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

Question 1. Factorise: \( a^2 + 5a + 6 \)
Answer:
\( a^2 + 5a + 6 \)
\( = a^2 + 2a + 3a + 6 \)
(We split 5a into 2a and 3a because \( 2 \times 3 = 6 \) and \( 2 + 3 = 5 \).)
\( = a(a + 2) + 3(a + 2) \)
\( = (a + 2)(a + 3) \)
Factoring trinomials like this helps simplify algebraic expressions, making them easier to work with in equations or other mathematical problems.
In simple words: We need to break down the middle term (5a) into two parts (2a and 3a) so we can group parts of the expression and find common factors. This way, we change a long sum into a shorter multiplication.

🎯 Exam Tip: When factoring trinomials of the form \(x^2 + bx + c\), always look for two numbers that multiply to 'c' and add up to 'b'.

Question 2. Factorise: \( a^2 + 6a + 8 \)
Answer:
\( a^2 + 6a + 8 \)
\( = a^2 + 2a + 4a + 8 \)
(We split 6a into 2a and 4a because \( 2 \times 4 = 8 \) and \( 2 + 4 = 6 \).)
\( = a(a + 2) + 4(a + 2) \)
\( = (a + 2)(a + 4) \)
This method is often called splitting the middle term because we break the middle term into two parts that help us find common factors.
In simple words: Just like before, we split the middle number (6) into two numbers (2 and 4) that multiply to 8. Then we group the terms and take out what's common to get the answer.

🎯 Exam Tip: Ensure that the signs of the numbers you choose are correct for both multiplication and addition, especially when negative numbers are involved.

Question 3. Factorise: \( p^2 + 10p + 16 \)
Answer:
\( p^2 + 10p + 16 \)
\( = p^2 + 2p + 8p + 16 \)
(We split 10p into 2p and 8p because \( 2 \times 8 = 16 \) and \( 2 + 8 = 10 \).)
\( = p(p + 2) + 8(p + 2) \)
\( = (p + 2)(p + 8) \)
This technique works well for quadratic trinomials where the leading coefficient is 1, simplifying the process of finding the two numbers.
In simple words: We need two numbers that multiply to 16 and add up to 10. These numbers are 2 and 8. Using them, we group the terms to find the final two brackets.

🎯 Exam Tip: Clearly show the grouping step, as it demonstrates your understanding of extracting common factors.

Question 4. Factorise: \( a^2 + 13a + 42 \)
Answer:
\( a^2 + 13a + 42 \)
\( = a^2 + 7a + 6a + 42 \)
(We split 13a into 7a and 6a because \( 7 \times 6 = 42 \) and \( 7 + 6 = 13 \).)
\( = a(a + 7) + 6(a + 7) \)
\( = (a + 7)(a + 6) \)
Recognising the correct pair of factors for the constant term is crucial to successfully splitting the middle term.
In simple words: We look for two numbers that multiply to 42 and add up to 13. Once we find 6 and 7, we can rewrite the middle term and factor by grouping.

🎯 Exam Tip: Always double-check your factoring by multiplying the binomials back together to see if you get the original expression.

Question 5. Factorise: \( a^2 + 25a - 54 \)
Answer:
\( a^2 + 25a - 54 \)
(We split 25a into 27a and -2a because \( 27 \times (-2) = -54 \) and \( 27 - 2 = 25 \).)
\( = a(a + 27) - 2(a + 27) \)
\( = (a + 27)(a - 2) \)
When the constant term is negative, the two numbers you pick must have opposite signs, and their difference will give the middle term's coefficient.
In simple words: For a negative number like -54, one number must be positive and the other negative. We found 27 and -2 work because they add to 25 and multiply to -54. Then, we factor by grouping.

🎯 Exam Tip: If the constant term is negative, the two numbers must have opposite signs. The sign of the middle term will tell you which number is larger in absolute value.

Question 6. Factorise: \( x^2 + 5x - 176 \)
Answer:
\( x^2 + 5x - 176 \)
\( = x^2 + 16x - 11x - 176 \)
(We split 5x into 16x and -11x because \( 16 \times (-11) = -176 \) and \( 16 - 11 = 5 \).)
\( = x(x + 16) - 11(x + 16) \)
\( = (x + 16)(x - 11) \)
Finding factors for larger numbers like 176 might require a bit more trial and error, but remembering that one factor is positive and one is negative helps.
In simple words: We need two numbers that multiply to -176 and add up to 5. We use 16 and -11 to split the middle term, then group the parts to factor the expression.

🎯 Exam Tip: For larger numbers, systematically list factor pairs of the constant term and check their sums/differences until you find the correct pair.

Question 7. Factorise: \( y^2 - 18y + 65 \)
Answer:
\( y^2 - 18y + 65 \)
\( = y^2 - 13y - 5y + 65 \)
(We split -18y into -13y and -5y because \( (-13) \times (-5) = 65 \) and \( (-13) + (-5) = -18 \).)
\( = y(y - 13) - 5(y - 13) \)
\( = (y - 13)(y - 5) \)
When the constant term is positive and the middle term is negative, both factors used for splitting the middle term will be negative.
In simple words: We need two negative numbers that multiply to 65 and add to -18. These numbers are -13 and -5. We use them to rewrite the middle part and then factor by grouping.

🎯 Exam Tip: Pay close attention to the signs. If the constant term is positive and the middle term is negative, both factors must be negative.

Question 8. Factorise: \( m^2 - 29m + 204 \)
Answer:
\( m^2 - 29m + 204 \)
\( = m^2 - 17m - 12m + 204 \)
(We split -29m into -17m and -12m because \( (-17) \times (-12) = 204 \) and \( (-17) + (-12) = -29 \).)
\( = m(m - 17) - 12(m - 17) \)
\( = (m - 17)(m - 12) \)
For larger numbers like 204, it can be helpful to list out pairs of factors and check their sums to find the correct combination.
In simple words: Find two numbers that multiply to 204 and add up to -29. These numbers are -17 and -12. Use them to split the middle term and then factor the expression by grouping.

🎯 Exam Tip: If the numbers are large, try prime factorization of the constant term to efficiently find its factor pairs.

Question 9. Factorise: \( b^2 - 2b - 48 \)
Answer:
\( b^2 - 2b - 48 \)
\( = b^2 - 8b + 6b - 48 \)
(We split -2b into -8b and 6b because \( (-8) \times 6 = -48 \) and \( (-8) + 6 = -2 \).)
\( = b(b - 8) + 6(b - 8) \)
\( = (b - 8)(b + 6) \)
With a negative constant term, one factor will be positive and the other negative, and their sum should equal the middle term.
In simple words: We need two numbers that multiply to -48 and add to -2. These numbers are -8 and 6. After splitting the middle term, we can factor by grouping.

🎯 Exam Tip: Remember that the common factor you pull out from each pair must result in the same remaining binomial for successful grouping.

Question 10. Factorise: \( x^2 - 11x - 102 \)
Answer:
\( x^2 - 11x - 102 \)
\( = x^2 - 17x + 6x - 102 \)
(We split -11x into -17x and 6x because \( (-17) \times 6 = -102 \) and \( (-17) + 6 = -11 \).)
\( = x(x - 17) + 6(x - 17) \)
\( = (x - 17)(x + 6) \)
When factoring, it is useful to check that the two factors chosen not only multiply to the constant but also add up to the coefficient of the middle term.
In simple words: We need two numbers that multiply to -102 and add up to -11. We find these numbers are -17 and 6. Using them to split the middle term helps us factor the expression easily.

🎯 Exam Tip: A quick mental check of the multiplication and addition of your chosen factors can save time and prevent errors.

Question 11. Factorise: \( 3 - 4t + t^2 \)
Answer:
\( 3 - 4t + t^2 \)
\( = 3 - 3t - t + t^2 \)
(We split -4t into -3t and -t because \( (-3) \times (-1) = 3 \) and \( (-3) + (-1) = -4 \).)
\( = 3(1 - t) - t(1 - t) \)
\( = (1 - t)(3 - t) \)
Even if the terms are not in the standard \(at^2 + bt + c\) order, rearranging them mentally or on paper can make factoring simpler.
In simple words: We need two numbers that multiply to 3 and add up to -4. These numbers are -3 and -1. We use them to break down the middle term, then group and factor out common parts.

🎯 Exam Tip: If the terms are not in standard form \(ax^2+bx+c\), it's often helpful to rearrange them before attempting to factor.

Question 12. Factorise: \( 51 - 20k + k^2 \)
Answer:
\( 51 - 20k + k^2 \)
\( = 51 - 17k - 3k + k^2 \)
(We split -20k into -17k and -3k because \( (-17) \times (-3) = 51 \) and \( (-17) + (-3) = -20 \).)
\( = 17(3 - k) - k(3 - k) \)
\( = (3 - k)(17 - k) \)
The order of terms in the factors (e.g., \(3-k\) versus \(k-3\)) matters for consistency, so pay attention to the signs when extracting common factors.
In simple words: We are looking for two numbers that multiply to 51 and add up to -20. We find that -17 and -3 work. After splitting the middle term, we factor by taking out common parts.

🎯 Exam Tip: Be careful with signs when factoring out a negative term; for example, \( -k(3-k) \) ensures the sign inside the bracket remains correct.

Question 13. Factorise: \( 2x^2 - 10x + 12 \)
Answer:
\( 2x^2 - 10x + 12 \)
\( = 2(x^2 - 5x + 6) \)
(First, we take out the common factor 2.)
\( = 2[x^2 - 2x - 3x + 6] \)
(Now, we split the middle term inside the bracket: \( (-2) \times (-3) = 6 \) and \( (-2) + (-3) = -5 \).)
\( = 2[x(x - 2) - 3(x - 2)] \)
\( = 2(x - 2)(x - 3) \)
When a common factor exists for all terms in a polynomial, always factor it out first to simplify the expression before applying other factoring techniques.
In simple words: First, we take out the common number 2 from all parts. Then, inside the bracket, we find two numbers that multiply to 6 and add to -5, which are -2 and -3. Finally, we group and factor what's left.

🎯 Exam Tip: Always look for the greatest common factor (GCF) at the start of any factoring problem. This simplifies the expression and makes subsequent steps easier.

Question 14. Factorise: \( 3x^3 - 33x^2 + 84x \)
Answer:
\( 3x^3 - 33x^2 + 84x \)
\( = 3x[x^2 - 11x + 28] \)
(First, we take out the common factor \( 3x \).)
\( = 3x[x^2 - 7x - 4x + 28] \)
(Now, we split the middle term inside the bracket: \( (-7) \times (-4) = 28 \) and \( (-7) + (-4) = -11 \).)
\( = 3x[x(x - 7) - 4(x - 7)] \)
\( = 3x(x - 7)(x - 4) \)
Factoring out the greatest common factor at the beginning significantly reduces the complexity of the polynomial, making subsequent steps easier.
In simple words: First, we pull out the common factor \(3x\) from all terms. Inside the remaining bracket, we find two numbers that multiply to 28 and add up to -11. These are -7 and -4. We then use them to factor by grouping.

🎯 Exam Tip: Always ensure that the common factor (like \(3x\)) is included in your final factored answer.

Question 15. Factorise: \( 5y^2 - 45y - 110 \)
Answer:
\( 5y^2 - 45y - 110 \)
\( = 5(y^2 - 9y - 22) \)
(First, we take out the common factor 5.)
\( = 5[y^2 - 11y + 2y - 22] \)
(Now, we split the middle term inside the bracket: \( (-11) \times 2 = -22 \) and \( (-11) + 2 = -9 \).)
\( = 5[y(y - 11) + 2(y - 11)] \)
\( = 5(y - 11)(y + 2) \)
The process of factoring out a common number first helps in dealing with larger coefficients and makes the quadratic expression inside the bracket simpler.
In simple words: We start by taking out the number 5 which is common to all terms. Then, for the expression inside the bracket, we find two numbers that multiply to -22 and add up to -9. These are -11 and 2. We use them to finish factoring.

🎯 Exam Tip: Remember to carry the common factor (5 in this case) through all steps of the solution until the final answer.

Question 16. Factorise: \( x^4 - 13x^2 + 36 \)
Answer:
\( x^4 - 13x^2 + 36 \)
\( = x^4 - 9x^2 - 4x^2 + 36 \)
(We split \(-13x^2\) into \(-9x^2\) and \(-4x^2\) because \( (-9) \times (-4) = 36 \) and \( (-9) + (-4) = -13 \).)
\( = x^2(x^2 - 9) - 4(x^2 - 9) \)
\( = (x^2 - 9)(x^2 - 4) \)
\( = (x^2 - 3^2)(x^2 - 2^2) \)
(Both terms are a difference of squares, \( a^2 - b^2 = (a-b)(a+b) \).)
\( = (x + 3)(x - 3)(x + 2)(x - 2) \)
This type of expression, which is a quadratic in form, often factors into a difference of squares after the initial trinomial factoring step, leading to four linear factors.
In simple words: We treat \(x^2\) like a single variable. We find two numbers that multiply to 36 and add to -13, which are -9 and -4. After factoring by grouping, we see two "difference of squares" patterns, \(x^2 - 9\) and \(x^2 - 4\), which we factor further into two pairs of brackets each.

🎯 Exam Tip: Always look for further factoring opportunities, especially the "difference of squares" pattern, until the expression is fully factored into its simplest parts.

Question 17. Factorise: \( x^2 + 3xy - 88y^2 \)
Answer:
\( x^2 + 3xy - 88y^2 \)
\( = x^2 + 11xy - 8xy - 88y^2 \)
(We split \( 3xy \) into \( 11xy \) and \( -8xy \) because \( 11 \times (-8) = -88 \) and \( 11 - 8 = 3 \).)
\( = x(x + 11y) - 8y(x + 11y) \)
\( = (x + 11y)(x - 8y) \)
When dealing with two variables, the middle term splitting still applies, but the factors will also include the second variable.
In simple words: We need two terms that multiply to \(-88y^2\) and add up to \(3xy\). We use \(11xy\) and \(-8xy\) to split the middle term. Then, we group the terms and factor out the common parts to get the final answer.

🎯 Exam Tip: For trinomials with two variables, ensure that your chosen factors account for the variable parts of both the middle and constant terms.

Question 18. Factorise: \( x^4 - x^2y^2 - 72y^4 \)
Answer:
\( x^4 - x^2y^2 - 72y^4 \)
\( = x^4 - 9x^2y^2 + 8x^2y^2 - 72y^4 \)
(We split \(-x^2y^2\) into \(-9x^2y^2\) and \(8x^2y^2\) because \( (-9) \times 8 = -72 \) and \( (-9) + 8 = -1 \).)
\( = x^2(x^2 - 9y^2) + 8y^2(x^2 - 9y^2) \)
\( = (x^2 - 9y^2)(x^2 + 8y^2) \)
\( = (x^2 - (3y)^2)(x^2 + 8y^2) \)
(The term \( x^2 - 9y^2 \) is a difference of squares.)
\( = (x + 3y)(x - 3y)(x^2 + 8y^2) \)
When expressions have even powers and multiple variables, substitute \(a = x^2\) and \(b = y^2\) mentally to simplify the view to a standard quadratic form, like \(a^2 - ab - 72b^2\).
In simple words: We treat \(x^2\) and \(y^2\) as basic units. We find two terms that multiply to \(-72y^4\) and add to \(-x^2y^2\). These are \(-9x^2y^2\) and \(8x^2y^2\). After factoring by grouping, we see a "difference of squares" pattern, which we factor further.

🎯 Exam Tip: Recognize patterns like "quadratic in form" (\(a x^{2n} + b x^n + c\)) where substitution can simplify the initial factoring process.

Question 19. Factorise: \( a^3b^3 - 9a^2b^2 + 20ab \)
Answer:
\( a^3b^3 - 9a^2b^2 + 20ab \)
\( = ab[a^2b^2 - 9ab + 20] \)
(First, we take out the common factor \( ab \).)
\( = ab[a^2b^2 - 4ab - 5ab + 20] \)
(Now, we split the middle term inside the bracket: \( (-4) \times (-5) = 20 \) and \( (-4) + (-5) = -9 \).)
\( = ab[ab(ab - 4) - 5(ab - 4)] \)
\( = ab(ab - 4)(ab - 5) \)
When facing a polynomial with terms that are powers of a product (like \(ab\)), treat \(ab\) as a single variable, making the factoring process more straightforward.
In simple words: First, we take out the common factor \(ab\) from all terms. Then, inside the bracket, we need two numbers that multiply to 20 and add up to -9. These are -4 and -5. We use them to split the middle term and complete the factoring.

🎯 Exam Tip: For expressions where a common term appears in different powers (e.g., \( (ab)^3, (ab)^2, ab \)), consider substituting a single variable for the common term to simplify the appearance of the polynomial.

Question 20. Factorise: \( (x^2 + x)^2 + 4(x^2 + x) - 21 \)
Answer:
\( (x^2 + x)^2 + 4(x^2 + x) - 21 \)
Let \( x^2 + x = a \). Then, the expression becomes:
\( a^2 + 4a - 21 \)
\( = a^2 + 7a - 3a - 21 \)
(We split 4a into 7a and -3a because \( 7 \times (-3) = -21 \) and \( 7 - 3 = 4 \).)
\( = a(a + 7) - 3(a + 7) \)
\( = (a + 7)(a - 3) \)
Now, substitute back \( x^2 + x \) for \( a \):
\( = (x^2 + x + 7)(x^2 + x - 3) \)
Using a substitution like \(a = x^2 + x\) simplifies complex expressions into a more familiar quadratic form, making factoring much easier before substituting back the original expression.
In simple words: We can make this problem easier by calling \(x^2 + x\) simply 'a'. Then we factor the simpler expression with 'a'. After finding the factors, we put \(x^2 + x\) back where 'a' was to get the final answer.

🎯 Exam Tip: When an expression contains a repeated complex term, using a substitution variable (like 'a') can make it look like a simpler quadratic, which is easier to factor.