Class 9 Exam  >  Class 9 Notes  >  Mathematics (Maths) Class 9  >  Long Answer Type Questions: Polynomials

Class 9 Maths Chapter 2 Question Answers - Polynomials

Question 1. Factorise: Class 9 Maths Chapter 2 Question Answers - Polynomials

Solution: Class 9 Maths Chapter 2 Question Answers - Polynomials  
Class 9 Maths Chapter 2 Question Answers - Polynomials

Thus,    Class 9 Maths Chapter 2 Question Answers - Polynomials

 

Question 2. Factorise: (x6 – y6)
Solution: 
x6 – y= (x3)2 – (y3)
= (x3 – y3)(x3 + y3)  [∵ a2 – b= (a + b)(a – b)]

= [(x – y)(x2 + xy + y2)][(x + y)(x– xy + y2)]  

[∵ a3 + b3 = (a2 + b2 – ab)(a + b) and a– b3 = (a+ ab + b2)(a – b)]

= (x – y)(x + y)(x+ xy + y2)(x2 – xy + y2)

Thus, x– y6 = (x – y)(x + y)(x2 + xy + y2)(x2 – xy + y2)


Question 3. If the polynomials 2x3 + 3x2 – a and ax3 – 5x + 2 leave the same remainder when each is divided by x – 2, find the value of ‘a’
Solution: 
Let p(x) = 2x3 + 3x2 – a and f(x) = ax– 5x + 2

when p(x) is divided by x – 2 then
remainder = p(2)
since p(2) = 2(2)3 + 3(2)2 – a
= 2(8) + 3(4) – a = 16 + 12 – a
∴ Remainder = 28 – a
when f(x) is divided by x – 2, then
remainder = f(2)
since, f(2) = a(2)3 – 5(2) + 2
= a(8) – 10 + 2
= 8a – 8
∴ Remainder = 8a – 8

 28 – a = 8a – 8
⇒ 8a + a = 28 + 8
⇒ 9a = 36
Class 9 Maths Chapter 2 Question Answers - Polynomials
Thus , a = 4


Question 4. Find the values of ‘p’ and ‘q’, so that (x – 1) and (x + 2) are the factors of x3 + 10x2 + px + q.
Solution:
Here f(x) = x+ 10x2 + px + q
Since, x + 2 = 0 [∵ x + 2 is a factor of f(x)]
⇒ x= –2 If x + 2 is a factor f(x),
then f(–2) = 0 i.e. (–2)+ 10(–2)2 + p(–2) + q = 0 [∵Factor theorem]
⇒ –8 + 40 + (–2p) + q = 0 ⇒ 32 – 2p + q = 0 ...(1)
⇒ 2p – q = 32 Also x – 1 = 0 ⇒ x = 1
If (x – 1) is a factor of f(x), then f(1) must be equal to 0. [∵Factor theorem]
i.e. (1)+ 10(1)2 + p(1) + q = 0
⇒ 1 + 10 + p + q = 0
⇒ 11 + p + q = 0
⇒ p + q = –11                                 ...(2)
Now, by adding (1) and (2), we get

Class 9 Maths Chapter 2 Question Answers - Polynomials

Now we put p = 7 in (2), we have  7 + q = –11
⇒ q = –11 – 7 = –18
Thus, the required value of p and q are 7 and –18 respectively.


Question 5. If (x2 – 1) is a factor of the polynomial px+ qx3 + rx+ sx + t, then prove that p + r + t = q + s = 0.
Solution:
We have f(x) = px4 + qx3 + rx+ sx + t
Since, (x2 – 1) is a factor of f(x), [∵ x2 – 1 = (x + 1)(x – 1)]
then (x + 1) and (x – 1) are also factors of f(x).
∴ By factor theorem, we have   f(1) = 0 and f(–1) = 0
For f(1) = 0, p(1)4 + q(1)3 + r(1)2 + s(1) + t = 0
⇒ p + q + r + s + t = 0                       ...(1)
For f(–1) = 0, p(–1)4 + q(–1)+ r(–1)+ s(–1) + t = 0
⇒ p – q + r – s + t = 0                     ...(2)

Class 9 Maths Chapter 2 Question Answers - Polynomials  

Class 9 Maths Chapter 2 Question Answers - Polynomials

From (4) and (3), we get p + r + t = q + s = 0


Question 6. If x + y = 12 and xy = 27, find the value of x3 + y3.
Solution: 
Since, (x + y)3 = x3 + y3 + 3xy (x + y)
∴ Substituting x + y = 12 and xy = 27,
we have: (12)3 =x+ y3 + 3 (27) (12)
⇒ x3 + y3 = 81(12) – 123
= [92 – 122] × 12
= [(9 + 12) (9 – 12)] × 12
= 21 × 3 × 12 = 756


Question 7. If a + b + c = 5 and ab + bc + ca = 10, Then prove that a+ b3 + c3 – 3abc = –25.
Solution:
Since, a3 + b3 + c– 3abc
= (a + b + c) (a+ b2 + c– ab – bc – ca)
∴ a3 + b3 + c3 – 3 abc = (a + b + c) [(a2 + b2 + c2 + 2ab+ 2bc+ 2cb) − 3ab − 3bc − 3ca]
= (a + b + c) [(a + b + c)2 − 3 (ab + bc + ca)]
= 5 [ 52 − 3(10)]
= 5[25 – 30]
= 5[–5] = –25

Question 8. If a, b, c are all non-zero and a + b + c = 0, prove thatClass 9 Maths Chapter 2 Question Answers - Polynomials
Solution:
Since, a + b + c = 0
∴ a3 + b3 + c3 = 3abc                         ..... (1)
Now, in  Class 9 Maths Chapter 2 Question Answers - Polynomials = 3, we have

Class 9 Maths Chapter 2 Question Answers - Polynomials                   [Multiplying and dividing by ‘abc’]

Class 9 Maths Chapter 2 Question Answers - Polynomials                   ..... (2)

From (1) and (2), we have

Class 9 Maths Chapter 2 Question Answers - Polynomials

The document Class 9 Maths Chapter 2 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 Question Answers - Polynomials

1. What is a polynomial and how is it defined?
Ans. A polynomial is a mathematical expression consisting of variables, coefficients, and exponents, combined using addition, subtraction, and multiplication operations. It is defined as the sum of terms, where each term is a product of a coefficient and a variable raised to a non-negative integer exponent.
2. How do we classify polynomials based on the number of terms they have?
Ans. Polynomials can be classified based on the number of terms they have. A polynomial with one term is called a monomial, with two terms is called a binomial, and with three terms is called a trinomial. Polynomials with more than three terms are generally referred to as polynomials.
3. What is the degree of a polynomial and how is it determined?
Ans. The degree of a polynomial is the highest power of the variable present in the polynomial expression. To determine the degree, we look at the exponents of the variables in each term and find the highest exponent. For example, if the highest exponent is 3, the degree of the polynomial is 3.
4. Can a polynomial have negative exponents or fractional exponents?
Ans. No, a polynomial cannot have negative exponents or fractional exponents. The exponents in a polynomial must be non-negative integers. If a term has a negative exponent or a fractional exponent, it is not considered a polynomial. However, it may still be a valid mathematical expression, but not a polynomial.
5. How can polynomials be used in real-life applications?
Ans. Polynomials have various real-life applications. They can be used in physics to model the motion of objects, in economics to analyze supply and demand curves, in engineering to design circuits and control systems, and in computer graphics to create smooth curves and surfaces. Polynomials provide a powerful mathematical tool for solving problems in a wide range of disciplines.
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