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NCERT Class 9 Mathematics Exercise Set 4.5 Solutions: Exploring Algebraic Identities

Step-by-step working for every question of Exercise Set 4.5 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.

1. Ex 4.5 Q1

   Simplify (denominators not zero): (i) (3p² − 3pq − 18q²)/(p² + 3pq − 10q²); (ii) (n³ − 3n²m + 3nm² − m³)/(5m² − 10mn + 5n²); (iii) (w³ − v³ + x³ + 3wvx)/(w² + v² + x² − 2wv − 2vx + 2wx); (iv) (4y² − 20yz + 25z²)/(25z² − 4y²); (v) (x² + x − 6)(x² − 7x + 12)/((x² − 6x + 8)(x² − 9)); (vi) (p⁴ − 16)/(p² − 4p + 4).

   1. (i) Factor the numerator 3p² − 3pq − 18q² as 3(p² − pq − 6q²) = 3(p − 3q)(p + 2q).
   2. (i) Factor the denominator p² + 3pq − 10q² as (p + 5q)(p − 2q).
   3. (i) The factors p − 3q, p + 2q, p + 5q and p − 2q are all different, so nothing cancels: the simplest form is 3(p − 3q)(p + 2q)/((p + 5q)(p − 2q)).
   4. (ii) Recognize the numerator n³ − 3n²m + 3nm² − m³ as the expansion of (n − m)³.
      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.

   5. (ii) Factor the denominator 5m² − 10mn + 5n² as 5(m² − 2mn + n²) = 5(m − n)².
      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. (ii) Since (m − n)² = (n − m)², the fraction is (n − m)³/(5(n − m)²); cancel (n − m)² to get (n − m)/5.
      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.

   7. (iii) Use a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca) with a = w, b = −v, c = x.
   8. (iii) Then a³ + b³ + c³ − 3abc = w³ − v³ + x³ + 3wvx, the numerator, and it equals (w − v + x)(w² + v² + x² + wv + vx − wx).
   9. (iii) The denominator is (a + b + c)² = (w − v + x)² = w² + v² + x² − 2wv − 2vx + 2wx.
   10. (iii) Cancel one factor (w − v + x) to get (w² + v² + x² + wv + vx − wx)/(w − v + x).
   11. (iv) Factor the numerator 4y² − 20yz + 25z² as (2y − 5z)².
   12. (iv) Factor the denominator 25z² − 4y² as (5z − 2y)(5z + 2y).
   13. (iv) Rewrite the numerator as (5z − 2y)² and cancel one (5z − 2y) to get (5z − 2y)/(5z + 2y).
   14. (v) Factor x² + x − 6 as (x + 3)(x − 2) and x² − 7x + 12 as (x − 3)(x − 4).
   15. (v) Factor x² − 6x + 8 as (x − 2)(x − 4) and x² − 9 as (x − 3)(x + 3).
   16. (v) Cancel all common factors (x + 3), (x − 2), (x − 3), and (x − 4) to get 1.
   17. (vi) Factor the numerator p⁴ − 16 as (p² − 4)(p² + 4) = (p − 2)(p + 2)(p² + 4).
   18. (vi) Factor the denominator p² − 4p + 4 as (p − 2)².
   19. (vi) Cancel one (p − 2) to get (p² + 4)(p + 2)/(p − 2).

   Answer: (i) 3(p − 3q)(p + 2q)/((p + 5q)(p − 2q)); (ii) (n − m)/5; (iii) (w² + v² + x² + wv + vx − wx)/(w − v + x); (iv) (5z − 2y)/(5z + 2y); (v) 1; (vi) (p² + 4)(p + 2)/(p − 2)

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