Question 1. Show that (x + 3) is a factor of x^{3} + x^{2} – 4x + 6.
Solution: ∵ p(x) = x^{3 }+ x^{2} – 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 x^{3 }+ x^{2} – 4x + 6
Question 2. Show that (x – 5) is a factor of: x^{3} – 3x^{2 } – 13x + 15
Solution: ∵ p(x) = x^{3} – 3x^{2} – 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 x^{3 } – 3x^{2} – 13x + 15.
Question 3. Find the value of a such that (x + α ) is a factor of the polynomial
f(x) = x^{4} – α 2x^{2} + 2x + α + 3.
Solution: Here f(x) = x^{4 } – α^{2} x^{2} + 2x + α + 3
Since, (x + a) is a factor of f(x)
∴ f ( – α) = 0
⇒ ( – α)^{4} – α^{2} ( – α)^{2} + 2( – α) + a + 3 = 0
⇒ α^{4} – α^{4} – 2α + α + 3= 0
⇒ – α + 3 = 0
⇒ α = 3
Question 4. Show that (x – 2) is a factor of 3x^{3} + x^{2} – 20x + 12.
Solution: f(x) = 3x^{3} + x^{2} – 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)
= a^{2} – a – b^{2} + b
= a^{2} – b^{2} – (a – b)
(Rearranging the terms) = [(a + b)(a – b)] – (a – b)
[∵ x^{2 }– y^{2} = (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 x^{3} + y^{3} = (x + y )(x^{2} – xy + y^{2}).
Solution: Since (x + y)^{3} = x^{3} + y^{3 }+ 3xy(x + y)
∴ x^{3} + y^{3} = [(x + y)^{3}] –3xy(x + y)
= [(x + y)(x + y)^{2}] – 3xy(x + y)
= (x + y)[(x + y)^{2} – 3xy]
= (x + y)[(x^{2 }+ y^{2} + 2xy) – 3xy]
= (x + y)[x^{2} + y^{2} – xy]
= (x + y)[x^{2} + y^{2} – xy]
Thus, x^{3} + y^{3} = (x + y)(x^{2} – xy + y^{2})
Question 8. Show that x^{3} – y^{3} = (x – y)(x^{2 }+ xy + y^{2}).
Solution: Since (x – y)^{3} = x^{3} – y^{3} – 3xy(x – y)
∴ x^{3} – y^{3} = (x – y)^{3} + 3xy(x – y)
= [(x – y)(x – y)^{2}] + 3xy(x – y)
= (x – y)[(x – y)^{2} + 3xy]
= (x – y)[(x^{2} + y^{2} – 2xy) + 3xy]
= (x – y)[x^{2} + y^{2} – 2xy + 3xy]
= (x – y)(x^{2} + y^{2} + xy)
Thus, x^{3} – y^{3} = (x – y)(x^{2} + xy + y^{2})
43 videos400 docs59 tests

Important Formulas: Polynomials Doc  2 pages 
NCERT Solutions: Polynomials (Exercise 2.2) Doc  3 pages 
NCERT Solutions  Polynomials (Exercise 2.3) Doc  4 pages 
1. What is a polynomial? 
2. How do you determine the degree of a polynomial? 
3. Can a polynomial have negative exponents? 
4. What is the difference between a monomial and a polynomial? 
5. How do you add or subtract polynomials? 
Important Formulas: Polynomials Doc  2 pages 
NCERT Solutions: Polynomials (Exercise 2.2) Doc  3 pages 
NCERT Solutions  Polynomials (Exercise 2.3) Doc  4 pages 

Explore Courses for Class 9 exam
