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NCERT Class 9 Mathematics Exercise Set 4.4 Solutions: Exploring Algebraic Identities

AI-written, checked on 23 Oct 2026

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Step-by-step working for every question of Exercise Set 4.4 in NCERT's Class 9 Mathematics book, chapter “Exploring Algebraic Identities”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.

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  1. Ex 4.4 Q1

    Fill the blanks: (i) s² − 11s + 24 = (__)(__); (ii) (__)(x + 1) = 3x² − 4x − 7; (iii) 10x² − 11x − 6 = (2x − __)(__ + 2); (iv) 6x² + 7x + 2 = (__)(__).

    1. (i) We need two numbers that multiply to 24 and add to −11, which are −3 and −8.
    2. (i) So s² − 11s + 24 = (s − 3)(s − 8).
    3. (ii) Split −4x using two numbers with product 3 × (−7) = −21 and sum −4, namely −7 and 3: 3x² − 4x − 7 = 3x² + 3x − 7x − 7.
    4. (ii) Group: 3x(x + 1) − 7(x + 1) = (3x − 7)(x + 1), so the blank is 3x − 7.
    5. (iii) Split −11x using two numbers with product 10 × (−6) = −60 and sum −11, namely −15 and 4.
    6. (iii) So 10x² − 11x − 6 = 10x² − 15x + 4x − 6 = 5x(2x − 3) + 2(2x − 3) = (2x − 3)(5x + 2).
    7. (iv) Split 7x using two numbers with product 6 × 2 = 12 and sum 7, namely 4 and 3.
    8. (iv) So 6x² + 7x + 2 = 6x² + 4x + 3x + 2 = 2x(3x + 2) + 1(3x + 2) = (2x + 1)(3x + 2).

    Answer: (i) (s − 3)(s − 8); (ii) (3x − 7)(x + 1); (iii) (2x − 3)(5x + 2), blanks 3 and 5x; (iv) (2x + 1)(3x + 2)

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  2. Ex 4.4 Q2

    Choose an identity to find, without direct multiplication: (i) 41², (ii) 27², (iii) 23 × 17, (iv) 135², (v) 97², (vi) 18 × 29, (vii) 34 × 43, (viii) 205².

    1. (i) Write 41 as (40 + 1) and use the identity (a + b)² = a² + 2ab + b².
      why?

      Identities such as (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b² and a² − b² = (a + b)(a − b) are true for every value of the letters, so they can replace long multiplication.

    2. (i) Calculate 40² + 2(40)(1) + 1² to get 1600 + 80 + 1 = 1681.
    3. (ii) Write 27 as (30 − 3) and use the identity (a − b)² = a² − 2ab + b².
      why?

      Identities such as (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b² and a² − b² = (a + b)(a − b) are true for every value of the letters, so they can replace long multiplication.

    4. (ii) Calculate 30² − 2(30)(3) + 3² to get 900 − 180 + 9 = 729.
    5. (iii) Write 23 × 17 as (20 + 3)(20 − 3) and use the identity (a + b)(a − b) = a² − b².
      why?

      Identities such as (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b² and a² − b² = (a + b)(a − b) are true for every value of the letters, so they can replace long multiplication.

    6. (iii) Calculate 20² − 3² to get 400 − 9 = 391.
    7. (iv) Write 135 as (100 + 35) and use the identity (a + b)² = a² + 2ab + b².
      why?

      Identities such as (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b² and a² − b² = (a + b)(a − b) are true for every value of the letters, so they can replace long multiplication.

    8. (iv) Calculate 100² + 2(100)(35) + 35² to get 10000 + 7000 + 1225 = 18225.
    9. (v) Write 97 as (100 − 3) and use the identity (a − b)² = a² − 2ab + b².
      why?

      Identities such as (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b² and a² − b² = (a + b)(a − b) are true for every value of the letters, so they can replace long multiplication.

    10. (v) Calculate 100² − 2(100)(3) + 3² to get 10000 − 600 + 9 = 9409.
    11. (vi) Write 18 × 29 as (20 − 2)(20 + 9) and expand using (a − b)(a + c) = a² + a(c − b) − bc.
      why?

      Expanding multiplies every term in one bracket by every term in the other: (a + b)(c + d) = ac + ad + bc + bd.

    12. (vi) Calculate 20² + 20(9 − 2) − 2(9) to get 400 + 140 − 18 = 522.
    13. (vii) Write 34 × 43 as (40 − 6)(40 + 3) and expand using (a − b)(a + c) = a² + a(c − b) − bc.
      why?

      Expanding multiplies every term in one bracket by every term in the other: (a + b)(c + d) = ac + ad + bc + bd.

    14. (vii) Calculate 40² + 40(3 − 6) − 6(3) to get 1600 − 120 − 18 = 1462.
    15. (viii) Write 205 as (200 + 5) and use the identity (a + b)² = a² + 2ab + b².
      why?

      Identities such as (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b² and a² − b² = (a + b)(a − b) are true for every value of the letters, so they can replace long multiplication.

    16. (viii) Calculate 200² + 2(200)(5) + 5² to get 40000 + 2000 + 25 = 42025.

    Answer: (i) 1681; (ii) 729; (iii) 391; (iv) 18225; (v) 9409; (vi) 522; (vii) 1462; (viii) 42025

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  3. Ex 4.4 Q3

    Factor: (i) 9a² + b² + 4c² − 6ab + 12ac − 4bc, (ii) 16s² + 25t² − 40st, (iii) r² − r − 42, (iv) 49g² + 14gh + h², (v) 64u² + 121v² + 4w² − 176uv − 32uw + 44vw.

    1. (i) Identify the terms as squares: (3a)², (−b)², (2c)².
    2. (i) Check the cross terms: 2(3a)(−b) = −6ab, 2(−b)(2c) = −4bc, 2(3a)(2c) = 12ac.
    3. (i) Apply the identity (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx.
    4. (i) The expression factors to (3a − b + 2c)².
    5. (ii) Identify the terms as squares: (4s)², (−5t)².
    6. (ii) Check the cross term: 2(4s)(−5t) = −40st.
    7. (ii) Apply the identity (x − y)² = x² + y² − 2xy.
      why?

      Identities such as (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b² and a² − b² = (a + b)(a − b) are true for every value of the letters, so they can replace long multiplication.

    8. (ii) The expression factors to (4s − 5t)².
    9. (iii) Find two numbers that multiply to −42 and add to −1.
    10. (iii) The numbers are −7 and +6.
    11. (iii) Split the middle term: r² − 7r + 6r − 42.
      why?

      Splitting the middle term into two parts whose product equals (first coefficient × constant) lets the expression be grouped into two brackets; a product is zero only when one of its factors is zero.

    12. (iii) Group and factor: r(r − 7) + 6(r − 7).
    13. (iii) The expression factors to (r − 7)(r + 6).
    14. (iv) Identify the terms as squares: (7g)², h².
    15. (iv) Check the cross term: 2(7g)(h) = 14gh.
    16. (iv) Apply the identity (x + y)² = x² + y² + 2xy.
      why?

      Identities such as (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b² and a² − b² = (a + b)(a − b) are true for every value of the letters, so they can replace long multiplication.

    17. (iv) The expression factors to (7g + h)².
    18. (v) Identify the terms as squares: (8u)², (−11v)², (−2w)².
    19. (v) Check the cross terms: 2(8u)(−11v) = −176uv, 2(8u)(−2w) = −32uw, 2(−11v)(−2w) = 44vw.
    20. (v) Apply the identity (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx.
    21. (v) The expression factors to (8u − 11v − 2w)².

    Answer: (i) (3a − b + 2c)²; (ii) (4s − 5t)²; (iii) (r − 7)(r + 6); (iv) (7g + h)²; (v) (8u − 11v − 2w)²

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