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.

 

Helping Topics

Polynomials

Worksheet Class 9

Algebraic expressions

 

Leave a comment