RS Aggarwal Class 8 Mathematics Solutions Chapter 7 Factorisation

Exercise 7A

Question 1. Factorise 12x + 15
Answer: \( 3(4x + 5) \)
In simple words: Find the greatest common number that divides both 12x and 15. That number is 3. Take it out in front, and write what's left inside the brackets.

Exam Tip: Always identify the highest common factor first before writing the factored form.

Question 2. Factorise 14m - 21
Answer: \( 7(2m - 3) \)
In simple words: The biggest number that goes into both 14m and 21 is 7. Pull out the 7 and see what remains inside.

Exam Tip: Check your answer by expanding the brackets - you should get back the original expression.

Question 3. Factorise 9n - 12n²
Answer: \( 3n(3 - 4n) \)
In simple words: The common factor is 3n. Once you extract it, what's left is (3 - 4n).

Exam Tip: Look for common factors in both the coefficient and the variable parts.

Question 4. Factorise 16a² and 24ab is 8a
Answer: \( 16a^2 - 24ab = 8a(2a - 3b) \)
In simple words: The highest common factor of 16a² and 24ab is 8a. When you remove it, you get 2a - 3b left.

Exam Tip: For expressions with multiple variables, check the highest common factors of both the numbers and the variable parts separately.

Question 5. Factorise 15ab² and 20a²b is 5ab
Answer: \( 15ab^2 - 20a^2b = 5ab(3b - 4a) \)
In simple words: The highest common factor is 5ab. Pull it out, and what's left is 3b - 4a.

Exam Tip: When factoring terms with multiple variables, take out the smallest power of each variable present in all terms.

Question 6. Factorise 12x²y³ and 21x³y² is 3x²y²
Answer: \( 12x^2y^3 - 21x^3y^2 = 3x^2y^2(4y - 7x) \)
In simple words: The shared factor is 3x²y². Once you take it out, 4y - 7x remains inside the brackets.

Exam Tip: Verify by multiplying back: 3x²y² times (4y - 7x) should give your original expression.

Question 7. Factorise 9x³ - 6x² + 12x is 3x
Answer: \( 9x^3 - 6x^2 + 12x = 3x(3x^2 - 2x + 4) \)
In simple words: The common factor from all three terms is 3x. Remove it to get 3x² - 2x + 4 inside.

Exam Tip: With three or more terms, always extract the greatest common factor before attempting other factoring methods.

Question 8. Factorise 8x³ - 72xy + 12x is 4x
Answer: \( 8x^3 - 72xy + 12x = 4x(2x^2 - 18y + 3) \)
In simple words: The common factor is 4x. Once you factor it out, you're left with 2x² - 18y + 3.

Exam Tip: Don't forget to include all terms when factoring - missing even one changes your answer completely.

Question 9. Factorise 18a³b³ and 27a²b³ and 36a³b² is 9a²b²
Answer: \( 18a^3b^3 - 27a^2b^3 + 36a^3b^2 = 9a^2b^2(2ab - 3b + 4a) \)
In simple words: The highest common factor across all three terms is 9a²b². Extract it to get 2ab - 3b + 4a remaining.

Exam Tip: Always find the greatest common factor of the coefficients and use the lowest power of each variable that appears in all terms.

Question 10. Factorise (x + 5)² - 4(x + 5)
Answer: We recognize that (x + 5) is a common term.
\( (x + 5)^2 - 4(x + 5) = (x + 5)[(x + 5) - 4] = (x + 5)(x + 1) \)
In simple words: Treat (x + 5) like a single block. It appears in both parts, so factor it out. Then simplify what's left: (x + 5) - 4 = x + 1.

Exam Tip: When an entire bracket appears more than once, treat it as a single common factor just like a variable would be.

Question 11. Factorise 3(a - 2b)² - 5(a - 2b)
Answer: \( 3(a - 2b)^2 - 5(a - 2b) = (a - 2b)[3(a - 2b) - 5] = (a - 2b)(3a - 6b - 5) \)
In simple words: The bracket (a - 2b) is common to both terms. Extract it, then work out what stays: 3(a - 2b) - 5 simplifies to 3a - 6b - 5.

Exam Tip: Simplify the remaining expression inside the brackets after factoring to get your final form.

Question 12. Factorise 2a + 6b - 3(a + 3b)²
Answer: We have:
\( 2a + 6b - 3(a + 3b)^2 = (a + 3b)[2 - 3(a + 3b)] = (a + 3b)(2 - 3a - 9b) \)
In simple words: You can write 2a + 6b as 2(a + 3b). Now (a + 3b) is common to both parts. Pull it out and see what remains: 2 - 3(a + 3b), which becomes 2 - 3a - 9b.

Exam Tip: When terms don't look obviously related, try to rewrite them so a common factor emerges.

Question 13. Factorise 16(2p - 3q)² - 4(2p - 3q)
Answer: \( 16(2p - 3q)^2 - 4(2p - 3q) = (2p - 3q)[16(2p - 3q) - 4] = (2p - 3q)(32p - 48q - 4) \)
In simple words: The bracket (2p - 3q) is in both parts. Factor it out. Then simplify: 16(2p - 3q) - 4 gives 32p - 48q - 4.

Exam Tip: Always expand inside the brackets after factoring to verify your final answer is in simplest form.

Question 14. Factorise x(a - 3) + y(3 - a)
Answer: We recognize that (3 - a) = -(a - 3).
\( x(a - 3) + y(3 - a) = x(a - 3) - y(a - 3) = (a - 3)(x - y) \)
In simple words: The second bracket is (3 - a), which is the negative of (a - 3). Rewrite it, and now (a - 3) is common to both parts. Factor it out to get (a - 3)(x - y).

Exam Tip: Watch for brackets that are negatives of each other - flip the sign of one to reveal the common factor.

Question 15. Factorise 12(2x - 3y)² - 16(3y - 2x)
Answer: We see that (3y - 2x) = -(2x - 3y).
\( 12(2x - 3y)^2 - 16(3y - 2x) = 12(2x - 3y)^2 + 16(2x - 3y) = (2x - 3y)[12(2x - 3y) + 16] = (2x - 3y)(24x - 36y + 16) \)
In simple words: Rewrite (3y - 2x) as -(2x - 3y), which changes the minus sign. Now (2x - 3y) is common. Factor it out and simplify what's left.

Exam Tip: When you change the sign of a bracket, remember it affects the sign of the term in front - this is a common source of mistakes.

Question 16. Factorise (x + y)(2x + 5) - (x + 3)
Answer: We expand and reorganise:
\( (x + y)(2x + 5) - (x + y)(x + 3) = (x + y)[(2x + 5) - (x + 3)] = (x + y)(x + 2) \)
In simple words: Pull out the common bracket (x + y). What's left is (2x + 5) - (x + 3), which simplifies to x + 2.

Exam Tip: Be careful with subtraction - distribute the minus sign properly before simplifying.

Question 17. Factorise ar + br + at + bt
Answer: By grouping the terms:
\( ar + br + at + bt = (ar + br) + (at + bt) = r(a + b) + t(a + b) = (a + b)(r + t) \)
In simple words: Group the first two terms and the last two terms. Factor out r from the first pair and t from the second. Now (a + b) is common, so factor it out.

Exam Tip: This method is called factoring by grouping - look for common factors in pairs of terms.

Question 18. Factorise x² - ax - bx + ab
Answer: By arranging the terms suitably:
\( x^2 - ax - bx + ab = (x^2 - bx) - (ax - ab) = x(x - b) - a(x - b) = (x - b)(x - a) \)
In simple words: Group terms so that a common factor emerges. From the first pair, factor out x; from the second, factor out a. Then (x - b) is common to both.

Exam Tip: Rearrange terms to make grouping easier - the order doesn't matter as long as you follow the algebra correctly.

Question 19. Factorise ab² - bc² - ab + c²
Answer: By arranging the terms suitably:
\( ab^2 - bc^2 - ab + c^2 = (ab^2 - ab) - (bc^2 - c^2) = ab(b - 1) - c^2(b - 1) = (b - 1)(ab - c^2) \)
In simple words: Group and factor: the first pair gives ab(b - 1) and the second gives c²(b - 1). The bracket (b - 1) is now common.

Exam Tip: When grouping, make sure both groups yield a common factor - if they don't, try a different grouping.

Question 20. Factorise x² - xz + xy - yz
Answer: By arranging the terms suitably:
\( x^2 - xz + xy - yz = (x^2 + xy) - (xz + yz) = x(x + y) - z(x + y) = (x + y)(x - z) \)
In simple words: Reorder so pairs share a factor. From the first pair factor out x, from the second factor out z. The bracket (x + y) becomes common.

Exam Tip: Try different groupings if the first attempt doesn't yield a common factor - persistence pays off.

Question 21. Factorise 6ab - b² + 12ac - 2bc
Answer: By arranging the terms suitably:
\( 6ab - b^2 + 12ac - 2bc = (6ab + 12ac) - (b^2 + 2bc) = 6a(b + 2c) - b(b + 2c) = (b + 2c)(6a - b) \)
In simple words: Group the first two terms and the last two terms. Factor 6a from the first pair and b from the second. The bracket (b + 2c) is now common.

Exam Tip: For four-term expressions, always try grouping first before looking for other patterns.

Question 22. Factorise (x - 2y)² + 4x - 8y
Answer: We have:
\( (x - 2y)^2 + 4x - 8y = (x - 2y)^2 + 4(x - 2y) = (x - 2y)[(x - 2y) + 4] = (x - 2y)(x - 2y + 4) \)
In simple words: Rewrite 4x - 8y as 4(x - 2y). Now (x - 2y) appears in both parts. Factor it out to get your answer.

Exam Tip: When linear terms don't factor easily, check if they are a multiple of a bracketed expression elsewhere in the polynomial.

Question 23. Factorise y² - xy(1 - x) - x³
Answer: We have:
\( y^2 - xy(1 - x) - x^3 = y^2 - xy + x^2y - x^3 = (y^2 - xy) + (x^2y - x^3) = y(y - x) + x^2(y - x) = (y - x)(y + x^2) \)
In simple words: Expand the middle term, then group. The first pair gives y(y - x) and the second gives x²(y - x). The bracket (y - x) is common.

Exam Tip: Always expand expressions first if they help you see groupings more clearly.

Question 24. Factorise (ax + by)² + (bx - ay)²
Answer: We have:
\( (ax + by)^2 + (bx - ay)^2 = (a^2x^2 + b^2y^2 + 2axby) + (b^2x^2 + a^2y^2 - 2bxay) = a^2x^2 + a^2y^2 + b^2x^2 + b^2y^2 + 2axby - 2axby = a^2(x^2 + y^2) + b^2(x^2 + y^2) = (x^2 + y^2)(a^2 + b^2) \)
In simple words: Expand both squares, combine like terms, and group by common factors. You'll find (x² + y²) is common to both groups.

Exam Tip: When expanding squares of brackets, remember the middle term has a coefficient of 2 - don't forget it.

Question 25. Factorise ab² + (a - 1)b - 1
Answer: We have:
\( ab^2 + (a - 1)b - 1 = ab^2 + ab - b - 1 = (ab^2 + ab) - (b + 1) = ab(b + 1) - 1(b + 1) = (b + 1)(ab - 1) \)
In simple words: Expand the middle term. Group the first two terms and the last two. Factor ab from the first pair and 1 from the second. The bracket (b + 1) is now common.

Exam Tip: Sometimes the second group factors out a number like 1 - don't skip this step.

Question 26. Factorise x³ - 3x² + x - 3
Answer: We have:
\( x^3 - 3x^2 + x - 3 = (x^3 - 3x^2) + (x - 3) = x^2(x - 3) + 1(x - 3) = (x - 3)(x^2 + 1) \)
In simple words: Group the first two terms and the last two. Factor out x² from the first pair and 1 from the second. The bracket (x - 3) is common.

Exam Tip: For four-term cubics, try grouping in pairs - it often reveals a nice factorisation.

Question 27. Factorise a²b(x + y) + 1(x + y)
Answer: We have:
\( a^2b(x + y) + 1(x + y) = (x + y)(a^2b + 1) \)
In simple words: The bracket (x + y) is already common to both terms. Simply pull it out.

Exam Tip: Sometimes the factorisation is already half-done for you - look for obvious common brackets first.

Question 28. Factorise ab(x² + y²) - xy(a² + b²)
Answer: We have:
\( ab(x^2 + y^2) - xy(a^2 + b^2) = abx^2 + aby^2 - xya^2 - xyb^2 = abx^2 - xya^2 + aby^2 - xyb^2 = ax(bx - ay) + by(ay - bx) = ax(bx - ay) - by(bx - ay) = (bx - ay)(ax - by) \)
In simple words: Expand everything, reorder, and group carefully. Factor ax from the first pair and by from the second (with a sign change). The bracket (bx - ay) becomes common.

Exam Tip: When you have a bracket and its negative, flip one side to reveal the common factor.

Exercise 7B

Question 1. Factorise x² - 36
Answer: This fits the difference of squares pattern \( a^2 - b^2 = (a + b)(a - b) \).
\( x^2 - 36 = x^2 - 6^2 = (x + 6)(x - 6) \)
In simple words: Both x² and 36 are perfect squares. Use the rule: difference of squares equals the sum times the difference.

Exam Tip: Always check if both terms are perfect squares before trying other methods.

Question 2. Factorise 4a² - 9
Answer: We recognise that:
\( 4a^2 - 9 = (2a)^2 - (3)^2 = (2a + 3)(2a - 3) \)
In simple words: 4a² is (2a)² and 9 is 3². So this is a difference of squares, giving (2a + 3)(2a - 3).

Exam Tip: Remember that the first term can also be a perfect square of an expression, not just a simple variable.

Question 3. Factorise 81 - 49x²
Answer: We have:
\( 81 - 49x^2 = (9)^2 - (7x)^2 = (9 + 7x)(9 - 7x) \)
In simple words: 81 = 9² and 49x² = (7x)². Apply the difference of squares rule.

Exam Tip: Don't forget that coefficients in front of variables can also be part of a perfect square.

Question 4. Factorise 4x² - 9y²
Answer: We have:
\( 4x^2 - 9y^2 = (2x)^2 - (3y)^2 = (2x + 3y)(2x - 3y) \)
In simple words: Both terms are perfect squares: 4x² = (2x)² and 9y² = (3y)². Use the difference of squares formula.

Exam Tip: When you have multiple variables, check each part separately to confirm it's a perfect square.

Question 5. Factorise 16a² - 225b²
Answer: We have:
\( 16a^2 - 225b^2 = (4a)^2 - (15b)^2 = (4a + 15b)(4a - 15b) \)
In simple words: 16a² = (4a)² and 225b² = (15b)². Apply the difference of squares pattern.

Exam Tip: Check your factors by expanding - (4a + 15b)(4a - 15b) should give you back the original.

Question 6. Factorise 9a²b² - 25
Answer: We have:
\( 9a^2b^2 - 25 = (3ab)^2 - (5)^2 = (3ab + 5)(3ab - 5) \)
In simple words: 9a²b² = (3ab)² and 25 = 5². This is a difference of squares.

Exam Tip: Products of variables can be grouped together as a single perfect square term.

Question 7. Factorise 16a² - 144
Answer: We have:
\( 16a^2 - 144 = (4a)^2 - (12)^2 = (4a + 12)(4a - 12) = 4(a + 3) \cdot 4(a - 3) = 16(a + 3)(a - 3) \)
In simple words: First use the difference of squares: (4a + 12)(4a - 12). Then notice each bracket has a common factor: (4a + 12) = 4(a + 3) and (4a - 12) = 4(a - 3). So the result is 16(a + 3)(a - 3).

Exam Tip: After factoring using difference of squares, always check if the resulting brackets can be factored further.

Question 8. Factorise 63a² - 112b²
Answer: We have:
\( 63a^2 - 112b^2 = 7(9a^2 - 16b^2) = 7[(3a)^2 - (4b)^2] = 7(3a + 4b)(3a - 4b) \)
In simple words: First, pull out the common factor 7. Then you have 9a² - 16b², which is (3a)² - (4b)². Apply difference of squares to get 7(3a + 4b)(3a - 4b).

Exam Tip: Always look for a common factor before applying other factoring methods - it makes the rest easier.

Question 9. Factorise 20a² - 45b²
Answer: First, extract the common factor:
\( 20a^2 - 45b^2 = 5(4a^2 - 9b^2) = 5[(2a)^2 - (3b)^2] = 5(2a + 3b)(2a - 3b) \)
In simple words: Pull out 5. What's left is 4a² - 9b² = (2a)² - (3b)², a difference of squares.

Exam Tip: Factoring out the GCF first often reveals a simpler difference of squares underneath.

Question 10. Factorise 20a² - 45b²
Answer: First, remove the common factor:
\( 20a^2 - 45b^2 = 5(4a^2 - 9b^2) = 5[(2a)^2 - (3b)^2] = 5(2a + 3b)(2a - 3b) \)
In simple words: Factor out 5 to get 5(4a² - 9b²). Then 4a² - 9b² is the difference of (2a)² and (3b)².

Exam Tip: Once you remove the GCF, the remaining expression often becomes much easier to factor.

Question 11. Factorise x³ - 64x
Answer: We have:
\( x^3 - 64x = x(x^2 - 64) = x[(x)^2 - (8)^2] = x(x + 8)(x - 8) \)
In simple words: Pull out x first. What's left is x² - 64 = x² - 8², a difference of squares.

Exam Tip: Always look for a monomial common factor before trying other techniques.

Question 12. Factorise 16x⁵ - 144x³
Answer: We have:
\( 16x^5 - 144x^3 = 16x^3(x^2 - 9) = 16x^3[(x)^2 - (3)^2] = 16x^3(x + 3)(x - 3) \)
In simple words: Factor out 16x³. Then x² - 9 = x² - 3² is a difference of squares.

Exam Tip: When extracting common factors with variables, take the lowest power that appears in all terms.

Question 13. Factorise 3x⁵ - 48x³
Answer: We have:
\( 3x^5 - 48x^3 = 3x^3(x^2 - 16) = 3x^3[(x)^2 - (4)^2] = 3x^3(x + 4)(x - 4) \)
In simple words: Pull out 3x³. What's left is x² - 16 = x² - 4², which factors as (x + 4)(x - 4).

Exam Tip: Check your answer by multiplying back through all the factors.

Question 14. Factorise 16p³ - 4p
Answer: We have:
\( 16p^3 - 4p = 4p(4p^2 - 1) = 4p[(2p)^2 - (1)^2] = 4p(2p + 1)(2p - 1) \)
In simple words: Factor out 4p. Then 4p² - 1 = (2p)² - 1² is a difference of squares.

Exam Tip: Watch for coefficients that are perfect squares - 4, 9, 16, 25, etc.

Question 15. Factorise 63a²b² - 7
Answer: We have:
\( 63a^2b^2 - 7 = 7(9a^2b^2 - 1) = 7[(3ab)^2 - (1)^2] = 7(3ab + 1)(3ab - 1) \)
In simple words: Pull out 7. Then 9a²b² - 1 = (3ab)² - 1² is a difference of squares.

Exam Tip: The second term doesn't always have variables - it might just be a number that's a perfect square.

Question 16. Factorise 1 - (b - c)²
Answer: This is a difference of squares where one term is a bracket:
\( 1 - (b - c)^2 = (1)^2 - (b - c)^2 = [1 + (b - c)][1 - (b - c)] = (1 + b - c)(1 - b + c) \)
In simple words: You have 1 - (b - c)², which is 1² - (b - c)². Use the difference of squares rule, being careful with the signs in the brackets.

Exam Tip: When a whole bracket is squared, treat it as a single unit for the difference of squares formula.

Question 17. Factorise (2a + 3b)² - 16c²
Answer: We have:
\( (2a + 3b)^2 - 16c^2 = (2a + 3b)^2 - (4c)^2 = [(2a + 3b) + 4c][(2a + 3b) - 4c] = (2a + 3b + 4c)(2a + 3b - 4c) \)
In simple words: The first term is (2a + 3b)² and the second is (4c)². Apply the difference of squares formula, keeping the brackets intact.

Exam Tip: When expanding the difference of squares result, distribute carefully - don't lose any signs.

Question 18. Factorise (l + m)² - (l - m)²
Answer: We have:
\( (l + m)^2 - (l - m)^2 = [(l + m) + (l - m)][(l + m) - (l - m)] = (2l)(2m) = 4lm \)
In simple words: Both terms are perfect squares. When you apply the difference of squares rule and simplify the brackets, (l + m) + (l - m) = 2l and (l + m) - (l - m) = 2m, giving 4lm.

Exam Tip: Simplify inside the brackets carefully - you may find terms cancel or combine unexpectedly.

Question 19. Factorise (2x + 5y)² - 1
Answer: We have:
\( (2x + 5y)^2 - 1 = (2x + 5y)^2 - (1)^2 = [(2x + 5y) + 1][(2x + 5y) - 1] = (2x + 5y + 1)(2x + 5y - 1) \)
In simple words: This is (2x + 5y)² - 1², a difference of squares. Apply the rule to get (2x + 5y + 1)(2x + 5y - 1).

Exam Tip: Don't forget that 1 = 1² - it's a perfect square too.

Question 20. Factorise 36c² - (5a + b)²
Answer: We have:
\( 36c^2 - (5a + b)^2 = (6c)^2 - (5a + b)^2 = [6c + (5a + b)][6c - (5a + b)] = (6c + 5a + b)(6c - 5a - b) \)
In simple words: 36c² = (6c)² and you have (5a + b)² as the second term. Use the difference of squares formula, keeping the bracket (5a + b) intact.

Exam Tip: When subtracting a bracket, distribute the negative sign: 6c - (5a + b) = 6c - 5a - b.

Question 21. Factorise 6ab - b² + 12ac - 2bc
Answer: By arranging the terms suitably:
\( 6ab - b^2 + 12ac - 2bc = (6ab + 12ac) - (b^2 + 2bc) = 6a(b + 2c) - b(b + 2c) = (b + 2c)(6a - b) \)
In simple words: Group the first two and last two terms. Factor 6a from the first group and b from the second. The bracket (b + 2c) is now common.

Exam Tip: This uses both difference of squares and grouping techniques - be ready to combine methods.

Question 1. Factorize \( x^2 + 8x + 16 \).
Answer: \( x^2 + 8x + 16 = x^2 + 2 \times x \times 4 + (4)^2 = (x + 4)^2 \)
In simple words: Recognize the expression as a perfect square trinomial where the first term, the last term, and twice the product of their square roots match the middle term.

Exam Tip: Always check if an expression fits the pattern \( a^2 + 2ab + b^2 = (a + b)^2 \) before attempting other factorization methods.

Question 2. Factorize \( x^2 + 14x + 49 \).
Answer: \( x^2 + 14x + 49 = x^2 + 2 \times x \times 7 + (7)^2 = (x + 7)^2 \)
In simple words: This is a perfect square trinomial matching the form \( a^2 + 2ab + b^2 \), which factors as \( (a + b)^2 \).

Exam Tip: Check whether the middle term equals twice the product of the square roots of the first and last terms.

Question 3. Factorize \( 1 + 2x + x^2 \).
Answer: \( 1 + 2x + x^2 = x^2 + 2x + 1 = x^2 + 2 \times x \times 1 + (1)^2 = (x + 1)^2 \)
In simple words: Rearrange the terms in descending order of powers. Then identify it as a perfect square with the structure \( (a + b)^2 \).

Exam Tip: When a trinomial is given in non-standard order, rearrange it first to identify the pattern clearly.

Question 5. Factorize \( x^2 + 6ax + 9a^2 \).
Answer: \( x^2 + 6ax + 9a^2 = x^2 + 2 \times x \times 3a + (3a)^2 = (x + 3a)^2 \)
In simple words: When you have a trinomial involving multiple variables, still look for the perfect square pattern where the middle term is twice the product of the square roots of the outer terms.

Exam Tip: Be careful with coefficients and variables; verify that \( (2 \times x \times 3a) = 6ax \) before finalizing the factorization.

Question 6. Factorize \( 4y^2 + 20y + 25 \).
Answer: \( 4y^2 + 20y + 25 = (2y)^2 + 2 \times 2y \times 5 + (5)^2 = (2y + 5)^2 \)
In simple words: Identify the square root of the first term as \( 2y \) and the square root of the last term as \( 5 \). Verify that their doubled product gives the middle term.

Exam Tip: When the first term has a coefficient, always extract its square root correctly - here, \( \sqrt{4y^2} = 2y \), not just \( y \).

Question 7. Factorize \( 36a^2 + 36a + 9 \).
Answer: \( 36a^2 + 36a + 9 = (6a)^2 + 2 \times 6a \times 3 + (3)^2 = (6a + 3)^2 \)
In simple words: Extract the square roots: \( \sqrt{36a^2} = 6a \) and \( \sqrt{9} = 3 \). Check that \( 2 \times 6a \times 3 = 36a \), confirming the perfect square pattern.

Exam Tip: Show each step of recognizing the pattern clearly; this demonstrates your understanding of perfect square trinomials.

Question 8. Factorize \( 9m^2 + 24m + 16 \).
Answer: \( 9m^2 + 24m + 16 = (3m)^2 + 2 \times 3m \times 4 + (4)^2 = (3m + 4)^2 \)
In simple words: Find the square root of the first term (\( 3m \)) and the last term (\( 4 \)). Verify that twice their product equals the middle term.

Exam Tip: Perfect square trinomials always result in a binomial squared; look for the repeated factor after expanding.

Question 9. Factorize \( z^2 + z + \frac{1}{4} \).
Answer: \( z^2 + z + \frac{1}{4} = z^2 + 2 \times z \times \frac{1}{2} + \left(\frac{1}{2}\right)^2 = \left(z + \frac{1}{2}\right)^2 \)
In simple words: The square root of \( \frac{1}{4} \) is \( \frac{1}{2} \). Verify that \( 2 \times z \times \frac{1}{2} = z \), matching the middle term exactly.

Exam Tip: Fractions follow the same perfect square rules; always simplify before checking the pattern.

Question 10. Factorize \( 49a^2 + 84ab + 36b^2 \).
Answer: \( 49a^2 + 84ab + 36b^2 = (7a)^2 + 2 \times 7a \times 6b + (6b)^2 = (7a + 6b)^2 \)
In simple words: Both variables appear in this perfect square trinomial. Extract \( 7a \) from the first term and \( 6b \) from the last, then check that their doubled product yields the middle term.

Exam Tip: When multiple variables are present, treat them as a single unit within the square root operation.

Question 12. Factorize \( p^2 - 10p + 25 \).
Answer: \( p^2 - 10p + 25 = p^2 - 2 \times p \times 5 + (5)^2 = (p - 5)^2 \)
In simple words: This is a perfect square trinomial with a minus sign. The pattern \( a^2 - 2ab + b^2 = (a - b)^2 \) applies, so the factors are \( (p - 5) \) squared.

Exam Tip: Watch the sign carefully - when the middle term is negative, the factorization results in a difference, not a sum.

Question 13. Factorize \( 1 - 6x + 9x^2 \).
Answer: \( 1 - 6x + 9x^2 = 9x^2 - 6x + 1 = (3x)^2 - 2 \times 3x \times 1 + (1)^2 = (3x - 1)^2 \)
In simple words: Rearrange in standard form first. Then notice it matches the pattern \( a^2 - 2ab + b^2 \), which factors as \( (a - b)^2 \).

Exam Tip: Reordering terms by descending powers makes the pattern much easier to recognize.

Question 14. Factorize \( 9y^2 - 12y + 4 \).
Answer: \( 9y^2 - 12y + 4 = (3y)^2 - 2 \times 3y \times 2 + (2)^2 = (3y - 2)^2 \)
In simple words: The first term's square root is \( 3y \) and the last term's is \( 2 \). Confirm that \( 2 \times 3y \times 2 = 12y \) before writing the final factorization.

Exam Tip: Always double-check your square roots - \( \sqrt{9y^2} = 3y \) and \( \sqrt{4} = 2 \).

Question 15. Factorize \( 16x^2 - 24x + 9 \).
Answer: \( 16x^2 - 24x + 9 = (4x)^2 - 2 \times 4x \times 3 + (3)^2 = (4x - 3)^2 \)
In simple words: Recognize that \( \sqrt{16x^2} = 4x \) and \( \sqrt{9} = 3 \). Verify the pattern by checking \( 2 \times 4x \times 3 = 24x \).

Exam Tip: Breaking down each component separately prevents calculation errors in finding square roots.

Question 16. Factorize \( m^2 - 4mn + 4n^2 \).
Answer: \( m^2 - 4mn + 4n^2 = m^2 - 2 \times m \times 2n + (2n)^2 = (m - 2n)^2 \)
In simple words: Extract \( m \) from the first term and \( 2n \) from the last term. Check that their doubled product equals the middle term.

Exam Tip: With multiple variables, treat coefficients and variables together when computing square roots.

Question 17. Factorize \( a^2b^2 - 6abc + 9c^2 \).
Answer: \( a^2b^2 - 6abc + 9c^2 = (ab)^2 - 2 \times ab \times 3c + (3c)^2 = (ab - 3c)^2 \)
In simple words: The first term's square root is \( ab \) and the last is \( 3c \). Confirm that \( 2 \times ab \times 3c = 6abc \) before finalizing.

Exam Tip: For expressions with multiple variables and coefficients, carefully track each component in the pattern.

Question 19. Factorize \( m^4 + 2m^2n^2 + n^4 \).
Answer: \( m^4 + 2m^2n^2 + n^4 = (m^2)^2 + 2 \times m^2 \times n^2 + (n^2)^2 = (m^2 + n^2)^2 \)
In simple words: Treat \( m^2 \) and \( n^2 \) as single units. The expression becomes a perfect square of \( (m^2 + n^2) \).

Exam Tip: Sometimes higher powers can be rewritten as squares of lower powers to reveal the perfect square pattern.

Question 20. Factorize \( (l + m)^2 - 4lm \).
Answer: \( (l + m)^2 - 4lm = (l^2 + m^2 + 2lm) - 4lm = l^2 + m^2 - 2lm = (l)^2 + (m)^2 - 2 \times l \times m = (l - m)^2 \)
In simple words: Expand the squared binomial first. Combine like terms to get a perfect square trinomial with a negative middle term, which factors as \( (l - m)^2 \).

Exam Tip: Always expand squared expressions and simplify before looking for factorization patterns.

Factorisation

Ex 7D

Question 1. Factorize \( x^2 + 5x + 6 \).
Answer: The given expression is \( x^2 + 5x + 6 \). We need to find two numbers that add to 5 and multiply to 6. The numbers are 3 and 2. Therefore: \( x^2 + 5x + 6 = x^2 + 3x + 2x + 6 = x(x + 3) + 2(x + 3) = (x + 3)(x + 2) \)
In simple words: Look for two numbers whose sum is the middle coefficient and whose product is the last term. Use these to split the middle term and factor by grouping.

Exam Tip: This is the most common factorization method for trinomials with a leading coefficient of 1 - always check the sum and product conditions first.

Question 2. Factorize \( y^2 + 10y + 24 \).
Answer: The given expression is \( y^2 + 10y + 24 \). We need two numbers that sum to 10 and have a product of 24. These numbers are 6 and 4. Therefore: \( y^2 + 10y + 24 = y^2 + 6y + 4y + 24 = y(y + 6) + 4(y + 6) = (y + 6)(y + 4) \)
In simple words: Find the pair of numbers satisfying both conditions - they must add to 10 and multiply to 24 - then use the grouping method.

Exam Tip: Write out the pair of numbers clearly before splitting the middle term - this prevents arithmetic errors.

Question 3. Factorize \( z^2 + 12z + 27 \).
Answer: The given expression is \( z^2 + 12z + 27 \). We need two numbers that sum to 12 and multiply to 27. These numbers are 9 and 3. Therefore: \( z^2 + 12z + 27 = z^2 + 9z + 3z + 27 = z(z + 9) + 3(z + 9) = (z + 9)(z + 3) \)
In simple words: Select the pair of factors of 27 (namely 9 and 3) that also add to 12, then proceed with splitting and grouping.

Exam Tip: For the product, list all factor pairs of the constant term and check which pair has the correct sum.

Question 5. Factorize \( p^2 + 6p + 8 \).
Answer: The given expression is \( p^2 + 6p + 8 \). We need two numbers that sum to 6 and multiply to 8. These numbers are 4 and 2. Therefore: \( p^2 + 6p + 8 = p^2 + 4p + 2p + 8 = p(p + 4) + 2(p + 4) = (p + 4)(p + 2) \)
In simple words: Identify the two numbers (4 and 2) that both add to 6 and multiply to 8, then factor by grouping.

Exam Tip: Factor pairs of the constant term are limited - list them systematically to find the correct pair quickly.

Question 6. Factorize \( x^2 + 15x + 56 \).
Answer: The given expression is \( x^2 + 15x + 56 \). We need two numbers that sum to 15 and have a product of 56. These numbers are 8 and 7. Therefore: \( x^2 + 15x + 56 = x^2 + 8x + 7x + 56 = x(x + 8) + 7(x + 8) = (x + 8)(x + 7) \)
In simple words: The factor pair 8 and 7 satisfies both conditions - they add to 15 and multiply to 56 - so use them to split the middle term.

Exam Tip: Verify by multiplication: \( (x + 8)(x + 7) = x^2 + 7x + 8x + 56 = x^2 + 15x + 56 \) ✓

Question 7. Factorize \( x^2 + 13x + 40 \).
Answer: The given expression is \( x^2 + 13x + 40 \). We need two numbers that sum to 13 and multiply to 40. These numbers are 8 and 5. Therefore: \( x^2 + 13x + 40 = x^2 + 8x + 5x + 40 = x(x + 8) + 5(x + 8) = (x + 8)(x + 5) \)
In simple words: Find the factors of 40 that also sum to 13 - in this case, 8 and 5 - and use the grouping method to factor.

Exam Tip: When factoring \( 40 \), remember: \( 1 \times 40, 2 \times 20, 4 \times 10, 5 \times 8 \) - check which pair sums correctly.

Question 8. Factorize \( q^2 - 10q + 21 \).
Answer: The given expression is \( q^2 - 10q + 21 \). We need two numbers that sum to -10 and multiply to 21. These numbers are -7 and -3. Therefore: \( q^2 - 10q + 21 = q^2 - 7q - 3q + 21 = q(q - 7) - 3(q - 7) = (q - 7)(q - 3) \)
In simple words: When the middle term is negative, look for two negative numbers that satisfy the conditions. Here, -7 and -3 add to -10 and multiply to 21.

Exam Tip: Pay close attention to signs - both numbers must be negative if the middle term is negative and the constant is positive.

Question 10. Factorize \( p^2 + 6p - 16 \).
Answer: The given expression is \( p^2 + 6p - 16 \). We need two numbers that sum to 6 and multiply to -16. These numbers are 8 and -2. Therefore: \( p^2 + 6p - 16 = p^2 + 8p - 2p - 16 = p(p + 8) - 2(p + 8) = (p + 8)(p - 2) \)
In simple words: When the constant is negative, one number is positive and one is negative. Here, 8 and -2 add to 6 and multiply to -16.

Exam Tip: With a negative constant, look for one positive and one negative factor pair that gives the correct sum and product.

Question 11. Factorize \( x^2 - 23x + 42 \).
Answer: The given expression is \( x^2 - 23x + 42 \). We need two numbers that sum to -23 and multiply to 42. These numbers are -21 and -2. Therefore: \( x^2 - 23x + 42 = x^2 - 21x - 2x + 42 = x(x - 21) - 2(x - 21) = (x - 21)(x - 2) \)
In simple words: The two negative numbers -21 and -2 both add to -23 and have a product of 42, so use them to split and group the terms.

Exam Tip: When both the middle and constant terms are large, systematically check factor pairs of the constant to find the matching pair.

Question 12. Factorize \( x^2 - 17x + 16 \).
Answer: The given expression is \( x^2 - 17x + 16 \). We need two numbers that sum to -17 and multiply to 16. These numbers are -16 and -1. Therefore: \( x^2 - 17x + 16 = x^2 - 16x - x + 16 = x(x - 16) - 1(x - 16) = (x - 16)(x - 1) \)
In simple words: The pair -16 and -1 meets both conditions - they sum to -17 and multiply to 16 - so apply the splitting method.

Exam Tip: Always verify: \( (x - 16)(x - 1) = x^2 - x - 16x + 16 = x^2 - 17x + 16 \) ✓

Question 13. Factorize \( y^2 - 21y + 90 \).
Answer: The given expression is \( y^2 - 21y + 90 \). We need two numbers that sum to -21 and have a product of 90. These numbers are -15 and -6. Therefore: \( y^2 - 21y + 90 = y^2 - 15y - 6y + 90 = y(y - 15) - 6(y - 15) = (y - 15)(y - 6) \)
In simple words: Find the negative factor pair of 90 (namely -15 and -6) that also sums to -21, then use the grouping method.

Exam Tip: Factor pairs of 90 include: 1 - 90, 2 - 45, 3 - 30, 5 - 18, 6 - 15. Check which pair, when negated, sums to -21.

Question 15. Factorize \( x^2 - 22x + 117 \).
Answer: The given expression is \( x^2 - 22x + 117 \). We need two numbers that sum to -22 and have a product of 117. These numbers are -13 and -9. Therefore: \( x^2 - 22x + 117 = x^2 - 13x - 9x + 117 = x(x - 13) - 9(x - 13) = (x - 13)(x - 9) \)
In simple words: The factor pair -13 and -9 satisfy both conditions - their sum is -22 and their product is 117 - so apply grouping.

Exam Tip: Verify by checking: \( (x - 13)(x - 9) = x^2 - 9x - 13x + 117 = x^2 - 22x + 117 \) ✓

Question 16. Factorize \( x^2 + x - 132 \).
Answer: The given expression is \( x^2 + x - 132 \). We need two numbers that sum to 1 and have a product of -132. These numbers are 12 and -11. Therefore: \( x^2 + x - 132 = x^2 + 12x - 11x - 132 = x(x + 12) - 11(x + 12) = (x + 12)(x - 11) \)
In simple words: One number is positive and one is negative (since the product is negative). Here, 12 and -11 add to 1 and multiply to -132.

Exam Tip: When the constant is negative and the middle term is small, the two numbers being multiplied must have a large difference in absolute value.

Question 17. Factorize \( x^2 + 5x - 104 \).
Answer: The given expression is \( x^2 + 5x - 104 \). We need two numbers that sum to 5 and multiply to -104. These numbers are 13 and -8. Therefore: \( x^2 + 5x - 104 = x^2 + 13x - 8x - 104 = x(x + 13) - 8(x + 13) = (x + 13)(x - 8) \)
In simple words: The positive number is 13 and the negative number is -8. They satisfy both the sum condition (13 - 8 = 5) and the product condition (\( 13 \times (-8) = -104 \)).

Exam Tip: For products with a negative constant, always check factor pairs where one is positive and one is negative.

Question 18. Factorize \( y^2 + 7y - 144 \).
Answer: The given expression is \( y^2 + 7y - 144 \). We need two numbers that sum to 7 and have a product of -144. These numbers are 16 and -9. Therefore: \( y^2 + 7y - 144 = y^2 + 16y - 9y - 144 = y(y + 16) - 9(y + 16) = (y + 16)(y - 9) \)
In simple words: The pair 16 and -9 adds to 7 and multiplies to -144, so use the splitting and grouping method to factor.

Exam Tip: List factor pairs of 144 systematically: 1 - 144, 2 - 72, 3 - 48, 4 - 36, 6 - 24, 8 - 18, 9 - 16. Then identify which pair (with opposite signs) sums to 7.

Question 20. Factorize \( z^2 + 19z - 150 \).
Answer: The given expression is \( z^2 + 19z - 150 \). We need two numbers that sum to 19 and have a product of -150. These numbers are 25 and -6. Therefore: \( z^2 + 19z - 150 = z^2 + 25z - 6z - 150 = z(z + 25) - 6(z + 25) = (z + 25)(z - 6) \)
In simple words: The numbers 25 and -6 both add to 19 and multiply to -150, so split the middle term using these values and group accordingly.

Exam Tip: Verify by expansion: \( (z + 25)(z - 6) = z^2 - 6z + 25z - 150 = z^2 + 19z - 150 \) ✓

Question 21. Factorize \( y^2 + y - 72 \).
Answer: The given expression is \( y^2 + y - 72 \). We need two numbers that sum to 1 and have a product of -72. These numbers are 9 and -8. Therefore: \( y^2 + y - 72 = y^2 + 9y - 8y - 72 = y(y + 9) - 8(y + 9) = (y + 9)(y - 8) \)
In simple words: The numbers 9 and -8 satisfy both conditions - they add to 1 and multiply to -72 - so use the grouping method.

Exam Tip: When the middle term coefficient is small (like 1), look for factor pairs with a large difference in absolute value.

Question 22. Factorize \( p^2 - 4p - 77 \).
Answer: The given expression is \( p^2 - 4p - 77 \). We need two numbers that sum to -4 and have a product of -77. These numbers are -11 and 7. Therefore: \( p^2 - 4p - 77 = p^2 - 11p + 7p - 77 = p(p - 11) + 7(p - 11) = (p - 11)(p + 7) \)
In simple words: The negative number -11 and positive number 7 add to -4 and multiply to -77, so split and group using these values.

Exam Tip: Always arrange the grouping so that the common factor emerges clearly in both terms.

Question 23. Factorize \( x^2 - 7x - 30 \).
Answer: The given expression is \( x^2 - 7x - 30 \). We need two numbers that sum to -7 and have a product of -30. These numbers are -10 and 3. Therefore: \( x^2 - 7x - 30 = x^2 - 10x + 3x - 30 = x(x - 10) + 3(x - 10) = (x - 10)(x + 3) \)
In simple words: The pair -10 and 3 add to -7 and multiply to -30, so apply the splitting and grouping method.

Exam Tip: Check: \( (x - 10)(x + 3) = x^2 + 3x - 10x - 30 = x^2 - 7x - 30 \) ✓

Question 24. Factorize \( x^2 - 11x - 42 \).
Answer: The given expression is \( x^2 - 11x - 42 \). We need two numbers that sum to -11 and have a product of -42. These numbers are -14 and 3. Therefore: \( x^2 - 11x - 42 = x^2 - 14x + 3x - 42 = x(x - 14) + 3(x - 14) = (x - 14)(x + 3) \)
In simple words: The numbers -14 and 3 satisfy both the sum condition (-14 + 3 = -11) and the product condition (\( -14 \times 3 = -42 \)), so use the grouping method.

Exam Tip: Factor pairs of 42 are: 1 - 42, 2 - 21, 3 - 14, 6 - 7. Test which pair (with opposite signs) gives the required sum.

Question 25. Factorize \( x^2 - 5x - 24 \).
Answer: The given expression is \( x^2 - 5x - 24 \). We need two numbers that sum to -5 and have a product of -24. These numbers are -8 and 3. Therefore: \( x^2 - 5x - 24 = x^2 - 8x + 3x - 24 = x(x - 8) + 3(x - 8) = (x - 8)(x + 3) \)
In simple words: The pair -8 and 3 adds to -5 and multiplies to -24, so split the middle term and factor by grouping.

Exam Tip: Always verify: \( (x - 8)(x + 3) = x^2 + 3x - 8x - 24 = x^2 - 5x - 24 \) ✓

Question 26. Factorize \( y^2 - 6y - 135 \).
Answer: The given expression is \( y^2 - 6y - 135 \). We need two numbers that sum to -6 and have a product of -135. These numbers are -15 and 9. Therefore: \( y^2 - 6y - 135 = y^2 - 15y + 9y - 135 = y(y - 15) + 9(y - 15) = (y - 15)(y + 9) \)
In simple words: The numbers -15 and 9 satisfy both conditions - they add to -6 and multiply to -135 - so use the grouping method.

Exam Tip: For larger constants like 135, list factor pairs carefully: 1 - 135, 3 - 45, 5 - 27, 9 - 15. Identify the pair that (with sign adjustment) gives the correct sum.

Question 27. Factorize \( z^2 - 12z - 45 \).
Answer: The given expression is \( z^2 - 12z - 45 \). We need two numbers that sum to -12 and have a product of -45. These numbers are -15 and 3. Therefore: \( z^2 - 12z - 45 = z^2 - 15z + 3z - 45 = z(z - 15) + 3(z - 15) = (z - 15)(z + 3) \)
In simple words: The pair -15 and 3 adds to -12 and multiplies to -45, so split and group accordingly.

Exam Tip: Check: \( (z - 15)(z + 3) = z^2 + 3z - 15z - 45 = z^2 - 12z - 45 \) ✓

Question 28. Factorize \( x^2 - 4x - 12 \).
Answer: The given expression is \( x^2 - 4x - 12 \). We need two numbers that sum to -4 and have a product of -12. These numbers are -6 and 2. Therefore: \( x^2 - 4x - 12 = x^2 - 6x + 2x - 12 = x(x - 6) + 2(x - 6) = (x - 6)(x + 2) \)
In simple words: The numbers -6 and 2 satisfy the conditions - they add to -4 and multiply to -12 - so use the grouping method.

Exam Tip: Factor pairs of 12 are: 1 - 12, 2 - 6, 3 - 4. Test which pair (with appropriate signs) yields the required sum.

Question 29. Factorize \( 3x^2 + 10x + 8 \).
Answer: The given expression is \( 3x^2 + 10x + 8 \). First, find two numbers that multiply to \( 3 \times 8 = 24 \) and add to 10. These numbers are 6 and 4. Therefore: \( 3x^2 + 10x + 8 = 3x^2 + 6x + 4x + 8 = 3x(x + 2) + 4(x + 2) = (x + 2)(3x + 4) \)
In simple words: For trinomials with a leading coefficient other than 1, multiply the first and last coefficients. Find two numbers that multiply to this product and add to the middle coefficient, then factor by grouping.

Exam Tip: This method (sometimes called "ac method") is essential when the leading coefficient is not 1. Always multiply \( a \times c \) first, not just use \( c \).

Question 30. Factorize \( 3y^2 + 14y + 8 \).
Answer: The given expression is \( 3y^2 + 14y + 8 \). Find two numbers that multiply to \( 3 \times 8 = 24 \) and add to 14. These numbers are 12 and 2. Therefore: \( 3y^2 + 14y + 8 = 3y^2 + 12y + 2y + 8 = 3y(y + 4) + 2(y + 4) = (y + 4)(3y + 2) \)
In simple words: Identify the product \( ac = 24 \) and the pair 12 and 2 that adds to 14. Split the middle term and factor by grouping.

Exam Tip: Verify: \( (y + 4)(3y + 2) = 3y^2 + 2y + 12y + 8 = 3y^2 + 14y + 8 \) ✓

Question 31. Factorize \( 3z^2 - 10z + 8 \).
Answer: The given expression is \( 3z^2 - 10z + 8 \). Find two numbers that multiply to \( 3 \times 8 = 24 \) and add to -10. These numbers are -6 and -4. Therefore: \( 3z^2 - 10z + 8 = 3z^2 - 6z - 4z + 8 = 3z(z - 2) - 4(z - 2) = (z - 2)(3z - 4) \)
In simple words: Multiply the leading and constant coefficients to get 24. Find two negative numbers (-6 and -4) that multiply to 24 and add to -10, then factor by grouping.

Exam Tip: Always remember to check the signs carefully. With a negative middle term and positive constant, both numbers in the pair are negative.

Question 32. Factorize \( 2x^2 + x - 45 \).
Answer: The given expression is \( 2x^2 + x - 45 \). Find two numbers that multiply to \( 2 \times (-45) = -90 \) and add to 1. These numbers are 10 and -9. Therefore: \( 2x^2 + x - 45 = 2x^2 + 10x - 9x - 45 = 2x(x + 5) - 9(x + 5) = (x + 5)(2x - 9) \)
In simple words: Calculate \( ac = 2 \times (-45) = -90 \). Find the pair 10 and -9 that multiplies to -90 and adds to 1. Split the middle term accordingly and factor.

Exam Tip: When the constant is negative, one number in the pair is positive and one is negative. The absolute value of the pair must multiply to 90.

Question 33. Factorize \( 6p^2 + 11p - 10 \).
Answer: The given expression is \( 6p^2 + 11p - 10 \). Find two numbers that multiply to \( 6 \times (-10) = -60 \) and add to 11. These numbers are 15 and -4. Therefore: \( 6p^2 + 11p - 10 = 6p^2 + 15p - 4p - 10 = 3p(2p + 5) - 2(2p + 5) = (2p + 5)(3p - 2) \)
In simple words: Compute \( ac = 6 \times (-10) = -60 \). Find the pair 15 and -4 that multiplies to -60 and adds to 11. Factor by grouping after splitting the middle term.

Exam Tip: Factor pairs of 60 are: 1 - 60, 2 - 30, 3 - 20, 4 - 15, 5 - 12, 6 - 10. Test which pair (with sign adjustment) gives the correct sum.

Question 34. Factorize \( 2x^2 - 17x - 30 \).
Answer: The given expression is \( 2x^2 - 17x - 30 \). Find two numbers that multiply to \( 2 \times (-30) = -60 \) and add to -17. These numbers are -20 and 3. Therefore: \( 2x^2 - 17x - 30 = 2x^2 - 20x + 3x - 30 = 2x(x - 10) + 3(x - 10) = (x - 10)(2x + 3) \)
In simple words: The product \( ac = -60 \) and the pair -20 and 3 adds to -17. Split and factor by grouping using these values.

Exam Tip: When one number is much larger than the other (in this case, -20 vs. 3), the pair often comes from a factor pair of \( |ac| \) where the larger number is negative.

Question 35. Factorize \( 7y^2 - 19y - 6 \).
Answer: The given expression is \( 7y^2 - 19y - 6 \). Find two numbers that multiply to \( 7 \times (-6) = -42 \) and add to -19. These numbers are -21 and 2. Therefore: \( 7y^2 - 19y - 6 = 7y^2 - 21y + 2y - 6 = 7y(y - 3) + 2(y - 3) = (y - 3)(7y + 2) \)
In simple words: Calculate \( ac = 7 \times (-6) = -42 \). The pair -21 and 2 multiplies to -42 and adds to -19, so use these to split the middle term.

Exam Tip: Verify: \( (y - 3)(7y + 2) = 7y^2 + 2y - 21y - 6 = 7y^2 - 19y - 6 \) ✓

Question 36. Factorize \( 28 - 31x - 5x^2 \).
Answer: The given expression is \( 28 - 31x - 5x^2 \). Rearrange as \( -5x^2 - 31x + 28 \), or factor out -1 to get \( -(5x^2 + 31x - 28) \). Find two numbers that multiply to \( 5 \times (-28) = -140 \) and add to 31. These numbers are 35 and -4. Therefore: \( -5x^2 - 31x + 28 = -(5x^2 + 35x - 4x - 28) = -(5x(x + 7) - 4(x + 7)) = -(x + 7)(5x - 4) = (x + 7)(4 - 5x) \)
In simple words: Rearrange in standard form with decreasing powers. Factor out -1 if needed, then apply the standard factorization method.

Exam Tip: Always arrange the expression in descending order of the variable's power before factoring.

Question 37. Factorize \( 3 + 23z - 8z^2 \).
Answer: The given expression is \( 3 + 23z - 8z^2 \). Rearrange as \( -8z^2 + 23z + 3 \), or factor out -1 to get \( -(8z^2 - 23z - 3) \). Find two numbers that multiply to \( 8 \times (-3) = -24 \) and add to -23. These numbers are -24 and 1. Therefore: \( -8z^2 + 23z + 3 = -(8z^2 - 24z + z - 3) = -(8z(z - 3) + 1(z - 3)) = -(z - 3)(8z + 1) = (3 - z)(8z + 1) \)
In simple words: Rearrange in standard descending form. Factor out -1 if the leading coefficient is negative, then proceed with the factorization method.

Exam Tip: When rearranging, ensure all terms are on the same side with the expression equal to zero, making standard form clear.

Question 38. Factorize \( 6x^2 - 5x - 6 \).
Answer: The given expression is \( 6x^2 - 5x - 6 \). Find two numbers that multiply to \( 6 \times (-6) = -36 \) and add to -5. These numbers are -9 and 4. Therefore: \( 6x^2 - 5x - 6 = 6x^2 - 9x + 4x - 6 = 3x(2x - 3) + 2(2x - 3) = (2x - 3)(3x + 2) \)
In simple words: The product \( ac = -36 \) and the pair -9 and 4 adds to -5. Split the middle term and factor by grouping.

Exam Tip: Always verify your result: \( (2x - 3)(3x + 2) = 6x^2 + 4x - 9x - 6 = 6x^2 - 5x - 6 \) ✓

Question 1. Factorise \( 7a^2 - 63b^2 \)
Answer: \( 7(a - 3b)(a + 3b) \)
In simple words: Take out the common factor 7, then use the difference of squares formula to break down what remains into two brackets.

Exam Tip: Always look for a common factor first before applying any other factorisation method - it simplifies the remaining expression.

Question 2. Factorise \( 2x - 32x^3 \)
Answer: \( 2x(1 - 4x)(1 + 4x) \)
In simple words: First pull out the common factor 2x, then recognise that what's left is a difference of two squares, which splits into two brackets.

Exam Tip: After factoring out the common element, check whether the remaining expression matches any standard algebraic identity like difference of squares.

Question 3. Factorise \( x^3 - 144x \)
Answer: \( x(x - 12)(x + 12) \)
In simple words: Remove x as a common factor, then recognise the difference of squares pattern in the remaining part.

Exam Tip: When you have a variable that appears in every term, factor it out completely before applying other identities.

Question 4. Factorise \( 2 - 50x^2 \)
Answer: \( 2(1 - 5x)(1 + 5x) \)
In simple words: Take out the common factor 2 first, then use the difference of squares formula on the brackets.

Exam Tip: Never skip the common factor step - it makes the rest of the problem much easier and cleaner.

Question 5. Factorise \( a^2 + bc + ab + ac \)
Answer: \( (a + c)(a + b) \)
In simple words: Rearrange the terms, group them in pairs with a common element, then factor out to get two brackets.

Exam Tip: When you have four terms, try grouping the first and second pair, and also the third and fourth pair, to find common factors.

Question 6. Factorise \( pq^2 + q(p - 1) - 1 \)
Answer: \( (pq - 1)(q + 1) \)
In simple words: Expand the expression, then group terms strategically to find two common factors that can be pulled out.

Exam Tip: After expanding, rearrange the terms so that you can easily identify which terms share a common element.

Question 7. Factorise \( ab - mn + an - bm \)
Answer: \( (a - m)(b + n) \)
In simple words: Reorder the terms, then group them into two pairs. Each pair will share a common factor, which can then be factored out.

Exam Tip: Rearranging the terms before grouping is crucial - the original order might not show the pattern immediately.

Question 8. Factorise \( ab - a - b + 1 \)
Answer: \( (a - 1)(b - 1) \)
In simple words: Group the first two terms and the last two terms separately, find the common factor in each pair, then extract the final common bracket.

Exam Tip: When grouping yields the same bracket in both groups, that bracket becomes one of your final factors.

Question 9. Factorise \( x^2 - xz + xy - yz \)
Answer: \( (x + y)(x - z) \)
In simple words: Group the terms into two pairs, extract a common element from each pair, and then pull out the matching bracket.

Exam Tip: Look for patterns within the grouping - if the same bracketed expression appears twice, you've found the right grouping.

Question 10. Factorise \( 12m^2 - 27 \)
Answer: \( 3(2m - 3)(2m + 3) \)
In simple words: Pull out the common factor 3, then notice that the remaining expression is a perfect difference of squares, which splits into two brackets.

Exam Tip: Always check if both terms are perfect squares before you finish - this is the quickest route to the difference of squares factorisation.

Question 11. Factorise \( x^3 - x \)
Answer: \( x(x - 1)(x + 1) \)
In simple words: Take out the common factor x, then apply the difference of squares rule to split what's left into two more brackets.

Exam Tip: Difference of squares works whenever you see one perfect square subtracted from another - recognise this pattern quickly.

Question 12. Factorise \( 1 - 2ab - (a^2 + b^2) \)
Answer: \( (1 - a - b)(1 + a + b) \)
In simple words: Rearrange and group the terms to recognise the pattern as a difference of two squares, where one "square" is \( (a + b)^2 \) and the other is 1.

Exam Tip: Sometimes you need to rewrite an expression to see the difference of squares pattern - expand bracketed terms if needed to spot it.

Question 13. Factorise \( x^2 + 6x + 8 \)
Answer: \( (x + 2)(x + 4) \)
In simple words: Find two numbers that add to 6 and multiply to 8 - those numbers are 2 and 4. Use them to build your two brackets.

Exam Tip: For a quadratic \( x^2 + bx + c \), look for two numbers whose sum is b and product is c - this is the foundation of trinomial factorisation.

Question 14. Factorise \( x^2 + 4x - 21 \)
Answer: \( (x - 3)(x + 7) \)
In simple words: Look for two numbers that add to 4 and multiply to -21. Since one must be negative, those numbers are -3 and 7.

Exam Tip: When the constant term is negative, one factor must be negative and one positive - their sum gives the middle coefficient.

Question 15. Factorise \( y^2 + 2y - 3 \)
Answer: \( (y - 1)(y + 3) \)
In simple words: Find two numbers that add to 2 and multiply to -3. Those numbers are -1 and 3, giving you the two brackets.

Exam Tip: Check your answer by expanding the brackets back out - if you get the original expression, your factorisation is correct.

Question 16. Factorise \( 40 + 3x - x^2 \)
Answer: \( (8 - x)(x + 5) \)
In simple words: Split the middle term into two parts with a common factor in each. Group and factor out to get your final answer.

Exam Tip: When the coefficient of \( x^2 \) is negative, be careful with your signs during factorisation - rewrite to make \( x^2 \) positive if it helps.

Question 17. Factorise \( 2x^2 + 5x + 3 \)
Answer: \( (2x + 3)(x + 1) \)
In simple words: Split the middle term 5x into two parts that, when grouped, reveal a common factor in each pair of terms.

Exam Tip: For trinomials where the leading coefficient isn't 1, find two numbers that multiply to (leading coefficient × constant term) and add to the middle coefficient.

Question 18. Factorise \( 6a^2 - 13a + 6 \)
Answer: \( (3a - 2)(2a - 3) \)
In simple words: Split -13a into two parts that multiply to \( 6 \times 6 = 36 \). Group the resulting terms and extract common factors from each pair.

Exam Tip: For harder trinomials, multiply the first and last coefficients to get the product you're looking for when splitting the middle term.

Question 19. Factorise \( 4z^2 - 8z + 3 \)
Answer: \( (2z - 1)(2z - 3) \)
In simple words: Find two numbers that multiply to 12 and add to -8. Split the middle term, group the pairs, and factor out the common elements.

Exam Tip: When all three terms have the same sign, both factors in the final answer will have the same sign as well.

Question 20. Factorise \( 3 + 23y - 8y^2 \)
Answer: \( (3 - y)(1 + 8y) \)
In simple words: Rearrange so \( y^2 \) comes first if needed. Split 23y into two parts that, when grouped, yield common factors in each group.

Exam Tip: Always arrange the expression in descending order of powers before you start - it makes the splitting process clearer and more systematic.