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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) = x+ x2 – 4x + 6
Since (x + 3) is a factor, then x + 3 = 0 ⇒ x = – 3
∴ p( – 3) = ( – 3)3 + ( – 3) – 4( – 3) + 6
= – 27 + 9 + 12 + 6
= 0
∴ (x + 3) is a factor of x+ x2 – 4x + 6


Question 2. Show that (x – 5) is a factor of: x3 – 3x – 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(5) – 13(5) + 15
= 125 – 75 – 65 + 15 = 140 – 140 = 0
∴ (x – 5) is a factor of x – 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) = xα2 x2 + 2x + α + 3
Since, (x + a) is a factor of f(x)
∴ f ( – α) = 0
⇒ ( – α)4α2 ( – α)2 + 2( – α) + a + 3 = 0
α4α4 – 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 Class 9 Maths Chapter 2 Practice Question Answers - Polynomials

Solution: We have  Class 9 Maths Chapter 2 Practice Question Answers - Polynomials

Class 9 Maths Chapter 2 Practice Question Answers - Polynomials

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)
[∵ x– 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 + y+ 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)[(x+ 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)(x+ 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) 

The document Class 9 Maths Chapter 2 Practice Question Answers - Polynomials is a part of the Class 9 Course Mathematics (Maths) Class 9.
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FAQs on Class 9 Maths Chapter 2 Practice Question Answers - Polynomials

1. What is a polynomial?
Ans. A polynomial is a mathematical expression that consists of variables, coefficients, and exponents, combined using addition, subtraction, multiplication, and non-negative integer exponents.
2. How do you determine the degree of a polynomial?
Ans. The degree of a polynomial is determined by finding the highest exponent of the variable in the polynomial expression. For example, if the highest exponent is 3, then the polynomial is a cubic polynomial and its degree is 3.
3. Can a polynomial have negative exponents?
Ans. No, a polynomial cannot have negative exponents. Negative exponents would result in fractions or irrational numbers, which are not allowed in polynomials. The exponents must be non-negative integers.
4. What is the difference between a monomial and a polynomial?
Ans. A monomial is a polynomial with only one term, while a polynomial can have multiple terms. In other words, a monomial is a single algebraic expression, whereas a polynomial is the sum or difference of multiple monomials.
5. How do you add or subtract polynomials?
Ans. To add or subtract polynomials, you combine like terms. Like terms are terms that have the same variables raised to the same exponents. Add or subtract the coefficients of these like terms while keeping the variables and exponents the same.
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