NCERT Solutions for Class 9 Maths Chapter 2:Polynomials

CBSE Class 9 MLQB

NCERT Solutions for Class 9 Maths Chapter 2 (Polynomials) PDF Download helps students understand basic algebraic expressions with clarity. These solutions follow the latest CBSE guidelines and cover all exercises from the NCERT textbook. Each Class 9 Maths Chapter 2 question answer is solved step-by-step to build strong problem-solving skills.

This chapter explains terms, coefficients, degrees, and types of polynomials class 9 questions and answers in a simple format. It includes clear explanations and accurate answers to help students revise quickly and score better in exams.

You can easily access and save the Polynomials Class 9th solutions in PDF format for offline study. These NCERT Solutions for Class 9 Maths Chapter 2 are ideal for daily practice, doubt-clearing, and homework help. Download the Class 9 Maths Polynomials question answer PDF now and prepare with confidence.

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    Chapter 2.

    Q. Find the value of (13)3 + (15)3 – (28)3.

    Ans. (13)3 + (15)3 – (28)3 =(13)3 + (15)3 + (– 28)3
    Let a = 13, b = 15 and c = – 28
    a + b + c = 13 + 15 – 28 = 0
    a3 + b3 + c3 = 3abc
    ⇒ (13)3 + (15)3 + (– 28)3 = 3(13) (15) (– 28)
    ⇒ (13)3 + (15)3 – (28)3 = – 3(13) (420)
    (13)3 + (15)3 – (28)3 = – 16380

    Q. Prove that : a3 + b3 + c3 – 3abc=1/22(a+b+c){(a-b)2+(b-c)2+(c-a)2}
    Ans. Since we know that, a3 + b3 + c3 – 3abc
    Q. Prove that : (a + b + c)3 – a3 – b3 – c3 = 3(a + b)(b + c) (c + a).
    Ans. L.H.S. = (a + b + c)3 – a3 – b3 – c3
    = {(a + b + c)3 – a3} – (b3 + c3)
    = [(a + b + c – a) {(a + b + c)2 + a(a + b + c) + a2}] – (b3 + c3)
    = (b + c) {a2 + b2 + c2 + 2ab + 2bc + 2ac + a2 + ab + ac + a2} – (b3 + c3)
    = (b + c) (3a2 + b2 + c2 + 3ab + 2bc + 3ac) – (b3 + c3)
    = (b + c) (3a2 + b2 + c2 + 3ab + 2bc + 3ac) – (b + c) (b2 – bc + c2)
    = (b + c) (3a2 + b2 + c2 + 3ab + 2bc + 3ac – b2 + bc – c2)
    = (b + c) (3a2 + 3ab + 3bc + 3ac)
    = 3(b + c) (a2 + ab + bc + ac)
    = 3(b + c) {a(a + b) + c (a + b)}
    = 3 (b + c) (a + b) (c + a)
    = 3 (a + b) (b + c) (c + a)
    = R.H.S.
    Q. Factorise : x3 + 13x2 + 32x + 20.

    Ans. Let p (x) = x3 + 13x2 + 32x + 20
    Put x = – 1
    = (– 1)3 + 13(– 1)2 + 32(– 1) + 20
    = – 1 + 13 – 32 + 20
    33 – 33
    = 0 (x + 1) is a factor of x3 + 13x2 + 32x + 20
    Now, divide x3 + 13x2 + 32x + 20 by x + 1, we get

    $$\space\space\space\space\space\space\space\space\space\space x^2+12x+20\\x+1)\overline{x^3+13x^2+32x+20}\\\space\space\space\space\space\space\space\space\space\space x^3+x^2\\\space\space\space\space\space\space\space\space\space\space (-)(-)\\\space\space\space\space\space\space\space\space\space\space \overline{12x^2+32x+20}\\\space\space\space\space\space\space\space\space\space\space 12x^2+12x\\\space\space\space\space\space\space\space\space\space\space (-)(-)\\\space\space\space\space\space\space\space\space\space\space \overline{20x+20}\\\space\space\space\space\space\space\space\space\space\space {20x+20}\\\space\space\space\space\space\space\space\space\space\space (-)(-)\\\space\space\space\space\space\space\space\space\space\space\space\space\overline{×}\\\underline{}$$

    x3 + 13x2 + 32x + 20 = (x + 1) (x2 + 12x + 20)
    = (x + 1) {x2 + 10x + 2x + 20}
    = (x + 1) {x (x + 10) + 2 (x + 10)}
    x3 + 13x2 + 32x + 20 = (x + 1) (x + 2) (x + 10).
    Q. Factorise : 2x3 – 3x2 – 17x + 30.
    Ans. p (x) = 2x3 – 3x2 – 17x + 30
    Put x = 2 p (2) = 2(2)3 – 3(2)2 – 17(2) + 30
    = 16 – 12 – 34 + 30
    = 16 – 16 = 0
    (x – 2) is a factor of p (x)

    $$\space\space\space\space\space\space\space\space\space\space 2x^2+x-15\\x-2)\overline{2x^3-3x^2-17x+30}\\\space\space\space\space\space\space\space\space\space\space 2x^3-4x^2\\\space\space\space\space\space\space\space\space\space\space (-)(+)\\\space\space\space\space\space\space\space\space\space\space \overline{x^2-17x+30}\\\space\space\space\space\space\space\space\space\space\space x^2-2x\\\space\space\space\space\space\space\space\space\space\space (-)(+)\\\space\space\space\space\space\space\space\space\space\space \overline{-15x+20}\\\space\space\space\space\space\space\space\space\space\space {-15x+20}\\\space\space\space\space\space\space\space\space\space\space (+)(-)\\\space\space\space\space\space\space\space\space\space\space \overline{×}\\\space\space\space\space\space\space\space\space\space\space \underline{}$$

    2x3 – 3x2 – 17x + 30
    = (x – 2) (2x2 + x – 15)
    = (x – 2) {2x2 + 6x – 5x – 15}
    = (x – 2) {2x (x + 3) – 5 (x + 3)}
    = (x – 2) (x + 3) (2x – 5) 2x3 – 3x2 – 17x + 30
    = (x – 2) (2x – 5) (x + 3).

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