x - 1) \)
Since \( (x - 1) \) is a common factor in all three terms, we factor it out.
\( = (x - 1) (x^2 + a + 1) \)
In simple words: We move the terms around to make groups, then take out common parts from each group. If the parts left in the brackets are all the same, we take that common bracket out to finish factoring.

🎯 Exam Tip: For expressions with many terms, look for patterns that can form common binomial factors, even if it requires rearranging terms.

Question 8. \( ab (c^2 + 1) + c (a^2 + b^2) \)
Answer: First, we expand the given expression by multiplying the terms.
\( ab (c^2 + 1) + c (a^2 + b^2) \)
\( = abc^2 + ab + a^2c + b^2c \)
Next, we rearrange the terms to group those with common factors. Group \( abc^2 \) with \( a^2c \) and \( b^2c \) with \( ab \).
\( = abc^2 + a^2c + b^2c + ab \)
Now, factor out \( ac \) from the first two terms and \( b \) from the last two terms.
\( = ac (bc + a) + b (bc + a) \)
We can see that \( (bc + a) \) is a common factor for both parts.
\( = (bc + a) (ac + b) \)
In simple words: We open the brackets, then rearrange the terms. We find common parts in groups of terms and pull them out. If a common bracket appears, we take that out too.

🎯 Exam Tip: Sometimes, expanding the terms first can reveal hidden common factors that are not obvious in the original form.

Question 9. \( a^3 - a^2 + xa + a - x - 1 \)
Answer: To factor this expression, we rearrange the terms to create groups that share common factors.
\( a^3 - a^2 + xa + a - x - 1 \)
Group \( a^3 \) with \( -a^2 \), \( xa \) with \( -x \), and \( a \) with \( -1 \).
\( = a^3 - a^2 + xa - x + a - 1 \)
Now, factor out common terms from each group: \( a^2 \) from \( a^3 - a^2 \), \( x \) from \( xa - x \), and \( 1 \) from \( a - 1 \).
\( = a^2 (a - 1) + x (a - 1) + 1 (a - 1) \)
Since \( (a - 1) \) is a common factor in all three terms, we factor it out.
\( = (a - 1) (a^2 + x + 1) \)
In simple words: We put the terms in a new order to find shared factors. Then, we take out the common parts from each small group. Finally, we take out the bracket that appears the same in all groups.

🎯 Exam Tip: When dealing with more than four terms, try to find a common binomial factor by grouping terms strategically, aiming for identical binomials.

Question 10. \( 6a^3b + 3a^2b^2 - 2a^2b - ab^2 \)
Answer: We begin by factoring out the greatest common factor from the first two terms and the last two terms.
\( 6a^3b + 3a^2b^2 - 2a^2b - ab^2 \)
From \( 6a^3b + 3a^2b^2 \), factor out \( 3a^2b \).
From \( -2a^2b - ab^2 \), factor out \( -ab \).
\( = 3a^2b (2a + b) - ab (2a + b) \)
Now, we notice that \( (2a + b) \) is a common factor in both terms. So, we factor it out.
\( = (2a + b) (3a^2b - ab) \)
Inside the second bracket, \( (3a^2b - ab) \), we can still factor out \( ab \).
\( = (2a + b) ab (3a - 1) \)
It's good practice to write the monomial factor first.
\( = ab (3a - 1) (2a + b) \)
In simple words: First, we find common parts in the beginning terms and the ending terms. Then, we take out a bigger common part from the two resulting groups. If there's still something to take out from one of the brackets, we do that too.

🎯 Exam Tip: Always perform a final check for common factors in the remaining expressions. Sometimes, you can factor further after the first step.

Question 11. \( a^3 + ab (1 - 2a) - 2b^2 \)
Answer: First, expand the term \( ab (1 - 2a) \).
\( a^3 + ab (1 - 2a) - 2b^2 \)
\( = a^3 + ab - 2a^2b - 2b^2 \)
Now, rearrange the terms to group them for factoring. Group \( a^3 \) with \( ab \) and \( -2a^2b \) with \( -2b^2 \).
\( = a^3 - 2a^2b + ab - 2b^2 \)
Factor out \( a \) from \( a^3 - 2a^2b \) and \( b \) from \( ab - 2b^2 \).
\( = a(a^2 - 2ab) + b(a - 2b) \)
This grouping doesn't lead to a common bracket immediately. Let's try another rearrangement: group \( a^3 \) with \( -2a^2b \) and \( ab \) with \( -2b^2 \).
\( = a^3 - 2a^2b + ab - 2b^2 \)
This is the same. Let's re-examine the original solution structure.
The provided solution uses:
\( a^3 + ab (1 - 2a) - 2b^2 \)
\( = a^3 + ab - 2a^2b - 2b^2 \)
Then it groups as \( a^3 + ab \) and \( -2a^2b - 2b^2 \).
\( = a(a^2 + b) - 2b(a^2 + b) \)
We can see that \( (a^2 + b) \) is a common factor.
\( = (a^2 + b) (a - 2b) \)
In simple words: We first open up the bracket. Then, we group the terms that share something in common. We take out the common part from each group. If the parts inside the brackets are now the same, we take that whole bracket out.

🎯 Exam Tip: Don't be afraid to try different groupings if your first attempt doesn't lead to a common binomial factor. Expanding first is often a good strategy.

Question 12. \( (p^2 + 1)q - p^2 - q^2 \)
Answer: First, expand the term \( (p^2 + 1)q \).
\( (p^2 + 1)q - p^2 - q^2 \)
\( = p^2q + q - p^2 - q^2 \)
Next, rearrange the terms to group those with common factors. Group \( p^2q \) with \( -p^2 \) and \( q \) with \( -q^2 \).
\( = p^2q - p^2 + q - q^2 \)
Now, factor out \( p^2 \) from the first pair and \( q \) from the second pair.
\( = p^2 (q - 1) - q (q - 1) \)
We observe that \( (q - 1) \) is a common factor in both terms.
\( = (q - 1) (p^2 - q) \)
In simple words: We first multiply out the bracket. Then, we put terms that look alike next to each other. We take out what's common from each pair, and if we get the same bracket, we pull that out too.

🎯 Exam Tip: When dealing with terms like \( -p^2 - q^2 \), remember that you can factor out \( -1 \) to potentially create a useful bracket, or group them carefully to find positive common factors.