NCERT Solutions for Class 9 Maths Chapter 2:Polynomials

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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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