Class 9 Maths Chapter 2 Practice Question Answers - Polynomials

Question 1. Show that (x + 3) is a factor of x³ + x² – 4x + 6.
 Solution: ∵ p(x) = x³ + x² – 4x + 6
Since (x + 3) is a factor, then x + 3 = 0 ⇒ x = – 3
∴ p( – 3) = ( – 3)³ + ( – 3)²  – 4( – 3) + 6
= – 27 + 9 + 12 + 6
= 0
∴ (x + 3) is a factor of x³ + x² – 4x + 6

Question 2. Show that (x – 5) is a factor of: x³ – 3x²  – 13x + 15
 Solution: ∵ p(x) = x³ – 3x² – 13x + 15
Since (x – 5) is a factor, then x – 5 = 0
⇒ x = 5
∴ p(5) = (5)³  – 3(5)²  – 13(5) + 15
= 125 – 75 – 65 + 15 = 140 – 140 = 0
∴ (x – 5) is a factor of x³  – 3x² – 13x + 15.

Question 3. Find the value of a such that (x + α ) is a factor of the polynomial
 f(x) = x⁴ – α 2x² + 2x + α + 3.
 Solution: Here f(x) = x⁴  – α² x² + 2x + α + 3
Since, (x + a) is a factor of f(x)
∴ f ( – α) = 0
⇒ ( – α)⁴ – α² ( – α)² + 2( – α) + a + 3 = 0
⇒ α⁴ – α⁴ – 2α + α + 3= 0
⇒ – α + 3 = 0
⇒ α = 3

Question 4. Show that (x – 2) is a factor of 3x³ + x² – 20x + 12.
 Solution: f(x) = 3x³ + x² – 20x + 12
For (x – 2) being a factor of f(x), then x – 2 = 0
⇒ x = 2
∴ f(2) must be zero.
Since f(2) = 3(2)³ + (2)² – 20(2) + 12
= 3(8) + 4 – 40 + 12
= 24 + 4 – 40 + 12
= 40 – 40 = 0
which proves that (x – 2) is a factor of f(x).

Question 5. Factorise the polynomial 

Solution: We have  

Question 6. Factorise a(a – 1) – b(b – 1).
 Solution: We have a(a – 1) – b(b – 1)
= a² – a – b² + b
= a² – b² – (a – b)
(Rearranging the terms) = [(a + b)(a – b)] – (a – b)
[∵ x² – y² = (x + y)(x – y)]
= (a – b)[(a + b) – 1]
= (a – b) (a + b – 1)
Thus, a(a – b) – b(b – 1) = (a – b)(a + b – 1) 

Question 7. Show that x³ + y³ = (x + y )(x² – xy + y²).
 Solution: Since (x + y)³ = x³ + y³ + 3xy(x + y)
∴ x³ + y³ = [(x + y)³] –3xy(x + y)
= [(x + y)(x + y)²] – 3xy(x + y)
= (x + y)[(x + y)² – 3xy]
= (x + y)[(x² + y² + 2xy) – 3xy]
= (x + y)[x² + y² – xy]
= (x + y)[x² + y² – xy]
Thus, x³ + y³ = (x + y)(x² – xy + y²) 

Question 8. Show that x³ – y³ = (x – y)(x² + xy + y²).
 Solution: Since (x – y)³ = x³ – y³ – 3xy(x – y)
∴   x³ – y³ = (x – y)³ + 3xy(x – y)
= [(x – y)(x – y)²] + 3xy(x – y)
= (x – y)[(x – y)² + 3xy]
= (x – y)[(x² + y² – 2xy) + 3xy]
= (x – y)[x² + y² – 2xy + 3xy]
= (x – y)(x² + y² + xy)
Thus, x³ – y³ = (x – y)(x² + xy + y²)