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-
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.
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.
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