Polynomials| NCERT Solutions| Class 9

NCERT Solutions of Chapter: Polynomials. NCERT Solutions along with worksheets and notes for Class 9.

 

By Long Division Method –

Hence, the remainder of given polynomial is 0.

By Remainder Theorem-

Dividend p(x) = x3 + 3x2 + 3x + 1

Divisor g(x) = x + 1

Zeroes of g(x)

x + 1 = 0

x = -1

Put x = -1 in p(x)

x3 + 3x2 + 3x + 1

= (-1)3 + 3 (-1)2 + 3(-1) + 1

= -1 +3 -3 +1

= 0

Hence, the remainder of given polynomial is 0.

 

By Long Division Method-

By Remainder Theorem-

 

 

Polynomials, exercise 2.3, Q 1, NCERT, class 9

By Long Division Method –

Hence, the remainder of the given polynomial is 1.

By Remainder Theorem-

Dividend p(x) = x3 + 3x2 + 3x + 1

Divisor g(x) = x

Zeroes of g(x)

x= 0

Put x = 0 in p(x)

x3 + 3x2 + 3x + 1

= (0)3 + 3(0)2 + 3(0) + 1

= 1

Hence, the remainder of the given polynomial is 1.

 

By Long Division Method-

Hence, the remainder of given polynomial is – π3 + 3 π2 – 3π  + 1.

 

By Remainder Theorem-

Dividend p(x) = x3 + 3x2 + 3x + 1

Divisor g(x) = x + π

Zeroes of g(x)

x + π =0

x = – π

Put x = – π in p(x)

x3 + 3x2 + 3x + 1

= (- π )3 + 3(-π )2 + 3(-π ) + 1

= – π3 + 3 π2 – 3π  + 1

Hence, the remainder of given polynomial is – π3 + 3 π2 – 3π  + 1.

Polynomials, exercise 2.3, Q 1, NCERT, class 9

By Long Division Method –

By Remainder Theorem-

Dividend p(x) = x3 + 3x2 + 3x + 1

Divisor g(x) = 2x + 5

Zeroes of g(x)

2x + 5 = 0

2x = -5

By Long Division Method –

Hence, remainder of the given polynomial is 5a.

 

By Remainder Theorem-

Dividend p(x) = x3 – ax2 + 6x – a

Divisor g(x) = x – a

Zero of g(x)

x – a = 0

x = a

Put x = a in p(x)

x3 – ax2 + 6x – a

= (a) 3 – a (a)2 + 6(a) – a

= a3 – a3 + 6a – a

= 5a

Hence, remainder of the given polynomial is 5a.

Method 2

(First, finding out one factor and then using splitting method)

Let p(x) = x3 – 2x2 –x + 2

By trial method,

p(1) = (1)3 – 2(1)2 – 1 + 2

= 1 – 2 – 1+2

= 0

Therefore, x-1 is a factor of the polynomial x3 – 2x2 –x + 2.

Arranging polynomial x3 – 2x2 –x + 2 in such a way that (x- 1) can be a common factor.

x3 – 2x2 –x + 2

= x3 – x2 – x2 – 2x + x +2

= x3 – x2 – x2 +x -2x + 2

=x2(x -1) – x(x -1) -2(x – 1)

=(x -1) (x2 – x -2)

Now, using splitting method

x2 – x -2

(ii) x3 – 3x2 – 9x – 5

Method 1

Let p(x) = x3 – 3x2 –9x – 5

By trial method put x = -1

p(-1) = -13 – 3(-1)2 -9(-1) – 5

= -1 – 3 + 9 – 5

= 0

Therefore x+1 is a factor of the polynomial x3 – 3x2 –9x – 5

Now, divide polynomial x3 – 3x2 –9x – 5 by factor x+1.

Method 2

Let p(x) = x3 – 3x2 –9x – 5

By trial method put x = -1

p(-1) = -13 – 3(-1)2 – 9(-1) – 5

= -1 – 3 + 9 – 5

= 0

Therefore x+1 is a factor of the polynomial x3 – 3x2 –9x – 5

Arranging polynomial x3 – 3x2 –9x – 5 in such a way that (x + 1) can be a common factor.

x3 – 3x2 –9x – 5

= x3 – 4x2 + x2 – 4x – 5x – 5

=x3 + x2 – 4x2 – 4x – 5x – 5

= x2(x + 1) – 4x (x + 1) – 5(x + 1)

=(x + 1) (x2 – 4x -5)

(iii) x3 + 13x2 + 32x + 20

Method 1

Let p(x) = x3 + 13x2 + 32x + 20

By trial method put x = -1

p(-1) = -13 + 13(-1)2 + 32(-1) + 20

= -1 + 13 – 32 + 20

= 0

Therefore, x+1 is a factor of the polynomial x3 +13x2 + 32x + 20

Now, divide polynomial x3 + 13x2 + 32x + 20 by factor x+1

Method 2

Let p(x) = x3 + 13x2 + 32x + 20

By trial method put x = -1

p(-1) = -13 + 13(-1)2 +32(-1) + 20

= -1 + 13 – 32 + 20

= 0

Therefore x+1 is a factor of the polynomial x3 +13x2 + 32x + 20

Arranging polynomial x3 +13x2 + 32x +20 in such a way that (x + 1) can be a common factor.

x3 +13x2 + 32x +20

= x3 + x2 + 12x2 + 12x + 20x + 20

= x2(x + 1) + 12x (x + 1) + 20(x + 1)

=(x + 1) (x2 + 12x + 20)

x2 + 12x + 20

Method 2

Let p(y) = 2y3 + y2 – 2y -1

By trial method put y = 1

p(1) = 2 + 1 – 2 – 1

= 0

Therefore, x + 1 is a factor of the polynomial 2y3 + y2 – 2y -1

Arranging polynomial 2y3 + y2 – 2y -1in such a way that (y – 1) can be a common factor.

2y3 + y2 – 2y – 1

= 2y3 – 2y2 + 3y2 -3y + y -1

= 2y2(y -1) + 3y(y -1) + 1 (y -1)

= (y – 1)( 2y2 + 3y + 1)

So,

2y2 + 3y + 1

We have to find two numbers p and q, such that

p + q = 3 and

pq = 1    2

Some factors of 2 are 1 and 2.

Therefore, 1 and 2 are the factors whose sum is 3.

2y2 + 3y + 1

= 2y2 + 2y +y + 1

= 2y(y + 1) + 1(y + 1)

= (y + 1) (2y + 1)

Therefore,

2y3 + y2 – 2y – 1 = (y -1)(y + 1) (2y + 1)

Hence, (y – 1) (y + 1) (2y + 1) are factors of the polynomial 2y3 + y2 – 2y -1.

 

(8) Factorise each of the following:

(i) 8a3 + b3 + 12a2b + 6ab2      (ii) 8a3 – b3 – 12a2b + 6ab2

(iii) 27 – 125a3 – 135a + 225a3 (iv)64a3 – 27b3 – 144a2b + 108ab2

(v) 27p3 –  – p2 + p

Ans-

(i) 8a3 + b3 + 12a2b + 6ab2

= (2a)3+ b3+3(2a)2b + 3(2a)b2

Using identity (x + y )3 = x3 + y3 + 3xy(x+y)

(x + y )3 = x3 + y3 + 3x2 y+3y2x

=(2a + b)3

=(2a + b)(2a + b) (2a + b)

Hence, (2a + b)(2a + b) (2a + b)are the factors of given polynomial.

(ii) 8a3 – b3 – 12a2b + 6ab2

=(2a)3 – b3 -3(2a)2 b + 3(2a) b2

Using identity (x – y )3 = x3 – y3 – 3xy(x-y)

(x + y )3 = x3 + y3 + 3x2 y+3y2x

=(2a – b)3

=(2a –b)(2a – b)(2a –b)

Hence, (2a –b)(2a – b)(2a –b) are the factors of the given polynomial.

(iii) 27 – 125a3 – 135a + 225a3

=33 – (5a)3 – 3(3)2 5 + 3(3) (5a)2

Using identity (x – y )3 = x3 – y3 – 3xy(x-y)

(x + y )3 = x3 + y3 + 3x2 y+3y2x

=(3 – 5a)3

=(3 – 5a)(3 – 5a)(3 – 5a)

Hence, (3 – 5a)(3 – 5a)(3 – 5a) are the factors of the given polynomial.

(iv) 64a3 – 27b3 – 144a2b + 108ab2

=(4a)3 –(3b)3 -3(4a)2 (3b) + 3(4a) (3b)2

Using identity (x – y )3 = x3 – y3 – 3xy(x-y)

(x + y )3 = x3 + y3 + 3x2 y+3y2x

=(4a – 3b)3

= (4a – 3b)(4a – 3b)(4a – 3b)

Hence, (4a – 3b)(4a – 3b)(4a – 3b) are the factors of the given polynomial.

(v) 27p3 –  – p2 + p

= (3p)3 – ( )3 – 3 (3p)2( ) + 3(3p) ( )2

= (3p – )3

=(3p – ) (3p – ) (3p – )

Hence, (3p – ) (3p – ) (3p – ) are the factors of given polynomial.

 

(10) Factories each of the following:

(i)27y3 + 125z3    (ii) 64m3 – 343n3

Ans-

(i)27y3 + 125z3

Using identity (x + y )3 = x3 + y3 + 3xy(x+y)

(x + y )3 – 3xy(x+y) = x3 + y3

x3 + y3 = (x + y )3 – 3xy(x+y)

Put values in the identity.

(3y)3 + (5z)3 = (3y + 5z)3 – 3 (3y)(5z)(3y + 5z)

= (3y + 5z)3  – 45yz(3y + 5z)

= (3y + 5z)[ 3y + 5z)2 – 45yz]

= (3y + 5z)[9y2 + 25z2 + 30yz – 45yz]

=(3y + 5z)[ 9y2 + 25z2 – 15yz]

= (3y + 5z)(9y2 + 25z2 – 15yz)

Hence, (3y + 5z)(9y2 + 25z2 – 15yz) are factors of given polynomial.

Helping Topics

Polynomials

Worksheet Class 9

Algebraic expressions

 

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