Class 9 Maths Chapter 2 Polynomials

Exercise 2.1

 Question 1.
Which of the following expressions are polynomials in one variable and which are not? State reason for your answer?

1. 4x2-3x + 7
2.  y2 +
3. 3 + t
4. y + 
5.  x10 + y3 +t50

Solution:
(1) 4x2– 3x +7 is an expression having only non-negative integral powers of x. So, it is a polynomial.
(2) y2 +  is an expression having only non-negative integral powers of So, it is a polynomial.
(3) 3 + t  is an expression in which one term namely 3  has rational power to t. So, it is not a polynomial.
(4) y +  is an expression in which one term namely y + 
=> i.e., 2y-3 has negative power of y. So, it is not a polynomial.
(5) x10 + y3 +t50 is an expression which has 3 variables.

 Question 2.
Write the coefficients of x2 in each of the following:
(i) 2 + x2+x
(ii) 2-x2 +xa
(iii) 
(iv)  x -1
Solution:
(i) The coefficient of x2 in 2 + x2 + x is 1.
(ii) The coefficient of x2 in 2 – x2 + x3 is -1.
(iii) The coefficient of x2 in   + x is .
(iv) The coefficient of x2 in x -1 is 0.

 Question 3.
Give one example each of a binomial of degree 35, and of a monomial of degree 100.
Solution:
(1) y35 + 2 is a binomial of degree 35.
(2) y100 is a monomial of degree 100.

 Question 4.
Write the degree of each of the following polynomials :
(i) 5x3 + 4x2    + 7x
(ii)  4 – y2
(iii) 5t-
(iv) 3
Solution:
(i) The highest power term is 5x3 and the exponent is 3. So, the degree is 3.
(ii) The highest power term is -y2 and the exponent is 2. So, the degree is 2.
(iii) The highest power term is 5t and the exponent is 1. So, the degree is 1.
(iv) The only term here is 3 which can be written as 3x° and so the exponent is 0.
Therefore, the degree is 0.

Question 5.
Classify the following as linear, quadratic and cubic polynomials:
(i) x2+x
(ii) x2-x
(iii) y + y2+4
(iv) 1 + x
(v) 3t
(vi) r2
(vii) 7r3
Solution:
(i) The highest degree of x2 + x is 2, so it is a quadratic polynomial.
(ii) The highest degree of x – x3 is 3, so it is a cubic polynomial.
(iii) The highest degree of y + y2 + 4 is 2, so it is a quadratic polynomial.
(iv) The highest degree of x in 1 + x is 1, so it is a linear polynomial.
(v) The highest degree oft in 3t is 1, so it is a linear polynomial.
(vi) The highest degree of r in r2 is 2, so it is a quadratic polynomial.
(vii) The highest degree of x in 7x3 is 3, so, it is a cubic polynomial.

 

Exercise 2.2  

Question 1.
Find the value of the polynomial 5x – 4x2 + 3 at
(i) x=0
(ii) x=-1
(iii) x=2
Solution:

 Question 2.
Find p(0), p(1) and p(2) for each of the following polynomials
(i) p(y)=y2-y+1
(ii) p(t) =2+t+2t2-t3
(iii) p(x) =x3
(iv) p (x)=(x-1)(x+1)
Solution:
(i) We have,p(y) = y2-y+1
∴ p(0)= (0)2-0+1 = 0-0+1 =1,
p(1)=(1)2 -1+1 = 1-1+1 = 1,
and  p(2) = (2)2 – 2 +1 = 4 – 2 +1 = 3

(ii) We have,    p(t) = 2 +1 + 2t2 -t3
∴  p(0) = 2 + 0 + 2(0)2 – (0)3
= 2 + 0+ 0-0 = 2,
p(1) = 2 +1 + 2(1)2 – (l)3
= 2+l+2-l = 5-1 = 4
p(2) = 2 + 2 + 2(2)2 – (2)3
= 2 + 2 + 8-8 = 4

(iii) We have, p(x) = x3
p(o) = (0 )3 = 0,
p(1)=(1)3 =1,
p(2) = (2)3 = 8

(iv) p(x) = (x-1) (x +1)
p(0) = (0 -1)(0 +1) = (-1)(1) = -1
p(1) = (2 -1) (1 +1) = (0)(2) = 0
p(2) = (2-1) (2 +1) = (1)(3) = 3

 Question 3.
Verify whether the following are zeroes of the polynomial, indicated against them.
(i) p(x) =3x+1,x=- 
(ii) p(x)=5x-π,x=
(iii) p(x) =x2 -1,x=1,-1
(iv) p(x)=(x+1)(x-2),x = -1,2
(v) p(x) =x2 ,x=0
(vi) p(x)=lx+m,x = – 
(vii) p(x) =3x2 -1,x= – ,
(viii) p(x) = 2x+1,x 
Solution:



 Question 4.
Find the zero of the polynomial in each of the following cases :
(i) p(x) = x + 5
(ii) p(x) = x – 5
(iii) p(x) = 2x + 5
(vi) p(x) = 3x-2
(v) p(x) = 3x
(vi) p(x) = ax, a≠0
(vii) p(x)  = cx + d, c≠0, c, d are real numbers.
Solution:
(i) We have, p(x) = x + 5
Now,   p(x) = 0      => x + 5 = 0 => x = -5
∴ 5 is a zero of the polynomial p(x).

 

Exercise 2.3 

Question 1.
Find the remainder when x3 + 3x2 + 3x + 1 is divided by
(i) x + 1
(ii) x- 
(iii) x
(iv) x+π
(v) 5+2x
Solution:
(1) By remainder theorem, the required remainder is equal to p(-1).
Now, p(x)= x3+3x2+3x+1
p(-1) = (-1)3 + 3(-1)2 + 3(-1) +1 =-1+3-3+1=0
Hence, the required remainder = p(-1) = 0

 Question 2.
Find the remainder when x3 – ax2 + 6x – a is divided by x – a.
Solution:
Let p(x) = x3-ax2+6x-a
By remainder theorem, when p(x) is divided by x – a,
Then,remainder = p(a)
p(a)= a3-a . a2+6a-a
=a3 – a3 +6a-a = 5a

Question 3.
Check, whether 7 + 3x is a factor of 3x3 + 7x.
Solution:
Let f(x) = 3x3 + 7x and g(x) = 7 + 3x
On putting g(x) = 0, we get
7 + 3x = 0 3x = – 7
=> x = 
Thus, zero of g(x) is x =.

Exercise 2.4  

Question 1.
Determine which of the following polynomials has (x +1) a factor:
(i) x3 +x2 + 1
(ii) x4 + x3+x2+x+1
(iii) x4 + 3x3 + 3x2 + x + 1
(iv) x3 -x2 -(2 + – )x + 
Solution:
The zero of x +1 is -1.
(i) Let p(x) = x3 + x2 + x +1
Then,   p(-1) =  (-1)3 + (-1)2 + (-1) +1
= -1+1-1+1 => p(-1)=   0
So, by the Factor theorem (x +1) is a factor of x3 + x2 + x+1.

(ii) Let p(x)= x4 + x3 + x2 + x +1
Then,   p(-1) = (-1)4 + (-1)3 + (-1)2 + (-1) +1
=1-1+1-1+1 =>      p(-1) =  1
So, by the.Factor theorem (x +1) is not a factor of x4 + x3 + x2 + x +1

(iii) Let p(x) = x4 + 3x3 + 3x2 + x +1
Then,  p(-l) = (-1)4 + 3(-1)3 + 3(-1)2 + (-1) +1
=1-3+3-1+1
=>  P(-1) =  1
So, by the Factor theorem (x +1) is not a factor of x4 + 3x3 + 3x2 + x +1

(iv) Let p(x) = x3-x2-(2 +  )x + 
Then, p(-1) = (-1)3 – (-1)2 – (2 +  )(-1) + 
= —1—1+2 +  + 
= 2
So, by the Factor theorem (x + 1) is not a factor ofx3 -x2 -(2 +  x + 

 Question 2.
Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases :
(i) p(x) – 2x3 + x2 -2x-1, g(x) = x + 1
(ii) p(x) = x3 + 3x2 + 3x + 1, g(x) = x + 2
(iii) p(x) = x3– 4x2 + x + 6, g(x) = x – 3
Solution:
(i) The zero of g(x) = x +1 is x = -1.
Then,  p(-1)= 2(-1)3 +(-1)2 -2(-1)-1
[∵  p(x) = 2x3 + x2 -2x-1]
= -2 +1+2-1 ⇒ p(-1) = 0
Hence, g(x) is a factor of p(x).

(ii) The zero of g(x) = x + 2 is -2.
Then, p(-2) = (-2)3 + 3(-2)2 + 3(-2) +1
[∵ p(x) = x3 + 3x2 + 3x +1]
= -8+12-6 +1 =>p(-2)=-1
Hence, g(x) is not a factor of p(x).

(iii) The zero of g(x) = x – 3 is 3.
Then,p(3) = 33 – 4(3)2 +3 + 6
[∵ p(x) = x3 – 4x2 + x + 6]
= 27-36 +3 + 6 =»       p(3) = 0
Hence, g(x) is a factor of p(x).

Question 3.
Find the value of k, if x -1 is a factor of p(x) in each of the following cases:
(i) p(x) = x2+x + k .
(ii) p(x) = 2x2 + kx + V2=
(iii) p(x) = kx2 -42x + 1
(iv) p(x) = kx2 -3x + k
Solution:

 Question 4.
Factorise:
(i) 12x2 – 7x+1
(ii) 2x2+7x+3
(iii) 6x2+5x-6
(iv) 3x2-x-4
Solution:

 Question 5.
Factorise:
(i) x3-2x2-x+2
(ii) x3-3x2-9x-5
(iii) x3-13x2+32x+20
(iv) 2y3+y2-2y-1
Solution:
(i) Let p(x) = x3-2x2-x+2, constant term of P(x) is 2.
Factors of 2 are ± 1 and ± 2.
Now,  p(1)=13-2(1)2-1+2
=1-2-1+2
By trial we find that p(l) = 0, so (x -1) is a factor of p(x).
So, x3-2x2-x+2= x3-x2-x2 + x-2x+2
= x2(x-1)-x(x-1)-2(x-1)
= (x—1)(x2 -x-2)
= (x -1)(x2 -2x + x -2)
= (x -1) [x(x -2) + l(x -2)]
= (x-1)(x-2)(x+1)

(ii) Let  p(x)= x3 -3x2 -9x-5
By trial, we find that p(5) = (5)3 – 3(5)2 – 9(5) – 5
= 125 – 75 – 45 – 5 = 0
So, (x – 5) is a factor of p(x).
So, x3-3x2-9x-5=x3-5x2+2x2 -10x + x – 5
= x2 (x – 5) + 2x(x – 5) + 1(x – 5)
= (x – 5) (x2+2x+1)
= (x – 5) (x2 + x + x +1)
= (x + 5) [x(x +1) + 1(x +1)]
= (x-5)(x +1) (x +1)
= (x – 5)(x +1)2

(iii) Let p(x)= x3 +13x2 +32x +20
By trial, we find that p(-1) = (-1)3 +13(-1)2 + 32(-1) + 20
= -1 +13 -32 + 20 = -33 + 33 = 0
So (x +1) is a factor of p(x).
So, x3 + 13x2 + 32x + 20 = x3 + x2 + 12x2 + 12x + 20x + 20
= x2 (x +1) + 12x(x +1) + 20(x +1) = (x+1)(x2 + 12x + 20)
= (x+1)(x2 + 10x + 2x + 20)
= (x +1)[x(x +10) + 2(x +10)]
= (x +l)(x +10)(x + 2)

(iv) Let p(y) = 2y3 + y2 – 2y -1
By trial we find that p(l) = 2(1)3 + (1)2 -2(1) -1,
=2+1-2-1=0 So (y -1) is a factor of p(y).
So, 2y3 + y2 -2y-1 = 2y3 -2y2 +3y2 -3y + y-1
= 2y2(y-1) + 3y(y-1) + 1(y-1)
= (y – 1)(2y2 +3y + 1)
= (y – 1)(2y2 + 2y + y + 1)
= (y -1) [2y(y +1) + l(y +1)
= (y-1)(y+ l)(2y+ 1)

 EXERCISE 2.5

 Question 1.
Use suitable identities to find the following products:
(i) (x+4)(x+10)
(ii) (x+8)(x-10)
(iii) (3x+4)(3x+2x)
(iv) ()( )
(v) (3-2x)(3+2x)
Solution:

Question 2.
Evaluate the following products without multiplying directly:
(i) 103 x 107
(ii) 95 x 96
(iii) 104 x 96
Solution:
(i) 103 x 107 = (100 + 3) (100 + 7)
= (100)2 + (3 + 7) (100) + 3 x 7
= 100 x 100 + (10)(100) + 21
= 10000 +1000 + 21 = 11021

(ii) 95 x 96 = (100 – 5) (100 – 4)
= (100)2 + (-5 – 4)(100) + (-5)(-4) – 100 x 100 + (-9)(100) + 20
= 10000 – 900 + 20 = 9120

(iii) 104 x 96 = (100 + 4)(100 – 4)
= (100)2 – (4)2
= 10000 -16
= 9984

 Question 3.
Factorise the following using appropriate identities :
(i) 9x2 + 6xy + y2
(ii) 4y2 – 4y +1
(iii) x2–
Solution:

 Question 4.
Expand each of the following, using suitable identities :
(i) (x + 2y + 4z)2
(ii) (2x -y +z)2
(iii) (-2x + 3y + 2z)2
(iv) (3a – 7b -c)2
(v) (-2x + 5y – 3z)2
(vi) 2
Solution:


Question 5.
Factorise:
(i) 4x2 + 9y2 + I622 + 12xy – 24yz -16xz
(ii) 2x2+ y2 + 822 – 2xy + 4yz – 8×2
Solution:
(i) 4x2 +9y2 +16z2 +12xy-24yz-16xz
= (2x)2 + (3y)2 + (-4z)2 + 2(2x)(3y) + 2(3y)(-4z) + 2(-4z)(2x)
= (2x +3y – 4z)2

(ii) 2x2 + y2 + 8z2 – 2xy + 4yz-8xz
= (-ax)2 + (y)2 + (2z)2 + (2 – x) (y) + 2(y) (2z) + 2(2z)(-x)
= (-x + y + 2z)2

 Question 6.
Write the following cubes in expanded form :
(i) (2x+1)3
(ii) (2a-3b)3
(iii) 3
(iv) 3
Solution:


 Question 7.
Evaluate the following using suitable identities :
(i) (99)3
(ii) (102)3
(iii) (998)3
Solution:

 Question 8.
Factorise each of the following:
(i) 8a3 + b3 + 12a26 + 6ab2
(ii) 8a3 – b3 -12a26 + 6a62
(iii) 27 – 125a3 – 135 a + 225 a2 64a3
(iv) 
Solution:


 Question 9.
Verify:
(1) x3 + y3 = Or + y)(x2-xy + y2)
(ii) x3-y3 = (x-y)(x2 + xy + y2)
Solution:
(i) We know that,
(x + y)3 = x3 + y3 + 3xy(x + y)
=> x3 + y3 = (x + y)3 -3xy(x + y)
= (x + y)[(x + y)2 -3xy]
= (x + y) [x2 + y2 + 2xy – 3x] = (x + y)[x2 + y2 – xy]
= RHS  Hence proved.

(ii) We know that, (x – y)3 = x3 – y3 -3xy(x – y)
=>x3 – y3 = (x – y)3 + 3xy(x – y)
= (x-y)[(x – y)2 +3xy]
= (x -y)[x2 + y2 -2xy + 3xy]
= (x — y)[x2 + y2 + xy]
= RHS  Hence proved.

 Question 10.
Factorise each of the following:
(i) 27y3 + 125z3
(ii) 64m3 -343n3 [Hint: See question 9]
Solution:
(i) 27 y3 +125z3 = (3y)3 + (5z)3
= (3y + 5z)[(3y)2 – (3y)(5z) + (5z)2]
= (3y + 5z) (9 y2 – 15yz + 25z2)

(ii) 64m3 -343n3 = (4m)3 -(7n)3
= (4m-7n)[(4m)2 + (4m)(7n) + (7n)2]
= (4m – 7n)[16m2 + 28mn + 49n2]

 Question 11.
Factorise :
27x3 + y3 + z3 – 9xyz
Solution:
27x3 +y3 +z3 -9xyz = (3x)3 + y3 +z3 -3(3x)(y)(z)
= (3x + y + z)[(3x)2 + y2 + z2 – (3x)y – yz -z(3x)]
= (3x + y + z)(9x2 + y2 + z2 -3xy – yz -3zx)

 Question 12.
Verify that
x3 + y3 +z3 -3xyz =  (x + y + z)[(x -y)2 +(y-z)2 +(z-x)2]
Solution:
We have, x3 + y3 + z3 – 3xyz
= (x + y + z) [x2 + y2 + z2 – xy – yz – zx]
= (x + y+ z)[2x2 +2y2 +2z2 -2xy-2yz -2zx]
= (x + y + z)[x2 + x2 + y2 + y2 + z2 + z2 -2xy-2yz-2zx]
= (x + y + z)[x2 + y2 – 2xy + y2 + z2 -2yz + z2 + x2 – 2zx]
= (x + y + z)[(x-y)2 + (y-z)2 +(z-x)2]

 Question 13.
If x + y + z = 0, show that x3 + y3 + z3 = 3xyz.
Solution:
We know that,
x3 +y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz-zx)
= 0(x2 + y2 + z2 – xy- yz-zx) (∵ x + y + z = 0 given)
= 0
=> x3 + y3 + z3 = 3xyz        Hence proved.

 Question 14.
Without actually calculating the cubes, find the value of each of the following: 
(i) (-12)3 + (7)3 + (5)3
(ii) (28)3 + (-15)3 + (-13)3.
Solution:
(i) We know that, x3 + y3 + z3 – 3xyz
= (x + y + z)(x2 + y2 + z2 – xy – yz – zx)
Also, we know that, if
x + y + z = 0
Then, x3 + y3 +z3 = 3 xyz
Given expression is (-12)3 + (7)3 + (5)3.
Here,            -12 + 7 + 5=0
∴ (-12)3 + (7)3 + (5)3 = 3 x (-12) x 7 x 5 = -1260

(ii) Given expression is (28)3 + (-15)3 + (-13)3
Here, 28 + (-15) + (-13) = 28 -15-13 = 0
∴  (28)3 + (-15)3 + (-13)3 = 3 x (28) x (-15) x (-13) = 16380

 Question 15.
Give possible expressions for the length and the breadth of each of the following rectangles, in which their areas are given :
(i) Area = 25a2 – 35a + 12
(ii) Area = 35y2 + 13y – 12.
Solution:
(i) We have,
Area of rectangle = 25a2 – 35a +12 [by splitting the middle term]
= 25a2 -20a-15a+12
= 5a(5a – 4) – 3(5a – 4)
= (5a-4)(5a-3)
Hence, possible expression for length = (5a – 3) and possible expression for breadth = (5a – 4)

(ii) We have,
Area of rectangle = 35y+ 13y -12
= 35y2 + (28 – 15)y -12
= 35y2 + 28y-15y-12
= (35y2 +28y)-(15y + 12)
= 7y(5y + 4) – 3(5y + 4)
= (7y – 3) (5y + 4)

 Question 16.
What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
(i) Volume = 3x2 – 12x
(ii) Volume = 12ky2 + 8ky – 20k.
Solution:
(i) We have,
Volume of cuboid = 3x2 -12x = 3x(x – 4)
So, the possible expressions for the dimensions of the cuboids are 3, x and x – 4.
[∵ volume of cuboid = length x breadth x height]

(ii) We have,
Volume of cuboid = 12ky2 + 8ky – 20k = 12ky2 + (20 -12)ky – 20k [by splitting the middle term]
= 12ky2 + 20 ky – 12ky – 20 k = 4ky(3y + 5) – 4k(3y + 5) = (3y + 5)(4ky – 4k)
= (3y + 5)4k(y -1) = 4k(3y + 5)(y -1)
So, the possible expressions for the dimensions of the cuboid are 4 k, 3y + 5 and y -1.