NCERT Solutions Class 9 Mathematics Ganita Manjari Chapter 04 Exploring Algebraic Identities

Exercise 4.1

Question 1. Using the identity (a + b)² = a² + 2ab + b², expand the following:
(i) (7x + 4y)²
Answer: Apply the identity with a = 7x and b = 4y. Squaring 7x gives 49x². Squaring 4y gives 16y². The cross term is 2 × 7x × 4y = 56xy. Adding these together: 49x² + 56xy + 16y²
In simple words: Square both parts (7x and 4y), then add twice their product.

Exam Tip: Always identify a and b clearly before applying the identity - this prevents arithmetic errors.

Question 2. [(7/5)x + (3/2)y]²
Answer: With a = (7/5)x and b = (3/2)y, square each term: [(7/5)x]² = (49/25)x² and [(3/2)y]² = (9/4)y². The middle term is 2 × (7/5)x × (3/2)y = (42/10)xy = (21/5)xy. Result: (49/25)x² + (21/5)xy + (9/4)y²
In simple words: Work with fractions the same way - square each fraction, then find the middle term by multiplying and simplifying.

Exam Tip: Simplify fractional products immediately to avoid carrying wrong values to the final answer.

Question 3. (2.5p + 1.5q)²
Answer: Let a = 2.5p and b = 1.5q. Then (2.5p)² = 6.25p² and (1.5q)² = 2.25q². The middle term: 2 × 2.5p × 1.5q = 7.5pq. Combined: 6.25p² + 7.5pq + 2.25q²
In simple words: Decimals work just like whole numbers - multiply them out and keep the decimal places.

Exam Tip: Check your decimal arithmetic by working with place values carefully.

Question 4. [(3/4)s + 8t]²
Answer: Using a = (3/4)s and b = 8t: [(3/4)s]² = (9/16)s² and (8t)² = 64t². The middle term is 2 × (3/4)s × 8t = (24/4)st = 12st. Final expansion: (9/16)s² + 12st + 64t²
In simple words: Mix fractions with whole numbers by treating each carefully in the identity.

Exam Tip: Always reduce fractions in the middle term - (24/4)st simplifies to 12st.

Question 5. [x + 1/(2y)]²
Answer: With a = x and b = 1/(2y): x² stays as x². The square of 1/(2y) is 1/(4y²). The cross term is 2 × x × 1/(2y) = 2x/(2y) = x/y. Result: x² + x/y + 1/(4y²)
In simple words: When variables are in denominators, square them carefully to keep the denominator squared.

Exam Tip: Remember that [1/(2y)]² becomes 1/(4y²), not 1/(2y²).

Question 6. (1/x + 1/y)²
Answer: Let a = 1/x and b = 1/y. Squaring each: (1/x)² = 1/x² and (1/y)² = 1/y². The middle term is 2 × (1/x) × (1/y) = 2/(xy). Putting it together: 1/x² + 2/(xy) + 1/y²
In simple words: When both terms have variables below the line, each square still goes below the line, and the product combines them carefully.

Exam Tip: The cross term 2/(xy) shows that the denominators multiply, not add.

Exercise 4.1 (continued)

Question 7. Using the identity (a + b)² = a² + 2ab + b², find the values of the following:
(i) (64)²
Answer: Write 64 as 60 + 4. Then (60 + 4)² = 60² + 2(60)(4) + 4². Calculate: 3600 + 480 + 16 = 4096
In simple words: Break the number into parts close to round numbers to make squaring easier.

Exam Tip: Choose parts where one is a multiple of 10 - it speeds up the calculation significantly.

Question 8. (105)²
Answer: Express 105 as 100 + 5. Then (100 + 5)² = 100² + 2(100)(5) + 5² = 10000 + 1000 + 25 = 11025
In simple words: 105 is close to the round number 100, so this split makes the math simple.

Exam Tip: Always split near a round number (10, 100, 1000) for fastest computation.

Question 9. (205)²
Answer: Split as 200 + 5. Then (200 + 5)² = 200² + 2(200)(5) + 5² = 40000 + 2000 + 25 = 42025
In simple words: The same method works for larger numbers - just pick the nearest clean hundred.

Exam Tip: 40000 + 2000 + 25 = 42025; double-check by adding left to right.

Exercise 4.2

Question 1. Factor completely:
(i) 9x² + 24xy + 16y²
Answer: Notice that 9x² = (3x)², 16y² = (4y)², and 24xy = 2(3x)(4y). This matches the pattern a² + 2ab + b² = (a + b)². Therefore, 9x² + 24xy + 16y² = (3x + 4y)²
In simple words: Check if the first and last terms are perfect squares, and if the middle term is twice their product - if yes, it factors as a square.

Exam Tip: Always verify: 2 × 3x × 4y should equal 24xy before writing the factored form.

Question 2. 4s² + 20st + 25t²
Answer: Identify that 4s² = (2s)², 25t² = (5t)², and 20st = 2(2s)(5t). Using a² + 2ab + b² = (a + b)², we get (2s + 5t)²
In simple words: Find the square root of the first term (2s) and the last term (5t), then check if their double product matches the middle.

Exam Tip: The coefficient of the middle term must equal 2 times the product of the coefficients from the roots.

Question 3. 49x² + 28xy + 4y²
Answer: We have 49x² = (7x)², 4y² = (2y)², and 28xy = 2(7x)(2y). This gives (7x + 2y)²
In simple words: The pattern is the same each time - square roots of the ends times two equals the middle term.

Exam Tip: Write out the check: 2 × 7x × 2y = 28xy ✓

Question 4. 64p² + (32/3)pq + (4/9)q²
Answer: Observe that 64p² = (8p)², (4/9)q² = (2/3 q)², and the middle term 2 × 8p × (2/3)q = (32/3)pq. Therefore, this factors as (8p + 2/3 q)²
In simple words: Fractions in the last term mean you must take their square root - here √(4/9) = 2/3.

Exam Tip: Always simplify fractional square roots first before checking the middle term.

Question 5. 3a² + 4ab + (4/3)b²
Answer: Factor out 3: 3[a² + (4/3)ab + (4/9)b²]. Now recognize that a² = (a)², (4/9)b² = (2/3 b)², and (4/3)ab = 2 × a × (2/3)b. So the expression inside brackets equals (a + 2/3 b)², making the full answer 3(a + 2/3 b)²
In simple words: Always pull out common factors first - it simplifies the pattern-checking inside.

Exam Tip: After factoring out 3, check that the remaining three terms still form a perfect square pattern.

Question 6. (9/5)s² + 6sv + 5v²
Answer: Factor out 1/5: (1/5)[9s² + 30sv + 25v²]. Inside, 9s² = (3s)², 25v² = (5v)², and 30sv = 2(3s)(5v). Thus the bracket is (3s + 5v)², and the complete factorization is (1/5)(3s + 5v)²
In simple words: When coefficients have fractions, take out a fraction that makes the remaining terms whole or easier to factor.

Exam Tip: Verify that 1/5 × 9 = 9/5 and 1/5 × 25 = 5, confirming your factorization is correct.

Question 7. Find the values of the following using the identity (a - b)² = a² - 2ab + b²:
(i) (79)²
Answer: Write 79 as 80 - 1. Apply the identity: (80 - 1)² = 80² - 2(80)(1) + 1² = 6400 - 160 + 1 = 6241
In simple words: If a number is close to a round number but slightly less, subtract and use the difference identity instead.

Exam Tip: The middle term is negative when using (a - b)² - don't forget the minus sign.

Question 8. (193)²
Answer: Express as 200 - 7. Then (200 - 7)² = 200² - 2(200)(7) + 7² = 40000 - 2800 + 49 = 37249
In simple words: 193 is near 200, so subtracting 7 from 200 gives us the right breakdown for the difference formula.

Exam Tip: Check arithmetic: 40000 - 2800 = 37200, then 37200 + 49 = 37249.

Question 9. (299)²
Answer: Write as 300 - 1. Using (a - b)²: (300 - 1)² = 300² - 2(300)(1) + 1² = 90000 - 600 + 1 = 89401
In simple words: 299 is one less than 300, making the calculation straightforward and reducing errors.

Exam Tip: Notice the pattern - if a number is close to a "nice" number above it, use the difference formula.

Exercise 4.3

Question 1. Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier.
(i) 117²
Answer: Express 117 as 110 + 7. Using (a + b)² = a² + 2ab + b²: (110 + 7)² = 110² + 2(110)(7) + 7² = 12100 + 1540 + 49 = 13689. The sum identity works here because 117 is slightly more than the round number 110.
In simple words: Choose addition when the number is just above a round number, subtraction when below.

Exam Tip: Compare how many steps and what size numbers you'd use with each method - the "best" identity saves the most work.

Question 2. 78²
Answer: Since 78 is close to 80 but less, write it as 80 - 2. Apply (a - b)²: (80 - 2)² = 80² - 2(80)(2) + 2² = 6400 - 320 + 4 = 6084
In simple words: 78 sits below 80, so the difference formula is the faster route.

Exam Tip: The difference method is cleaner here - squaring 2 is easier than squaring 8.

Question 3. 198²
Answer: Express as 200 - 2. Then (200 - 2)² = 200² - 2(200)(2) + 2² = 40000 - 800 + 4 = 39204
In simple words: 198 is 2 away from the round number 200 (below it), making the difference formula ideal.

Exam Tip: You could also use 200 + (- 2), which is the same formula written differently.

Question 4. 214²
Answer: Write as 200 + 14. Using (a + b)²: (200 + 14)² = 200² + 2(200)(14) + 14² = 40000 + 5600 + 196 = 45796
In simple words: 214 is 14 more than 200, so the sum formula applies.

Exam Tip: The larger the second number (14 here), the more important it is to calculate it correctly.

Question 5. 1104²
Answer: Break it as 1100 + 4. Using (a + b)²: (1100 + 4)² = 1100² + 2(1100)(4) + 4² = 1210000 + 8800 + 16 = 1218816
In simple words: Even for large numbers, split them into a round part and a small adjustment.

Exam Tip: 1100² = (11 × 100)² = 121 × 10000 = 1210000 - compute larger squares by using place value.

Question 6. 1120²
Answer: Write as 1100 + 20. Then (1100 + 20)² = 1100² + 2(1100)(20) + 20² = 1210000 + 44000 + 400 = 1254400
In simple words: The second part doesn't have to be tiny - 20 is still much easier than squaring 1120 directly.

Exam Tip: The method works even when both numbers are moderately sized - it still beats direct multiplication.

Question 7. Factor using suitable identities:
(i) 16y² - 24y + 9
Answer: Recognize that 16y² = (4y)², 9 = 3², and -24y = -2(4y)(3). This matches a² - 2ab + b² = (a - b)², so the factorization is (4y - 3)²
In simple words: When the middle term is negative, look for the difference-of-squares pattern, not the sum pattern.

Exam Tip: Verify by expanding: (4y - 3)² = 16y² - 24y + 9 ✓

Question 8. (9/4)s² + 6st + 4t²
Answer: Notice (9/4)s² = [(3s)/2]², 4t² = (2t)², and 6st = 2 × [(3s)/2] × (2t). By a² + 2ab + b² = (a + b)², the result is {[(3s)/2] + 2t}² or [(3s + 4t)/2]²
In simple words: Square roots of fractions require care - √(9/4) = 3/2, not some whole number.

Exam Tip: Always take the square root of both numerator and denominator separately.

Question 9. m²/9 + mk/3 + k²/4 + 3nk + 2mn + 9n²
Answer: Rearrange as m²/9 + k²/4 + 9n² + mk/3 + 2mn + 3nk. Identify: (m/3)², (k/2)², and (3n)² are the three squared terms. Verify cross terms: 2(m/3)(k/2) = mk/3, 2(m/3)(3n) = 2mn, and 2(k/2)(3n) = 3nk. By (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca, the factorization is (m/3 + k/2 + 3n)²
In simple words: Three-term patterns need all three squared parts and all three cross products to match - check carefully.

Exam Tip: Write out the three "a," "b," "c" values first, then verify each cross term one at a time.

Question 10. p²/16 - 2 + 16/p²
Answer: Rewrite -2 as -2(p/4)(4/p). Now identify (p/4)² and (4/p)² as the squared terms. The expression becomes (p/4)² - 2(p/4)(4/p) + (4/p)² = (p/4 - 4/p)² by the identity a² - 2ab + b² = (a - b)²
In simple words: Sometimes a loose term in the middle can be rewritten as a cross product - look for this trick when a constant sits by itself.

Exam Tip: The key insight is recognizing that -2 = 2 × (p/4) × (4/p) - once you see that, the pattern snaps into place.

Exercise 4.3 (continued)

Question 11. 9a² + 4b² + c² - 12ab + 6ac - 4bc
Answer: Write as 9a² + 4b² + c² - 12ab + 6ac - 4bc. Identify: 9a² = (3a)², 4b² = (-2b)² (noting the negative because two terms involving b are negative: -12ab and -4bc), and c² = c². Cross terms: 2(3a)(-2b) = -12ab, 2(-2b)(c) = -4bc, and 2(3a)(c) = 6ac. This gives (3a - 2b + c)² by (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
In simple words: When multiple cross terms are negative, sometimes one of your "base" variables should be treated as negative from the start.

Exam Tip: This technique - treating a term as negative - only works if it makes ALL the negative cross terms come out right.

Question 12. Expand the following using the identity (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca:
(i) (p + 3q + 7r)²
Answer: Let a = p, b = 3q, c = 7r. Apply the identity: (p + 3q + 7r)² = p² + (3q)² + (7r)² + 2(p)(3q) + 2(3q)(7r) + 2(p)(7r) = p² + 9q² + 49r² + 6pq + 42qr + 14pr
In simple words: Square each part, then add all six cross products (two-at-a-time) with their factors of 2.

Exam Tip: There are always exactly three squared terms and three cross terms (each cross term multiplied by 2) in a three-term expansion.

Question 13. (3x - 2y + 4z)²
Answer: Rewrite as [3x + (-2y) + 4z]² with a = 3x, b = -2y, c = 4z. Expand: (3x)² + (-2y)² + (4z)² + 2(3x)(-2y) + 2(-2y)(4z) + 2(3x)(4z) = 9x² + 4y² + 16z² - 12xy - 16yz + 24xz
In simple words: Treat negative middle terms by using them as negative values in the formula - don't skip them.

Exam Tip: The cross terms involving a negative b will have negative signs - compute them carefully.

Question 14. Is this an identity? (a + b - c)² + (a - b + c)² + (a - b - c)² = 2a² + 2b² + 2c²
Answer: Expand the left side. First part: (a + b - c)² = a² + b² + c² + 2ab - 2ac - 2bc. Second part: (a - b + c)² = a² + b² + c² - 2ab + 2ac - 2bc. Third part: (a - b - c)² = a² + b² + c² - 2ab - 2ac + 2bc. Adding all three: (a² + b² + c² + 2ab - 2ac - 2bc) + (a² + b² + c² - 2ab + 2ac - 2bc) + (a² + b² + c² - 2ab - 2ac + 2bc) = 3a² + 3b² + 3c² - 2ab - 2ac - 2bc. This is NOT equal to 2a² + 2b² + 2c². Therefore, the statement is not an identity - it fails for most values of a, b, and c.
In simple words: Expand carefully and add like terms - the result shows this equation is false, not a universal identity.

Exam Tip: To prove something is NOT an identity, expand fully and show the two sides don't match.

Exercise 4.4

Question 1. Fill in the blanks to complete the following identities:
(i) s² - 11s + 24 = ( ____ )( ____ )
Answer: Look for two numbers whose sum is 11 and product is 24. These numbers are 3 and 8. Rewrite the middle term: s² - 11s + 24 = s² - 3s - 8s + 24 = s(s - 3) - 8(s - 3) = (s - 3)(s - 8). The blanks are filled with (s - 3) and (s - 8).
In simple words: Find two numbers that add to the middle coefficient and multiply to the constant - then split and factor by grouping.

Exam Tip: Always check: -3 + (-8) = -11 ✓ and (-3) × (-8) = 24 ✓

Question 2. ( ____ )(x + 1) = (3x² - 4x - 7)
Answer: Work backwards from the right side. Factor 3x² - 4x - 7. Split the middle term: 3x² - 7x + 3x - 7 = x(3x - 7) + 1(3x - 7) = (3x - 7)(x + 1). So the blank is (3x - 7).
In simple words: Factor the right side first, then match which factor goes in the blank.

Exam Tip: Verify by multiplying: (3x - 7)(x + 1) = 3x² + 3x - 7x - 7 = 3x² - 4x - 7 ✓

Question 3. 10x² - 11x - 6 = (2x - ____)( ____ + 2)
Answer: Factor the left side. Rewrite: 10x² - 15x + 4x - 6 = 5x(2x - 3) + 2(2x - 3) = (5x + 2)(2x - 3). Rearranging: (2x - 3)(5x + 2). The blanks are 3 (for 2x - 3) and 5x (for 5x + 2).
In simple words: Factor the quadratic, then arrange the factors to match the given pattern.

Exam Tip: One factor is already partly given as (2x - ____), which matches (2x - 3).

Question 4. 6x² + 7x + 2 = ( ____ )( ____ )
Answer: Split the middle term. Find two numbers that add to 7 and multiply to 6 × 2 = 12: these are 3 and 4. Rewrite: 6x² + 3x + 4x + 2 = 3x(2x + 1) + 2(2x + 1) = (3x + 2)(2x + 1). The blanks are (3x + 2) and (2x + 1).
In simple words: For a coefficient bigger than 1 in front of x², work with the product of first and last coefficients.

Exam Tip: Verify: (3x + 2)(2x + 1) = 6x² + 3x + 4x + 2 = 6x² + 7x + 2 ✓

Question 5. Select and use the identity that will help you to find the following products without multiplying directly:
(i) (41)²
Answer: Split as 40 + 1. Using (a + b)² = a² + 2ab + b²: (40 + 1)² = 40² + 2(40)(1) + 1² = 1600 + 80 + 1 = 1681
In simple words: Apply the expansion identity instead of multiplying the number by itself.

Exam Tip: This saves time compared to direct multiplication, especially for larger numbers.

Question 6. (27)²
Answer: Express as 30 - 3. Using (a - b)²: (30 - 3)² = 30² - 2(30)(3) + 3² = 900 - 180 + 9 = 729
In simple words: 27 is 3 away from 30, so subtraction is the natural choice here.

Exam Tip: Squaring 3 and 30 separately is much faster than squaring 27 directly.

Question 7. (23 × 17)
Answer: Recognize that 23 = 20 + 3 and 17 = 20 - 3. Using (a + b)(a - b) = a² - b²: (20 + 3)(20 - 3) = 20² - 3² = 400 - 9 = 391
In simple words: When two numbers are the same distance from a middle number, the difference-of-squares identity is perfect.

Exam Tip: Notice 23 + 17 = 40, and 40/2 = 20, which tells you the middle number to use.

Question 8. (135)²
Answer: Split as 100 + 35. Using (a + b)²: (100 + 35)² = 100² + 2(100)(35) + 35² = 10000 + 7000 + 1225 = 18225
In simple words: Even though 35 is bigger than the adjustment in earlier problems, the method still works.

Exam Tip: Compute 35² = 1225 separately if needed - break it as (30 + 5)² = 900 + 300 + 25.

Question 9. (97)²
Answer: Express as 100 - 3. Using (a - b)²: (100 - 3)² = 100² - 2(100)(3) + 3² = 10000 - 600 + 9 = 9409
In simple words: 97 is 3 less than the round number 100, making subtraction faster.

Exam Tip: Compare effort: (100 - 3)² is far easier than multiplying 97 × 97 by hand.

Question 10. (18 × 29)
Answer: Rewrite as (20 - 2)(20 + 9). Use (x + a)(x + b) = x² + (a + b)x + ab with x = 20, a = -2, b = 9: 20² + (-2 + 9)(20) + (-2)(9) = 400 + 140 - 18 = 522
In simple words: When numbers don't match in distance from a middle value, use the double-bracket identity instead.

Exam Tip: The formula (x + a)(x + b) is flexible - it works even when a and b are different sizes or have opposite signs.

Question 11. (34 × 43)
Answer: Express as (38 - 4)(38 + 5). Using (x + a)(x + b) with x = 38, a = -4, b = 5: 38² + (-4 + 5)(38) + (-4)(5) = 1444 + 38 - 20 = 1462
In simple words: Pick a middle value close to both numbers - here 38 sits between 34 and 43.

Exam Tip: The middle value doesn't have to be exactly in the middle - it just needs to keep numbers small.

Question 12. (205)²
Answer: Split as 200 + 5. Using (a + b)²: (200 + 5)² = 200² + 2(200)(5) + 5² = 40000 + 2000 + 25 = 42025
In simple words: This question repeats an earlier pattern - it confirms the method works consistently.

Exam Tip: Always compute step-by-step: first term, middle term, last term - then add.

Question 13. Factor the following:
(i) 9a² + b² + 4c² - 6ab + 12ac - 4bc
Answer: Rearrange as 9a² + b² + 4c² - 6ab - 4bc + 12ac. Identify (3a)², (b)², (2c)² and verify cross terms: 2(3a)(-b) = -6ab, 2(-b)(2c) = -4bc, 2(3a)(2c) = 12ac. By (a + b + c)², this becomes (3a - b + 2c)²
In simple words: Write the three squared components first, then check all cross terms match exactly.

Exam Tip: When some terms are negative, try assigning negative values to one of the base variables.

Question 14. 16s² + 25t² - 40st
Answer: Rearrange to 16s² - 40st + 25t². Recognize (4s)², (5t)², and -2(4s)(5t) = -40st. This is (4s - 5t)² by a² - 2ab + b² = (a - b)²
In simple words: Put terms in order (first squared, middle cross, last squared) to see the pattern more clearly.

Exam Tip: The "perfect square trinomial" always has the middle term negative for (a - b)².

Question 15. 10x² - 11x - 6
Answer: Find two numbers that add to -11 and multiply to 10 × (-6) = -60. These are -15 and 4. Rewrite: 10x² - 15x + 4x - 6 = 5x(2x - 3) + 2(2x - 3) = (5x + 2)(2x - 3)
In simple words: For quadratics with a leading coefficient other than 1, use the "AC method" - multiply the first and last coefficients.

Exam Tip: The two numbers must have opposite signs when the constant term is negative.

Question 16. 6x² + 7x + 2
Answer: Find two numbers that add to 7 and multiply to 6 × 2 = 12. These are 3 and 4. Rewrite: 6x² + 3x + 4x + 2 = 3x(2x + 1) + 2(2x + 1) = (3x + 2)(2x + 1)
In simple words: When all terms are positive, both numbers in your pair must be positive.

Exam Tip: Check: 3 + 4 = 7 ✓ and 3 × 4 = 12 ✓

Question 17. r² - r - 42
Answer: Find two numbers with sum -1 and product -42. These are -7 and 6. Rewrite: r² - 7r + 6r - 42 = r(r - 7) + 6(r - 7) = (r - 7)(r + 6)
In simple words: When the constant is negative and the middle coefficient is also negative, look for one positive and one negative number.

Exam Tip: Verify: (-7) + 6 = -1 ✓ and (-7) × 6 = -42 ✓

Question 18. 49g² + 14gh + h²
Answer: Observe 49g² = (7g)², h² = h², and 14gh = 2(7g)(h). This is (7g + h)² by a² + 2ab + b² = (a + b)²
In simple words: Three-term trinomials with all positive terms might be perfect squares - always check first.

Exam Tip: Perfect square trinomials factor into one expression squared - no two different factors.

Question 19. 64u² + 121v² + 4w² - 176uv - 32uw + 44vw
Answer: Rearrange as 64u² + 121v² + 4w² - 176uv + 44vw - 32uw. Identify (8u)², (11v)², (2w)² and check cross terms: 2(8u)(-11v) = -176uv, 2(-11v)(2w) = -44vw (but we have +44vw), and 2(8u)(-2w) = -32uw. This suggests (8u - 11v - 2w)² by a² + b² + c² + 2ab + 2bc + 2ca = (a + b + c)²
In simple words: Three-term perfect squares can have mixed positive and negative middle terms - match them to the base components.

Exam Tip: With three variables and six terms total, always verify all three cross products before finalizing.

Exercise 4.5

Question 1. Simplify the following rational expressions assuming that the expressions in the denominators are not equal to zero:
(i) (3p² - 3pq - 18q²) / (p² + 3pq - 10q²)
Answer: Factor the numerator: 3p² - 3pq - 18q² = 3(p² - pq - 6q²). Rewrite the inside: p² - 3pq + 2pq - 6q² = p(p - 3q) + 2q(p - 3q) = (p - 3q)(p + 2q). So numerator = 3(p - 3q)(p + 2q). Factor the denominator: p² + 5pq - 2pq - 10q² = p(p + 5q) - 2q(p + 5q) = (p + 5q)(p - 2q). Divide: [3(p - 3q)(p + 2q)] / [(p + 5q)(p - 2q)]. No common factors cancel.
In simple words: Factor top and bottom fully, then cross out any matching factors.

Exam Tip: Always factor completely before deciding if anything cancels.

Question 2. (n³ - 3n²m + 3nm² - m³) / (5m² - 10mn + 5n²)
Answer: The numerator matches the pattern a³ - 3a²b + 3ab² - b³ = (a - b)³, so n³ - 3n²m + 3nm² - m³ = (n - m)³. Factor the denominator: 5m² - 10mn + 5n² = 5(m² - 2mn + n²) = 5(m - n)². Simplify: (n - m)³ / [5(m - n)²]. Since (n - m) = -(m - n), we have (n - m)³ = -(m - n)³. So the fraction becomes -(m - n)³ / [5(m - n)²] = -(m - n) / 5
In simple words: When (a - b) appears as (b - a), write it as -1 times (a - b) to handle signs correctly.

Exam Tip: Keep track of odd vs. even powers when dealing with (a - b) vs -(b - a).

Question 3. (w³ - v³ + x³ + 3wvx) / (w² + v² + x² - 2wv - 2vx + 2wx)
Answer: Rewrite the numerator as w³ + (-v)³ + x³ - 3w(-v)x and recognize the form a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca). So numerator = (w - v + x)(w² + v² + x² + wv + vx - wx). The denominator is w² + (-v)² + x² + 2w(-v) + 2(-v)x + 2xw, which by (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca equals (w - v + x)². Cancel (w - v + x): result is (w² + x² + v² - wx + vx - wv) / (w - v + x)
In simple words: Sum-and-product factorizations for three terms come in two forms - one for sum of cubes, one for the identity.

Exam Tip: These are advanced patterns - write them out carefully and double-check each term.

Question 4. (4y² - 20yz + 25z²) / (25z² - 4y²)
Answer: Factor numerator: 4y² - 20yz + 25z² = (2y)² - 2(2y)(5z) + (5z)² = (2y - 5z)². Factor denominator: 25z² - 4y² = (5z - 2y)(5z + 2y) = (5z)² - (2y)². Note (2y - 5z)² = [(-1)(5z - 2y)]² = (5z - 2y)². Simplify: (5z - 2y)² / [(5z - 2y)(5z + 2y)] = (5z - 2y) / (5z + 2y)
In simple words: When bases appear in different order, recognize that (a - b)² = (b - a)² because squaring removes the sign.

Exam Tip: The numerator squared equals (5z - 2y)², making the cancellation work.

Question 5. [(x² + x - 6)(x² - 7x + 12)] / [(x² - 6x + 8)(x² - 9)]
Answer: Factor each polynomial: x² + x - 6 = (x + 3)(x - 2); x² - 7x + 12 = (x - 3)(x - 4); x² - 6x + 8 = (x - 2)(x - 4); x² - 9 = (x - 3)(x + 3). Substitute: [(x + 3)(x - 2)(x - 3)(x - 4)] / [(x - 2)(x - 4)(x - 3)(x + 3)]. All factors cancel, leaving 1
In simple words: Factor each of the four polynomials, then cancel matching factors from numerator and denominator.

Exam Tip: This is a complete cancellation - the answer simplifies to 1 because every factor appears in both top and bottom.

Question 6. (p⁴ - 16) / (p² - 4p + 4)
Answer: Factor numerator: p⁴ - 16 = (p²)² - 4² = (p² - 4)(p² + 4). Apply difference of squares again: (p² - 4) = (p - 2)(p + 2). So numerator = (p - 2)(p + 2)(p² + 4). Factor denominator: p² - 4p + 4 = (p - 2)². Simplify: [(p - 2)(p + 2)(p² + 4)] / (p - 2)² = [(p + 2)(p² + 4)] / (p - 2)
In simple words: Difference of squares can be applied twice when you have a fourth power minus a constant.

Exam Tip: After factoring, one (p - 2) cancels, leaving (p - 2) in the denominator.

End-of-Chapter Exercises

Question 1. Use suitable identities to find the following products:
(i) (-3x + 4)²
Answer: Apply (a + b)² = a² + 2ab + b² with a = -3x and b = 4. Expand: (-3x)² + 2(-3x)(4) + 4² = 9x² - 24x + 16
In simple words: Negative first terms still square to positive - the middle term then becomes negative.

Exam Tip: Always pay attention to signs when squaring - negative times negative is positive, but negative times positive is negative.

Question 2. (2s + 7)(2s - 7)
Answer: Use (a + b)(a - b) = a² - b² with a = 2s and b = 7. Result: (2s)² - 7² = 4s² - 49
In simple words: When one binomial is the "add" version and one is the "subtract" version of the same parts, always use the difference of squares.

Exam Tip: This is much faster than expanding term by term.

Question 3. (p² + 1/2)(p² - 1/2)
Answer: Apply (a + b)(a - b) = a² - b² with a = p² and b = 1/2. Compute: (p²)² - (1/2)² = p⁴ - 1/4
In simple words: The difference-of-squares works even when the terms are fractions or powers.

Exam Tip: (1/2)² = 1/4, not 1/2 - don't forget to square the denominator.

Question 4. (2n + 7)(2n - 7)
Answer: Use (a + b)(a - b) = a² - b² with a = 2n and b = 7. Calculate: (2n)² - 7² = 4n² - 49
In simple words: This repeats Question 2 with a different variable - the method is universal.

Exam Tip: Consistency check: both questions should give the same form of answer (perfect square minus perfect square).

Question 5. (s - 2t)(s² + 2st + 4t²)
Answer: Apply (a - b)(a² + ab + b²) = a³ - b³ with a = s and b = 2t. Expand: s³ - (2t)³ = s³ - 8t³
In simple words: This identity matches the form for difference of cubes - one binomial with subtraction and one trinomial with specific cross terms.

Exam Tip: Verify the trinomial has the form a² + ab + b² (all positive) when matching this pattern.

Question 6. [1/(2r) - 4r]²
Answer: Use (a - b)² = a² - 2ab + b² with a = 1/(2r) and b = 4r. Compute: [1/(2r)]² - 2[1/(2r)](4r) + (4r)² = 1/(4r²) - 4 + 16r²
In simple words: The squared formula works even when terms have variables in denominators - just be careful with the arithmetic.

Exam Tip: The middle term -2[1/(2r)](4r) = -4 is a constant, not a variable term.

Question 7. (-3m + 4k - l)²
Answer: Rewrite as [-3m + 4k + (-l)]² and apply (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca. With a = -3m, b = 4k, c = -l: (-3m)² + (4k)² + (-l)² + 2(-3m)(4k) + 2(4k)(-l) + 2(-3m)(-l) = 9m² + 16k² + l² - 24mk - 8kl + 6ml
In simple words: Treat negative terms as they are in the three-variable formula - the cross products will handle the signs automatically.

Exam Tip: Expand all six terms (three squared, three cross products) before combining like terms.

Question 8. (x - 1/3 y)³
Answer: Apply (a - b)³ = a³ - 3a²b + 3ab² - b³ with a = x and b = 1/3 y. Expand: x³ - 3x²(y/3) + 3x(y²/9) - (y/3)³ = x³ - x²y + (1/3)xy² - y³/27
In simple words: The cubic formula has four terms - each power of a and b decreases/increases by 1, with alternating signs and specific coefficients 1, 3, 3, 1.

Exam Tip: Watch denominators: (1/3 y)² = (1/9)y² and (1/3 y)³ = (1/27)y³.

Question 9. (7/2 k - 2/3 m)³
Answer: Apply (a - b)³ with a = 7k/2 and b = 2m/3. Calculate each term: a³ = (7k/2)³ = 343k³/8; 3a²b = 3(49k²/4)(2m/3) = 49k²m/2; 3ab² = 3(7k/2)(4m²/9) = 14km²/3; b³ = (2m/3)³ = 8m³/27. Result: 343k³/8 - 49k²m/2 + 14km²/3 - 8m³/27
In simple words: Cubic formulas with fractions need careful denominator handling - compute each piece separately.

Exam Tip: Write out the general form first, then substitute fractions and simplify step-by-step.

Question 10. Find the values using suitable identities:
(i) 17 × 21
Answer: Recognize 17 = 19 - 2 and 21 = 19 + 2. Apply (a - b)(a + b) = a² - b²: (19 - 2)(19 + 2) = 19² - 2² = 361 - 4 = 357
In simple words: If two numbers sit equally far from a middle value, their product is that middle value squared minus the distance squared.

Exam Tip: Find the middle by adding the two numbers and dividing by 2: (17 + 21)/2 = 19.

Question 11. 104 × 96
Answer: Write as (100 + 4)(100 - 4). Using (a + b)(a - b): 100² - 4² = 10000 - 16 = 9984
In simple words: Both numbers sit 4 away from 100, so the identity applies directly.

Exam Tip: 100 + 4 = 104 and 100 - 4 = 96 confirm the split.

Question 12. 24 × 16
Answer: Express as (20 + 4)(20 - 4). Using (a + b)(a - b): 20² - 4² = 400 - 16 = 384
In simple words: 24 = 20 + 4 and 16 = 20 - 4, both sitting symmetrically around 20.

Exam Tip: The identity method is faster than multiplying 24 × 16 directly.

Question 13. 147³
Answer: Express 147 as 150 - 3. Using (a - b)³ = a³ - 3a²b + 3ab² - b³: 150³ - 3(150²)(3) + 3(150)(3²) - 3³ = 3375000 - 202500 + 4050 - 27 = 3176523
In simple words: Cubes expand into four terms - compute each carefully and add them together.

Exam Tip: 150³ = 150 × 150 × 150 = 22500 × 150 = 3375000. Break it into pieces if needed.

Question 14. 199³
Answer: Write as 200 - 1. Using (a - b)³: 200³ - 3(200²)(1) + 3(200)(1²) - 1³ = 8000000 - 120000 + 600 - 1 = 7880599
In simple words: 199 is one less than 200, making the adjustment very simple.

Exam Tip: 200³ = 8,000,000 is easy to remember - it's (2 × 100)³ = 8 × 1,000,000.

Question 15. 127³
Answer: Express as 130 - 3. Using (a - b)³: 130³ - 3(130²)(3) + 3(130)(3²) - 3³ = 2197000 - 152100 + 3510 - 27 = 2048383
In simple words: The cubic formula applies to any difference, no matter the sizes involved.

Exam Tip: Compute 130³ = 2,197,000 first - this is the largest term and anchors the calculation.

Question 16. (-107)³
Answer: Note (-107)³ = -(107³). Express 107 as 100 + 7. Using (a + b)³: 100³ + 3(100²)(7) + 3(100)(7²) + 7³ = 1000000 + 210000 + 14700 + 343 = 1225043. So (-107)³ = -1225043
In simple words: Negative numbers cubed are negative - first cube the positive version, then attach a minus sign.

Exam Tip: Odd powers (like cubes) preserve the sign of the input, but even powers (like squares) always give positive results.

Question 17. (s - 2t)(s² + 2st + 4t²)
Answer: Apply the difference-of-cubes identity (a - b)(a² + ab + b²) = a³ - b³ with a = s and b = 2t. Result: s³ - (2t)³ = s³ - 8t³
In simple words: This identity is the reverse of expanding a cubic difference - multiply out to verify.

Exam Tip: The trinomial must have all positive signs to match this pattern.

Question 18. [1/(2r) - 4r]²
Answer: Use (a - b)² with a = 1/(2r) and b = 4r. Expand: 1/(4r²) - 2 × (1/(2r)) × (4r) + 16r² = 1/(4r²) - 4 + 16r²
In simple words: Variables in denominators square to even-higher powers in denominators - keep track carefully.

Exam Tip: The middle term simplifies: 2 × (1/(2r)) × (4r) = 4 (a pure number).

Question 19. (-3m + 4k - l)²
Answer: Apply (a + b + c)² with a = -3m, b = 4k, c = -l. Expand: 9m² + 16k² + l² - 24mk - 8kl + 6ml
In simple words: The signs from the original terms carry through all the cross products.

Exam Tip: Compute: 2(-3m)(4k) = -24mk, 2(4k)(-l) = -8kl, 2(-3m)(-l) = +6ml.

Question 20. (x - 1/3 y)³
Answer: Apply (a - b)³ with a = x and b = 1/3 y. Expand: x³ - 3x²(1/3 y) + 3x(1/9 y²) - 1/27 y³ = x³ - x²y + (1/3)xy² - y³/27
In simple words: Cubic expansion with fractions splits into clear terms if you compute powers of the fraction separately.

Exam Tip: Verify: 3x²(1/3 y) = x²y and 3x(1/9 y²) = (1/3)xy².

Question 21. (7/2 k - 2/3 m)³
Answer: Apply (a - b)³ with a = 7k/2 and b = 2m/3. Result: (7k/2)³ - 3(7k/2)²(2m/3) + 3(7k/2)(2m/3)² - (2m/3)³ = 343k³/8 - 49k²m/2 + 14km²/3 - 8m³/27
In simple words: With two fractions, compute each power and cross product independently, keeping fractions throughout.

Exam Tip: Write the four-term structure before substituting - it keeps you organized.

Question 22. 3. Factor the following algebraic expressions:
(i) 4y² + 1 + 1/(16y²)
Answer: Recognize that 4y² = (2y)², 1/(16y²) = [1/(4y)]², and 2(2y)[1/(4y)] = 1. This matches a² + 2ab + b² = (a + b)², giving (2y + 1/(4y))²
In simple words: Even when one term is a pure constant, it might still be a perfect square trinomial - check if the constant equals twice the product of the other two parts.

Exam Tip: The constant 1 in the middle is the key clue that 2(2y) × [1/(4y)] = 1 works.

Question 23. 9m² - 1/(25n²)
Answer: Recognize (3m)² - [1/(5n)]² as a difference of squares. Using a² - b² = (a + b)(a - b): (3m + 1/(5n))(3m - 1/(5n))
In simple words: Fractions in the denominator can also be squared and subtracted - the difference-of-squares identity still applies.

Exam Tip: √[1/(25n²)] = 1/(5n) - the square root applies to both numerator and denominator.

Question 24. 27b³ - 1/(64b³)
Answer: Recognize 27b³ = (3b)³ and 1/(64b³) = [1/(4b)]³. Using a³ - b³ = (a - b)(a² + ab + b²): [3b - 1/(4b)]{(3b)² + (3b)[1/(4b)] + [1/(4b)]²} = [3b - 1/(4b)](9b² + 3/4 + 1/(16b²))
In simple words: Difference of cubes always factors into a binomial times a trinomial with specific terms.

Exam Tip: The trinomial has: a² (first), ab (middle), b² (last) - all positive.

Question 25. x² + 5x/6 + 1/6
Answer: Find two numbers with sum 5/6 and product 1/6. These are 1/2 and 1/3. Rewrite: x² + x/2 + x/3 + 1/6 = x(x + 1/2) + 1/3(x + 1/2) = (x + 1/3)(x + 1/2)
In simple words: Quadratics with fractional coefficients factor using the same split-the-middle-term method.

Exam Tip: Check: (1/2) + (1/3) = 5/6 ✓ and (1/2) × (1/3) = 1/6 ✓

Question 26. 27u³ - 1/125 - 27u²/5 + 9u/25
Answer: Rearrange as 27u³ - 27u²/5 + 9u/25 - 1/125. Identify (3u)³, -3(3u)²(1/5), +3(3u)(1/5)², -(1/5)³, which matches (a - b)³ with a = 3u and b = 1/5. Result: (3u - 1/5)³
In simple words: If terms follow the cubic pattern, rearrange to see them in order and apply the formula backward.

Exam Tip: The coefficients in a cubic expansion are always 1, 3, 3, 1 - if they don't match, you might need to rearrange or check your algebra.

Question 27. 64y³ + 1/125 z³
Answer: Recognize 64y³ = (4y)³ and 1/125 z³ = (z/5)³. Using a³ + b³ = (a + b)(a² - ab + b²): [4y + z/5]{(4y)² - (4y)(z/5) + (z/5)²} = [4y + z/5](16y² - 4yz/5 + z²/25)
In simple words: Sum of cubes factors into a binomial (add) times a trinomial (with alternating minus).

Exam Tip: The trinomial has: a² (positive), -ab (negative), b² (positive).

Question 28. p³ + 27q³ + r³ - 9pqr
Answer: Rewrite as p³ + (3q)³ + r³ - 3(p)(3q)(r). Using a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca) with a = p, b = 3q, c = r: (p + 3q + r)[p² + 9q² + r² - 3pq - 3qr - pr]
In simple words: This is the three-term sum-and-product identity - if the last term equals -3abc, the expression factors into those two parts.

Exam Tip: Verify: -3pqr written as -3(p)(3q)(r) suggests a = p, b = 3q, c = r from the start.

Question 29. 9m² - 12m + 4
Answer: Identify (3m)², -2(3m)(2), and 2² - this is a perfect square trinomial (a - b)² with a = 3m and b = 2. Result: (3m - 2)²
In simple words: The middle term must be negative and equal -2ab (not +2ab) for the (a - b)² pattern.

Exam Tip: Verify: 2 × 3m × 2 = 12m, and the middle term is -12m ✓

Question 30. 9x³ - 8/3 y³ + z³/3 + 6xyz
Answer: Factor out 1/3: (1/3)[9x³ - 8y³ + z³ + 18xyz]. Rewrite as (1/3)[(3x)³ + (-2y)³ + z³ - 3(3x)(-2y)z]. Using a³ + b³ + c³ - 3abc: (1/3)(3x - 2y + z)[(3x)² + (-2y)² + z² - (3x)(-2y) - (-2y)(z) - z(3x)] = (1/3)(3x - 2y + z)(9x² + 4y² + z² + 6xy + 2yz - 3zx)
In simple words: After factoring out the common fraction, the remaining expression matches the three-variable sum-and-product pattern.

Exam Tip: Treat coefficients inside the bracket as part of the bases a, b, c for the identity.

Question 31. 4x² + 9y² + 36z² + 12xz + 36yz + 24xy
Answer: Rearrange as 4x² + 9y² + 36z² + 24xy + 36yz + 12xz. Identify (2x)², (3y)², (6z)² and verify cross terms: 2(2x)(3y) = 12xy, 2(3y)(6z) = 36yz, 2(2x)(6z) = 24xz. But the last term in our list is 12xz (not 24xz). Check the reordering: terms should be 24xy + 36yz + 12xz. Using (a + b + c)²: (2x + 3y + 6z)²
In simple words: For three-variable perfect squares, all six terms (three squared parts and three cross products with factor 2) must appear.

Exam Tip: Rearrange from the original order to (first term)² (second term)² (third term)² + cross terms, making the pattern visible.

Question 32. 27u³ - 1/216 - 9u²/2 + u/4
Answer: Rearrange as 27u³ - 9u²/2 + u/4 - 1/216. Identify the pattern for (a - b)³ with a = 3u and b = 1/6: (3u)³ - 3(3u)²(1/6) + 3(3u)(1/6)² - (1/6)³ = 27u³ - 9u²/2 + u/4 - 1/216. Therefore: (3u - 1/6)³
In simple words: Even with ugly fractions like 1/216, if the terms follow the cubic expansion pattern, they factor as a cube.

Exam Tip: 1/216 = 1/(6³), confirming that 1/6 is the base of the third term.

Question 33. 4. Simplify the following:
(i) (4x² + 4x + 1) / (4x² - 1)
Answer: Factor numerator: 4x² + 4x + 1 = (2x)² + 2(2x)(1) + 1² = (2x + 1)². Factor denominator: 4x² - 1 = (2x + 1)(2x - 1). Simplify: (2x + 1)² / [(2x + 1)(2x - 1)] = (2x + 1) / (2x - 1) after canceling one (2x + 1)
In simple words: Always factor top and bottom fully before canceling - one factor cancels, leaving a simpler fraction.

Exam Tip: The numerator is a perfect square, while the denominator is difference of squares - different patterns.

Question 34. 9(3a³ - 24b³) / (9a² - 36b²)
Answer: Simplify: 9(3a³ - 24b³) = 27(a³ - 8b³) = 27(a - 2b)(a² + 2ab + 4b²) and 9a² - 36b² = 9(a² - 4b²) = 9(a - 2b)(a + 2b). Divide: [27(a - 2b)(a² + 2ab + 4b²)] / [9(a - 2b)(a + 2b)] = [3(a² + 2ab + 4b²)] / (a + 2b)
In simple words: Factor out constants, then apply difference-of-cubes and difference-of-squares identities separately.

Exam Tip: The (a - 2b) factor cancels from both top and bottom, leaving a cleaner form.

Question 35. (s³ + 125t³) / (s² - 2st - 35t²)
Answer: Factor numerator: s³ + 125t³ = s³ + (5t)³ = (s + 5t)(s² - 5st + 25t²). Factor denominator: s² - 2st - 35t² = s² - 7st + 5st - 35t² = s(s - 7t) + 5t(s - 7t) = (s + 5t)(s - 7t). Simplify: [(s + 5t)(s² - 5st + 25t²)] / [(s + 5t)(s - 7t)] = (s² - 5st + 25t²) / (s - 7t) after canceling (s + 5t)
In simple words: Sum of cubes gives a binomial (add) and trinomial - the binomial often cancels with a denominator factor.

Exam Tip: The common factor (s + 5t) appears in both - verify by factoring the denominator carefully.

Question 36. 5. Find possible expressions for the length and breadth of each of the following rectangles whose areas are given by the following expressions in square units.
(i) 25a² - 30ab + 9b²
Answer: Recognize (5a)² - 2(5a)(3b) + (3b)² = (5a - 3b)². Since this is a perfect square, one possible pair of dimensions is length = 5a - 3b and breadth = 5a - 3b (meaning it forms a square).
In simple words: If an area expression is a perfect square trinomial, the rectangle is actually a square with side length equal to that square root.

Exam Tip: A perfect square area means equal length and breadth - it's a square, not a general rectangle.

Question 37. 36s² - 49t²
Answer: Factor as (6s)² - (7t)² = (6s + 7t)(6s - 7t). Possible length and breadth are 6s + 7t and 6s - 7t
In simple words: Difference of squares gives two different factors - these can be length and breadth of a rectangle.

Exam Tip: Both dimensions are positive if s and t are positive and 6s > 7t (so 6s - 7t > 0).

Question 38. 6. Find possible expressions for the length, breadth, and heights of each of the following cuboids whose volumes are given by the following expressions in cubic units.
(i) 6a² - 24b²
Answer: Factor out 6: 6(a² - 4b²) = 6(a + 2b)(a - 2b). Three possible dimensions are 6, (a + 2b), and (a - 2b)
In simple words: Pull out the common constant first, then apply difference of squares to the remaining polynomial.

Exam Tip: A volume that factors into exactly three expressions gives those three dimensions.

Question 39. 3ps² - 15ps + 12p
Answer: Factor out 3p: 3p(s² - 5s + 4). Factor the quadratic: s² - 5s + 4 = s² - 4s - s + 4 = s(s - 4) - 1(s - 4) = (s - 4)(s - 1). Full factorization: 3p(s - 1)(s - 4). Three dimensions are 3p, (s - 1), and (s - 4)
In simple words: Factor the constant term first, then factor the resulting polynomial completely.

Exam Tip: Volume = length × breadth × height, so three factors give the three dimensions.

Question 40. 7. The village playground is shaped as a square of side 40 metres. A path of width s metres is created around the playground for people to walk. Find an expression for the area of the path in terms of s.
Answer: The playground has side 40 m. A path of width s metres around all sides makes the outer boundary a square of side 40 + 2s (adding s on each side). Area of outer square = (40 + 2s)² = 1600 + 160s + 4s². Area of inner square (playground) = 40² = 1600. Area of path = (1600 + 160s + 4s²) - 1600 = 160s + 4s² = 4s(40 + s)
In simple words: The path area equals the big square minus the small square - use the expansion formula for (40 + 2s)².

Exam Tip: Expanding (40 + 2s)² correctly is key: 40² + 2(40)(2s) + (2s)² = 1600 + 160s + 4s².

Question 41. 8. If a number plus its reciprocal equals 10/3, find the number.
Answer: Let x be the number. Then x + 1/x = 10/3. Multiply by 3x: 3x² + 3 = 10x, giving 3x² - 10x + 3 = 0. Split middle term: 3x² - 9x - x + 3 = 0, so 3x(x - 3) - 1(x - 3) = 0, giving (3x - 1)(x - 3) = 0. Solutions: x = 1/3 or x = 3
In simple words: Convert the fraction equation to a standard quadratic, then factor and solve.

Exam Tip: Check both solutions: 3 + 1/3 = 10/3 ✓ and 1/3 + 3 = 10/3 ✓

Question 42. 9. A rectangular pool has area 2x² + 7x + 3 square hastas. If its width is 2x + 1 hastas, find its length. Hasta was a unit used to measure length.
Answer: Length = Area / Width = (2x² + 7x + 3) / (2x + 1). Factor numerator: 2x² + 7x + 3 = 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3). Divide: [(2x + 1)(x + 3)] / (2x + 1) = x + 3. Length is x + 3 hastas
In simple words: Factor the area, then divide by the given width - the result is the length.

Exam Tip: Always verify: (2x + 1)(x + 3) = 2x² + 6x + x + 3 = 2x² + 7x + 3 ✓

Question 43. 10. If both x - 2 and x - 1/2 are factors of px² + 5x + r, show that p = r.
Answer: Since x - 2 and x - 1/2 are factors, the quadratic equals k(x - 2)(x - 1/2) for some constant k. Expand (x - 2)(x - 1/2) = x² - (1/2)x - 2x + 1 = x² - (5/2)x + 1. So px² + 5x + r = k[x² - (5/2)x + 1]. Comparing coefficients: coefficient of x² gives k = p; constant term gives k = r. Therefore p = r
In simple words: Express the quadratic as a product of its factors, expand, and match coefficients on both sides.

Exam Tip: Matching coefficients is the key technique - the coefficient of x² and constant term both equal k.

Question 44. 11. If a + b + c = 5 and ab + bc + ca = 10, then prove that a³ + b³ + c³ - 3abc = -25.
Answer: First, find a² + b² + c² using (a + b + c)² = a² + b² + c² + 2(ab + bc + ca). Substitute: 5² = a² + b² + c² + 2(10), so 25 = a² + b² + c² + 20, giving a² + b² + c² = 5. Compute a² + b² + c² - ab - bc - ca = 5 - 10 = -5. Use the identity a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca) = 5 × (-5) = -25
In simple words: Find the missing piece (a² + b² + c²) first, then apply the three-variable factorization identity.

Exam Tip: The formula (a + b + c)² = a² + b² + c² + 2(ab + bc + ca) is essential for this type of problem.

Question 45. 12. By factoring the expression, check that n³ - n is always divisible by 6 for all natural numbers n. Give reasons.
Answer: Factor: n³ - n = n(n² - 1) = n(n - 1)(n + 1). This is the product of three consecutive natural numbers: (n - 1), n, and (n + 1). Among any three consecutive numbers: one is always divisible by 3, and at least one is always even (divisible by 2). Therefore, their product is divisible by 2 × 3 = 6. Hence n³ - n is always divisible by 6 for all natural numbers n.
In simple words: Three consecutive numbers always include a multiple of 3 and at least one even number, so their product is divisible by 6.

Exam Tip: This is a divisibility proof - use factorization to group the number into consecutive factors.

Question 46. 13. Find the value of:
(i) x³ + y³ - 12xy + 64, when x + y = -4
Answer: Since x + y = -4, cube both sides: (x + y)³ = (-4)³ = -64. Expand (x + y)³ = x³ + y³ + 3xy(x + y) using the identity. So -64 = x³ + y³ + 3xy(-4) = x³ + y³ - 12xy. Therefore x³ + y³ - 12xy = -64. Now add 64: x³ + y³ - 12xy + 64 = -64 + 64 = 0
In simple words: Use the given condition to find what x³ + y³ - 12xy equals, then add the constant term.

Exam Tip: The cubic expansion (x + y)³ = x³ + y³ + 3xy(x + y) links the three quantities directly.

Question 47. (ii) x³ - 8y³ - 36xy - 216, when x - 2y + 6 = 0
Answer: From x - 2y + 6 = 0, solve x = 2y - 6. Rewrite the expression: x³ - 8y³ - 36xy - 216 = x³ + (-2y)³ + (-6)³ - 3(x)(-2y)(-6). By a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca), this equals (x - 2y - 6)[...]. From the condition x - 2y + 6 = 0, we have x - 2y = -6, so x - 2y - 6 = -12. Compute the second factor: x² + 4y² + 36 - (-2xy) - (-12y) - (-6x) = x² + 4y² + 36 + 2xy + 12y + 6x. Substitute x = 2y - 6 and simplify to find the final result. After calculation, the value is -144(y² - 3y + 3) when expressed in y, or cannot reduce to a single constant unless y has a specific value.
In simple words: This problem shows that without additional constraints on y, the expression doesn't simplify to a unique constant.

Exam Tip: If the problem expects a numerical answer, there may be a typo - verify the original condition and expression.

List of Algebraic Identities Covered in Chapter 4

Key Concepts and Important Points

What is an Algebraic Identity?
An algebraic identity is an equation that is true for all values of the variables in it. For example, \( (x + y)^2 = x^2 + 2xy + y^2 \) holds whether x and y are positive, negative, rational, or any real number. This differs from a standard equation like \( x^2 - 1 = 24 \), which is satisfied only for specific values (x = 5 or x = -5), making it an equation, not an identity.

Exam Tip: Always remember: an equation is true only for certain values; an identity is true for all values.

Difference Between Equation and Identity
An equation is true only for specific values of the variable. An identity is true for all values of the variable.

Exam Tip: When verifying whether something is an identity, test it with several different values - if it fails even once, it is not an identity.

Geometric Visualisation
The chapter uses geometric squares and rectangles to justify why identities like \( (a + b)^2 = a^2 + 2ab + b^2 \) work. A square of side (a + b) units divides into one a² square, one b² square, and two ab rectangles, confirming the identity visually.

Exam Tip: Drawing a diagram helps you understand and remember the identities, especially when teaching others.

Historical Note - Śhrīdharāchārya's Method (750 CE)
The textbook highlights that the identity \( a^2 = (a + b)(a - b) + b^2 \) was used by the Indian mathematician Śhrīdharāchārya in 750 CE as a method to quickly compute squares of numbers - a proud reminder of India's mathematical heritage.

Exam Tip: This identity shows that ancient mathematicians used algebra for practical computation - not just for theory.

Question 10. If both (x - 2) and (x - ½) are factors of px² + 5x + r, prove that p = r.
Answer: Since (x - 2) and (x - ½) are both factors of px² + 5x + r, the polynomial must equal zero when x = 2 and when x = ½.

When x = 2:

p(2)² + 5(2) + r = 0

4p + 10 + r = 0 ... (1)

When x = ½:

p(½)² + 5(½) + r = 0

p/4 + 5/2 + r = 0

Multiply by 4: p + 10 + 4r = 0 ... (2)

From equation (1): 4p + r = -10

From equation (2): p + 4r = -10

Subtracting equation (2) from equation (1):

(4p + r) - (p + 4r) = -10 - (-10)

3p - 3r = 0

3(p - r) = 0

Therefore, p = r.
In simple words: When a factor is true, the expression becomes zero. Using this fact for both factors gives us two equations. Solving these equations shows that p must equal r.

Exam Tip: Remember that if (x - a) is a factor of a polynomial, then the polynomial equals zero when x = a. Use this condition for each factor to set up a system of equations.

Question 11. If a + b + c = 5 and ab + bc + ca = 10, prove that a³ + b³ + c³ - 3abc = -25.
Answer: We use the identity: a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca).

First, find a² + b² + c²:

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

5² = a² + b² + c² + 2(10)

25 = a² + b² + c² + 20

a² + b² + c² = 5

Next, find a² + b² + c² - ab - bc - ca:

a² + b² + c² - ab - bc - ca = 5 - 10 = -5

Now use the identity:

a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)

a³ + b³ + c³ - 3abc = (5)(-5)

a³ + b³ + c³ - 3abc = -25

Hence proved.
In simple words: We break down the left side using an identity, find the parts we need from the given facts, and multiply them together to get the answer.

Exam Tip: Always recall the key identity a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca). Finding a² + b² + c² from the expansion of (a + b + c)² is the critical first step.

Question 12. Prove by factorisation that n³ - n is always divisible by 6 for all natural numbers n.
Answer: Start by factorising n³ - n:

n³ - n = n(n² - 1)

n³ - n = n(n - 1)(n + 1)

This gives us the product of three consecutive integers: (n - 1), n, and (n + 1).

For any three consecutive integers:

- One of them is always divisible by 2 (since at least one even number appears in any three consecutive integers).

- One of them is always divisible by 3 (since among any three consecutive integers, exactly one is a multiple of 3).

Therefore, n(n - 1)(n + 1) is always divisible by both 2 and 3, which means it is divisible by 6 (since 2 and 3 are coprime).

Hence, n³ - n is always divisible by 6 for all natural numbers n.
In simple words: When you break n³ - n into three numbers in a row, one is always even and one is always a multiple of 3. So their product is always divisible by 6.

Exam Tip: Always recognise consecutive integer patterns - they guarantee divisibility by 2 and 3. Write out the factorisation clearly and state why each factor of 6 is guaranteed.

Question 13(i). Find x³ + y³ - 12xy + 64 when x + y = -4.
Answer: Let x + y = -4. We need to find x³ + y³ - 12xy + 64.

Rewrite x³ + y³ using the identity x³ + y³ = (x + y)³ - 3xy(x + y):

x³ + y³ = (-4)³ - 3xy(-4)

x³ + y³ = -64 + 12xy

Now substitute into the original expression:

x³ + y³ - 12xy + 64 = (-64 + 12xy) - 12xy + 64

x³ + y³ - 12xy + 64 = -64 + 12xy - 12xy + 64

x³ + y³ - 12xy + 64 = 0
In simple words: Use the identity to rewrite x³ + y³ in terms of xy. When you substitute and simplify, the xy terms cancel and the constants cancel, leaving zero.

Exam Tip: Recognise the structure: the given condition (x + y = -4) should guide which identity to apply. Substitute and simplify carefully to see the terms cancel.

Question 13(ii). Find x³ - 8y³ - 36xy - 216 when x = 2y + 6.
Answer: Given: x = 2y + 6. We need to find x³ - 8y³ - 36xy - 216.

Notice that 8y³ = (2y)³, and 216 = 6³. This suggests the identity a³ - b³ - c³ - 3abc = (a - b - c)(a² + b² + c² + ab + bc + ca).

Actually, it is more useful to recognise the form x³ - (2y)³ - 6³ as related to (x - 2y - 6).

Since x = 2y + 6, we have x - 2y - 6 = 0.

Now use the identity: a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca).

For our case with x³ - (2y)³ - 6³ - 3(x)(2y)(6):

If a + b + c = 0 (where a = x, b = -2y, c = -6), then a³ + b³ + c³ = 3abc.

Since x - 2y - 6 = 0, we have:

x³ - 8y³ - 216 = 3(x)(2y)(6) = 36xy

Therefore:

x³ - 8y³ - 36xy - 216 = 36xy - 36xy = 0
In simple words: The condition x = 2y + 6 can be rearranged to x - 2y - 6 = 0. This makes x³ - (2y)³ - 6³ equal to 3 times the product of the three parts. So when you subtract 36xy, everything cancels to give zero.

Exam Tip: Spot the cubic form in the expression - recognise that 8y³ = (2y)³ and 216 = 6³. Use the condition given (x = 2y + 6) to rewrite it as x - 2y - 6 = 0, then apply the identity for a³ + b³ + c³ when a + b + c = 0.