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. When the polynomial kx3 + 9x2 + 4x – 8 is divided x + 3, then a remainder 7 is obtained.
Find the value of k.

Here, p(x) = kx3 + 9x+ 4x – 8
Since, Divisor = x + 3
∴ x + 3 = 0
⇒ x = –3
∴ p(–3) = 7
⇒ k(–3)3 + 9(–3)+ 4(–3) – 8 = 7
⇒ –27k + 81 – 12 – 8 = 7
⇒ –27k = 7 – 81 + 12 + 8
⇒ –27k = 27 – 81
⇒ –27k = –54
⇒ k= -(54/27) = 2
Thus, the required value of k = 2.


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 consisting of variables, coefficients, and exponents, combined using addition, subtraction, and multiplication. It can have one or more terms, and each term can have different degrees.
2. How do you identify the degree of a polynomial?
Ans. The degree of a polynomial is determined by the highest power of the variable present in the expression. To find the degree, look for the term with the highest exponent. For example, in the polynomial 3x^2 + 2x - 1, the degree is 2.
3. What is the leading coefficient of a polynomial?
Ans. The leading coefficient of a polynomial is the coefficient of the term with the highest degree. It is usually denoted by the letter "a" in standard form. For example, in the polynomial 4x^3 + 2x^2 - 3x + 1, the leading coefficient is 4.
4. Can a polynomial have a negative degree?
Ans. No, a polynomial cannot have a negative degree. The degree of a polynomial is always a non-negative integer. It represents the highest power of the variable in the expression. If a polynomial has no terms with a variable, its degree is considered to be zero.
5. 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. Monomials are simpler expressions, whereas polynomials are more complex and can have different degrees. For example, 3x^2 is a monomial, while 3x^2 + 2x - 1 is a polynomial with three terms.
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