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

Factorise:

Question 1. 8x + 16y
Answer: To factorise \( 8x + 16y \), we find the greatest common factor (GCF) of the terms. The GCF of 8 and 16 is 8. So, we can take 8 out as a common factor.
\( 8x + 16y = 8(x + 2y) \)
Factoring helps simplify expressions by showing common parts.
In simple words: Look for a number or letter that can divide all parts of the sum. Here, 8 divides both 8x and 16y. So, we pull 8 outside the bracket.

🎯 Exam Tip: Always double-check your factorization by multiplying the common factor back into the bracket to ensure it matches the original expression.

Question 2. ax - 2ay
Answer: To factorise \( ax - 2ay \), we look for a common factor in both terms. Here, 'a' is common in both \( ax \) and \( 2ay \). We take 'a' outside the brackets.
\( ax - 2ay = a(x - 2y) \)
This process makes the expression simpler to work with in further calculations.
In simple words: Both parts have 'a'. So, we take 'a' out, leaving what's left inside the bracket.

🎯 Exam Tip: Identify the common variables and constants across all terms before starting to factorize.

Question 3. 25xy – 50y
Answer: To factorise \( 25xy - 50y \), we find the greatest common factor (GCF). The numbers 25 and 50 both share 25. The variable 'y' is also common in both terms. So, the GCF is \( 25y \).
\( 25xy - 50y = 25y(x - 2) \)
This skill is fundamental for solving more complex algebraic equations.
In simple words: Both 25 and 50 can be divided by 25. Both terms have 'y'. So, \( 25y \) is the common part we pull out.

🎯 Exam Tip: Remember to include common variables like 'y' in your greatest common factor, not just the numbers.

Question 4. abc - ac
Answer: To factorise \( abc - ac \), we identify the common factors. Both terms, \( abc \) and \( ac \), contain 'a' and 'c'. So, \( ac \) is the greatest common factor.
\( abc - ac = ac(b - 1) \)
This method helps simplify polynomial expressions easily.
In simple words: Both parts have 'a' and 'c'. We take \( ac \) out. What's left from the first part is 'b', and from the second part (when \( ac \) is taken out) is 1.

🎯 Exam Tip: When an entire term is taken out as a common factor, always remember to leave '1' in its place inside the bracket.

Question 5. 121 - 11m²
Answer: To factorise \( 121 - 11m^2 \), we find the common factors. Both 121 and 11 can be divided by 11 (the H.C.F. of 121 and 11 is 11). So, we take 11 as the common factor.
\( 121 - 11m^2 = 11(11 - m^2) \)
This simplified form can be useful for further algebraic manipulations or finding roots.
In simple words: Both numbers, 121 and 11, can be divided by 11. So we take 11 out. We get 11 from 121, and \( m^2 \) from \( 11m^2 \).

🎯 Exam Tip: Ensure that you correctly identify the highest common factor for the numerical coefficients first, then for the variables.

Question 6. a²b + ab²
Answer: To factorise \( a^2b + ab^2 \), we look for common terms. The first term has \( a^2 \) and 'b', and the second term has 'a' and \( b^2 \). We can take 'a' and 'b' (specifically, the lowest power of each) as common factors.
\( a^2b + ab^2 = ab(a + b) \)
Factoring out common variables is a key step in simplifying algebraic expressions.
In simple words: Both parts have at least one 'a' and at least one 'b'. So, we take \( ab \) out of both. What's left from \( a^2b \) is 'a', and what's left from \( ab^2 \) is 'b'.

🎯 Exam Tip: When finding common factors for variables with powers, always choose the lowest power present in all terms (e.g., for \( a^2 \) and \( a \), the common factor is \( a \)).

Question 7. 24x³y – 30x²y
Answer: To factorise \( 24x^3y - 30x^2y \), we find the greatest common factor (GCF). The H.C.F. of 24 and 30 is 6. For variables, \( x^2 \) (lower power of x) and 'y' (lower power of y) are common. So the GCF is \( 6x^2y \).
\( 24x^3y - 30x^2y = 6x^2y(4x - 5) \)
Recognizing the GCF quickly helps in streamlining algebraic problem-solving.
In simple words: The biggest number that divides 24 and 30 is 6. The smallest \( x \) power is \( x^2 \), and 'y' is in both. So, we pull out \( 6x^2y \).

🎯 Exam Tip: When factoring out variables with different powers, take the variable with the smallest power that is common to all terms.

Question 8. – 9x³ – 12x² + 3x
Answer: To factorise \( -9x^3 - 12x^2 + 3x \), we look for the greatest common factor. The H.C.F. of 9, 12, and 3 is 3. Also, 'x' is common in all terms. Since the leading term is negative, it's often good practice to factor out a negative common factor, so we take out \( -3x \).
\( -9x^3 - 12x^2 + 3x = -3x(3x^2 + 4x - 1) \)
Factoring out a negative term changes the signs of the terms inside the parenthesis, making the expression simpler for future operations.
In simple words: All numbers (9, 12, 3) can be divided by 3, and all parts have at least one 'x'. Since the first number is negative, we take out \( -3x \). This flips the signs inside the bracket.

🎯 Exam Tip: If the first term of an expression is negative, it is usually best to factor out a negative common factor to make the expression inside the bracket start with a positive term.

Question 9. 25a²bc³ - 35ab²c³ + 45a³bc²
Answer: To factorise \( 25a^2bc^3 - 35ab^2c^3 + 45a^3bc^2 \), we identify the greatest common factor (GCF). The H.C.F. of 25, 35, and 45 is 5. For the variables, the lowest powers common to all terms are 'a', \( b^1 \) (or just b), and \( c^2 \). So, the GCF is \( 5abc^2 \).
\( 25a^2bc^3 - 35ab^2c^3 + 45a^3bc^2 = 5abc^2(5ac - 7bc + 9a^2) \)
Carefully identifying the lowest power of each variable across all terms is crucial for accurate factorization.
In simple words: Find the biggest number that divides 25, 35, and 45, which is 5. Then, take the smallest power of 'a', 'b', and 'c' that appears in all parts. Combine these to get \( 5abc^2 \) as the common factor.

🎯 Exam Tip: Be meticulous in checking the common factors for each variable across all terms, especially when dealing with multiple variables and powers.

Question 10. -x-y-z
Answer: To factorise \( -x - y - z \), we notice that all terms are negative. We can factor out -1 from each term.
\( -x - y - z = -1(x + y + z) \)
This is a fundamental step to make expressions with all negative terms easier to handle.
In simple words: Since all parts are minus, we can take out a minus one (\( -1 \)). This makes everything inside the bracket positive.

🎯 Exam Tip: When all terms in an expression are negative, factoring out -1 simplifies the expression and makes it easier to read and work with.

Question 11. x(x - y) + (x - y)
Answer: To factorise \( x(x - y) + (x - y) \), we can see that the term \( (x - y) \) is common in both parts. We can treat \( (x - y) \) as a single common factor.
\( x(x - y) + (x - y) = (x - y)(x + 1) \)
This technique, known as factoring by grouping, is very useful for more complex polynomials.
In simple words: The bracket \( (x - y) \) appears twice. We take it out as a common part. What's left is 'x' from the first part and '1' from the second part, which go into another bracket.

🎯 Exam Tip: When a bracketed expression is common in all terms, treat the entire bracket as a single common factor.

Question 12. (x + y) – (x + y)²
Answer: To factorise \( (x + y) - (x + y)^2 \), we observe that \( (x + y) \) is a common factor. We can take \( (x + y) \) out of both terms.
\( (x + y) - (x + y)^2 = (x + y) - (x + y)(x + y) \)
\( \implies (x + y)(1 - (x + y)) \)
\( \implies (x + y)(1 - x - y) \)
This shows how a common binomial factor can be extracted to simplify expressions.
In simple words: The group \( (x+y) \) is in both parts. We pull out one \( (x+y) \). From the first part, 1 is left. From the second part, another \( (x+y) \) is left, but it's subtracted.

🎯 Exam Tip: Remember to put parentheses around the remaining expression \( (1 - (x+y)) \) when factoring out a binomial, and then distribute the negative sign if one exists.

Question 13. a(x - y) – b(y – x)
Answer: To factorise \( a(x - y) - b(y - x) \), we first notice that \( (y - x) \) is the negative of \( (x - y) \). We can rewrite \( (y - x) \) as \( -(x - y) \).
So, \( a(x - y) - b(y - x) = a(x - y) - b(-(x - y)) \)
\( \implies a(x - y) + b(x - y) \)
Now, \( (x - y) \) is a common factor.
\( \implies (x - y)(a + b) \)
Understanding how to manipulate signs to create common factors is a crucial algebraic skill.
In simple words: Notice that \( (y-x) \) is the same as minus \( (x-y) \). So, change \( -b(y-x) \) to \( +b(x-y) \). Then, \( (x-y) \) is common, so pull it out.

🎯 Exam Tip: Always look for terms that are negatives of each other, like \( (y-x) \) and \( (x-y) \), as they can be transformed to create common factors.

Question 14. (a – b)⁴ + (a – b)³
Answer: To factorise \( (a - b)^4 + (a - b)^3 \), we find the common factor, which is \( (a - b) \) raised to the lowest power present, which is \( (a - b)^3 \).
\( (a - b)^4 + (a - b)^3 = (a - b)^3 [(a - b) + 1] \)
\( \implies (a - b)^3 (a - b + 1) \)
Factoring out a common binomial raised to a power is a direct application of GCF rules.
In simple words: Both parts have \( (a-b) \) raised to some power. We take out the smaller power, which is \( (a-b)^3 \). What's left from the first part is one \( (a-b) \), and from the second part, 1.

🎯 Exam Tip: When factoring expressions with common binomials raised to different powers, always factor out the common binomial with the lowest exponent.

Question 15. (x - y)⁶ + (y – x)⁵
Answer: To factorise \( (x - y)^6 + (y - x)^5 \), we first change \( (y - x) \) to \( -(x - y) \). Since \( (y - x)^5 \) has an odd power, it becomes \( (-(x - y))^5 = -(x - y)^5 \).
So, the expression becomes: \( (x - y)^6 - (x - y)^5 \)
Now, \( (x - y)^5 \) is a common factor.
\( \implies (x - y)^5((x - y) - 1) \)
\( \implies (x - y)^5(x - y - 1) \)
Understanding the properties of odd and even powers when changing signs is very important for this type of problem.
In simple words: \( (y-x) \) is the same as \( -(x-y) \). Because the power is 5 (an odd number), \( (y-x)^5 \) becomes \( -(x-y)^5 \). Then, we pull out the common part \( (x-y)^5 \).

🎯 Exam Tip: Remember that \( (b-a)^n = -(a-b)^n \) if 'n' is odd, and \( (b-a)^n = (a-b)^n \) if 'n' is even. This sign change is critical for correct factorization.