# Important Questions for Class 9 Chapter 2-Polynomials

Everyone keeps telling you that class 9th preparation also plays an important role in the class 10th board exams. Well, to certain extent, even that is true as some of the concepts in class 10th would be extension of concepts learned in class 9th. Your free set of important questions also carries step-by-step explanation and solution. The questions are based on each topic and chapter of the CBSE class 9th syllabus. Download your free PDF now and start practicing!

**Important questions based on NCERT syllabus for Class 9 Chapter 2 – Polynomials:**

*Question 1*: If x + x − 1 = √5, evaluate x^2+x−2 and x^4 + x − 4.

*Solution*:

x + x − 1 = √5,

(x + x^{−1})^2 = (√5)^2

⇒ x^2 + x^{–2} + 2 = 5

⇒ x^2 + x^{−2} = 5 – 2 = 3

(x^2 + x − 2)^2 = 3^2

⇒ x^4 + x^{−4} + 2 = 9

⇒ x^4 + x^{−4} = 9 – 2 = 7

*Question 2*: Find the remainder when x^3+3x^2+3x+1 is divided by:

(i)x+1

(ii)5+2x

*Solution*:

(i) Let p(x)=x^3+3x^2+3x+1 and q(x) = x + 1

According to Remainder Theorem, remainder is equal to p(a) when p(x) is divided by (x−a).

p(−1)=(−1)^3+3(−1)^2+3(−1)+1 = −1 + 3 − 3 + 1 = 0

Therefore, remainder is 0 when p(x)=x^3+3x^2+3x+1 is divided by x+1

(ii)Let p(x)=x^3 + 3x^2 + 3x + 1 and q(x) = 5 + 2x

p(−5/2) = (−5/2)^3 + 3(−5/2)^2 + 3(−5/2) + 1

= (−125/8) + (75/4) + (−15/2) + 1

= (−125 + 150 − 60 + 8)/8 = (−27/8)

Therefore, remainder is equal to (−278) when p(x)=x3+3x2+3x+1 is divided by 5+2x

*Question 3*: Factorise the following using appropriate identities:

a)9x^2 + 6xy + y^2

b)4y^2 − 4y + 1

*Solution*:

a)9x^2 + 6xy + y^2

=(3x)^2+2(3x)(y)+(y)^2

*a^2 + 2ab + b^2 = (a+b)^2*

a = 3x, b = y,

=(3x)^2 + 2(3x)(y) + (y)^2

= (3x+y)^2

= (3x+y)(3x+y).

b) 4y^2−4y+1

=(2y)^2−2(2y)(1)+(1)^2

*a^2−2ab+b^2 = (a−b)^2*

a = 2y,b = 1

=(2y)^2−2(2y)(1)+(1)^2

=(2y−1)^2

=(2y−1)(2y−1)