NCERT Solutions for Class 9 Maths Chapter 2:Polynomials

NCERT Solutions for Class 9 Mathematics Chapter 4 Free PDF Download

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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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