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

Class 9 Maths Chapter 2 Question Answers - Polynomials

Q1. Write the numerical co-efficient and degree of each term of: Class 9 Maths Chapter 2 Question Answers - Polynomials

Class 9 Maths Chapter 2 Question Answers - Polynomials

Q2. Find the remainder when x3 – ax2 + 4x – a is divided by (x – a).

p(x) = x– ax2 + 4x – a
(x – a) = 0
⇒ x = a
∴ p(a) = (a)3 – a(a)2 + 4(a) – a = a3 – a+ 4a – a = 4a – a = 3a
∴ The required remainder = 3a.


Q3. Factorise z2  – 4z –12

To factorize this expression, we need to find two numbers α and β such that α + β = -4 and αβ = -12

z2 – 6z + 2z – 12

z(z – 6) + 2(z – 6)

(z + 2)(z – 6)


Q4. (a) For what value of k, the polynomial x2 + (4 – k)x + 2 is divisible by x – 2?
(b) For what value of ‘m’ is x3 – 2mx2 + 16 is divisible by (x + 2)?
 

(a) Here p(x) = x+ 4x – kx + 2
If p(x) is exactly divisible by x – 2, then p(2) = 0
i.e. (2)2 + 4(2) – k(2) + 2 = 0
⇒ 4 + 8 – 2k + 2 = 0
⇒ 14 – 2k = 0
⇒ 2k = 14
⇒ k= 14/2 = 7
Thus, the required value of k is 7.

(b) Here, p(x) = x3 – 2mx2 + 16
∴ p(–2) = (–2)3 –2(–2)2m + 16
⇒ –8 –8m + 16
⇒ –8m + 8
Since, p(x) is divisible by x + 2
∴ p(–2) = 0
or –8m + 8 = 0
⇒ m = 1


Q5. Factorize x2 – x – 12.

We have  x2 – x – 12
⇒ x– 4x + 3x – 12
⇒ x(x – 4) + 3(x – 4)
⇒ (x – 4)(x + 3)
Thus, x– x – 12 = (x – 4)(x + 3)


Q6. If x + (1/2x)  = 5, then find the value of x2 + Class 9 Maths Chapter 2 Question Answers - Polynomials 

We have x + (1/2x)  = 5
Squaring both sides, we get:
Class 9 Maths Chapter 2 Question Answers - Polynomials
Class 9 Maths Chapter 2 Question Answers - Polynomials
Class 9 Maths Chapter 2 Question Answers - Polynomials
Class 9 Maths Chapter 2 Question Answers - Polynomials
Thus, the required value of Class 9 Maths Chapter 2 Question Answers - Polynomials

Q7. Check whether (x – 1) is a factor of the polynomial x3 – 27x2 + 8x + 18.

Here, p(x) = x– 27x2 + 8x + 18, (x – 1) will be a factor of p(x) only if (x – 1) divides p(x) leaving a remainder 0.
For  x – 1 = 0
⇒ x = 1
∴ p(1) = (1)3 – 27(1)2 + 8(1) + 18
⇒ 1 – 27 + 8 + 18
⇒ 27 – 27
⇒ 0
Since, p(1) = 0
∴ (x – 1) is a factor of p(x).
Thus, (x – 1) is a factor of x3 – 27x2 + 8x + 18.


Q8. Find the remainder when f(x) = Class 9 Maths Chapter 2 Question Answers - Polynomials

Here f(x) =  Class 9 Maths Chapter 2 Question Answers - Polynomials
Divisor = x+(2/3)
since, x+(2/3) = 0 ⇒ x = -2/3
∴ Remainder = f (-2/3)
i.e. Remainder
Class 9 Maths Chapter 2 Question Answers - Polynomials
Class 9 Maths Chapter 2 Question Answers - Polynomials
Thus, the required remainder =  (10/27)


Q9. Find the value of k, if (x – k) is a factor of x6 – kx5 + x4 – kx3 + 3x – k + 4.

Here, p(x) = x6 – kx5 + x– kx3 + 3x – k + 4
If (x – k) is a factor of p(x), then p(k) = 0
i.e (k)6 – k(k5) + k4 – k(k3) + 3k – k + 4 = 0
⇒ k– k6 + k4 – k4 + 3k – k + 4 = 0
⇒ 2k + 4 = 0
⇒ 2k = – 4
⇒ k = (-4/2) = –2
Thus, the required value of k is –2.


Q10. Factorize: 9a2 – 9b2 + 6a + 1

9a2 – 9b2 + 6a + 1
⇒ [9a2 + 6a + 1] – 9b
⇒ [(3a)2 + 2(3a)(1) + (1)2] – (3b)
⇒ (3a + 1)2 – (3b)2
⇒ [(3a + 1) + 3b][(3a + 1) – 3b] {using x– y2 = (x – y)(x + y)}
⇒ (3a + 1 + 3b)(3a + 1 – 3b)


Q11. Find the value of x3 + y3 – 12xy + 64, when x + y = –4.

x3 + y– 12xy + 64
⇒ (x)3 + (y)3 + (4)– 3(x)(y)(4)
⇒ [x2 + y2 + 42 – xy – y. 4 – 4 . x](x + y + 4)
⇒ [x2 + y2 + 16 – xy – 4y – 4x][x + y + 4] ...(1)
Since, x + y = –4 ∴ x + y + 4 = 0                                 ...(2)
From (1) and (2), we have x3 + y– 12xy + 64
⇒ [x+ y2 + 16 – xy – 4y – 4x][0] = 0
Thus, x3 + y3 – 12xy + 64 = 0.

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 ?
Ans.A polynomial is a mathematical expression that consists of variables raised to whole number powers and coefficients. It can be represented in the form \( P(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0 \), where \( a_n, a_{n-1}, ..., a_0 \) are constants and \( n \) is a non-negative integer.
2. How do you classify polynomials based on their degree ?
Ans.Polynomials can be classified based on their degree, which is the highest power of the variable in the expression. For example, a polynomial of degree 0 is a constant, degree 1 is a linear polynomial, degree 2 is a quadratic polynomial, degree 3 is a cubic polynomial, and so on.
3. What are the basic operations that can be performed on polynomials ?
Ans.The basic operations that can be performed on polynomials include addition, subtraction, multiplication, and division. These operations can be done by combining like terms, using the distributive property, or employing polynomial long division when necessary.
4. How can you determine the roots of a polynomial ?
Ans.The roots of a polynomial, also known as the solutions or zeros, can be determined by setting the polynomial equal to zero and solving for the variable. This can be done using various methods such as factoring, using the quadratic formula for second-degree polynomials, or applying numerical methods for higher-degree polynomials.
5. What is the importance of polynomials in real-life applications ?
Ans.Polynomials play a significant role in various fields including physics, engineering, economics, and computer science. They are used to model real-life situations, such as projectile motion, population growth, and financial forecasts, due to their ability to represent complex relationships in a simplified manner.
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