Lesson: Polynomials
Exercise 2.1 (1)
Question: 1
The graphs of are given in the following
y p x
figures for some polynomials . Find the number
p x
of zeroes of in each case.
p x
Solution:
(i) In the first figure, the graph does not intersect
x-axis at any point.
Therefore, the number of zeroes of is 0.
p x
(ii) In the second figure, the graph intersects x-axis
at only one point.
Therefore, the number of zeroes of is 1.
p x
(iii) In the third figure, the graph intersects x-axis
at three points.
Therefore, the number of zeroes of is 3.
p x
(iv) In the fourth figure, the graph intersects x-axis
at two points.
Therefore, the number of zeroes of is 2.
p x
(v) In the fifth figure, the graph intersects x-axis at
four points.
Therefore, the number of zeroes of is 4.
p x
(vi) In the sixth figure, the graph intersects x-axis
at three points.
Therefore, the number of zeroes of is 3.
p x
EXERCISE 2.2 (2)
Question: 1
Find the zeroes of the following quadratic
polynomials and verify the relationship between the
zeroes and the coefficients.
(i)
2
x 2x 8
(ii)
2
4s 4s 1
(iii)
2
6x 3 7x
(iv)
2
4u 8u
(v)
2
t 15
(vi)
2
3x x 4
Solution:
(i) Let
2
f x x 2x 8
By splitting, find the factors of quadratic
equation.
x 4 x 2 x 4
x 4 x 2
The zeroes of are
f x
f x 0
x 4 x 2 0
Now, equate these factors equal to 0.
or
x 4 0
x 2 0
or
x 4
x 2
Therefore, the zeroes of are 4 and
2
x 2x 8
2
.
Now, calculate the relationship between the
zeroes and the coefficients.
Sum of zeroes
4 2
2
2
1
2
Coefficient of x
Coefficient of x
Product of zeroes
4 2
8
2
1
2
Constant term
Coefficient of x
(ii) Let
2
f s 4s 4s 1
2
2
4s 4s 1 2s 1
The zeroes of is
f x
f x 0
2
2s 1 0
2s 1 0
1
s
2
Therefore, the zeroes of are and
2
4s 4s 1
1
2
.
1
2
Now, calculate the relationship between the
zeroes and the coefficients.
Sum of zeroes
1 1
2 2
1
4
4
2
Coefficient of s
Coefficient of s
Product of zeroes
1 1
2 2
1
4
2
Constant term
Coefficient of x
(iii) Let
2
f x 6x 3 7x
By splitting, find the factors of quadratic
equation.
2 2
6x 3 7x 6x 7x 3
2
6x 9x 2x 3
3x 2x 3 1 2x 3
3x 1 2x 3
The zeroes of is
f x
f x 0
3x 1 2x 3 0
Now, equate these factors equal to 0.
or
3x 1 0
2x 3 0
or
1
x
3
3
x
2
Therefore, the zeroes of are
2
6x 3 7x
1
3
and .
3
2
Now, calculate the relationship between the
zeroes and the coefficients.
Sum of zeroes
1 3
3 2
7
6
7
6
2
Coefficient of x
Coefficient of x
Product of zeroes
1 3
3 2
1
2
3
6
2
Constant term
Coefficient of x
(iv) Let
2
f u 4u 8u
By splitting, find the factors of quadratic
equation.
2 2
4u 8u 4u 8u 0
4u u 2
The zeroes of is,
f u
f u 0
4u u 2 0
Now, equate these factors equal to 0.
or
4u 0
u 2 0
or
u 0
u 2
Therefore, the zeroes of are 0 and .
2
4u 8u
2
Now, calculate the relationship between the
zeroes and the coefficients.
Sum of zeroes
0 2
2
8
4
2
Coefficient of u
Coefficient of u
Product of zeroes
0 2
0
0
4
2
Constant term
Coefficient of u
(v) Let
2
f t t 15
By splitting, find the factors of quadratic
equation.
2
2 2
t 15 t 15
t 15 t 15
The zeroes of is,
f t
f t 0
t 15 t 15 0
Now, equate these factors equal to 0.
or
t 15 0
t 15 0
or
t 15
t 15
Therefore, the zeroes of are and
2
t 15
15
.
15
Now, calculate the relationship between the
zeroes and the coefficients.
Sum of zeroes
15 15
0
0
1
2
Coefficient of t
Coefficient of t
Product of zeroes
15 15
15
15
1
2
Constant term
Coefficient of t
(vi) Let
2
f x 3x x 4
By splitting, find the factors of quadratic
equation.
2 2
3x x 4 3x 4x 3x 4
x 3x 4 1 3x 4
3x 4 x 1
The zeroes of is
f x
f x 0
3x 4 x 1 0
Now, equate these factors equal to 0.
or
3x 4 0
x 1 0
or
4
x
3
x 1
Therefore, the zeroes of are and
2
3x x 4
4
3
.
1
Now, calculate the relationship between the
zeroes and the coefficients.
Sum of zeroes
4
1
3
1
3
1
3
2
Coefficient of x
Coefficient of x
Product of zeroes
4
1
3
4
3
4
3
2
Constant term
Coefficient of x
Question: 2
Find a quadratic polynomial each with the given
numbers as the sum and product of its zeroes
respectively.
(i)
1
,
4
1
(ii)
2,
1
3
(iii)
0,
5
(iv)
1,
1
(v)
1
,
4
1
4
(vi)
4,
1
Solution:
(i)
1
, 1
4
2
ax bx c
and the zeroes be and .
1
4
b
a
1
4
4
c
a
Here, and .
a 4,
b 1
c 4
.
2
4x x 4
(ii)
2,
1
3
2
ax bx c
and the zeroes be and .
2
3 2
3
b
a
1
3

c
a
Here,
,
and .
3a
b 3 2
c 1
.
2
3x 3 2x 1
(iii)
0, 5
2
ax bx c
and the zeroes be and .
0
0
1
b
a
5
5
1
c
a
Here, and .
a 1,
b 0
c 5
2
x 5
.
(iv)
1,
1
2
ax bx c
and the zeroes be and .
1
1
1
b
a
1
1
1
c
a
Here, and .
a 1,
b 1
c 1
.
2
x x 1
(v)
1
,
4
1
4
2
ax bx c
and the zeroes be and .
1
4
b
a
1
4

c
a
Here, and .
a 4,
b 1
c 1
.
2
4x x 1
(vi)
4,1
2
ax bx c
and the zeroes be and .
4
4
1
b
a
1
1
1
c
a
Here, and .
a 1,
b 4
c 1
.
2
x 4x 1
Exercise 2.3 (5)
Question: 1
Divide the polynomial by the polynomial
p x
g x
and find the quotient and remainder in each of the
following:
(i)
3 2
p x x 3x 5x 3,
2
g x x 2
(ii)
4 2
p x x 3x 4x 5,
2
g x x 1 x
(iii)
4
p x x 5x 6,
2
g x 2 x
Solution:
(i)
3 2
p x x 3x 5x 3,
2
g x x 2
By long division method,
Quotient
x 3
Remainder
7x 9
(ii)
4 2
p x x 3x 4x 5,
2
g x x 1 x
4 2
p x x 3x 4x 5
4 3 2
x 0x 3x 4x 5
2 2
g x x 1 x x x 1
By long division method,
Quotient
2
x x 3
Remainder
8
(iii)
4
p x x 5x 6,
2
g x 2 x
4
p x x 5x 6
4 3 2
x 0x 0x 5x 6
2 2
g x 2 x x 2
By long division method,
Quotient
2
x 2
Remainder
5x 10
Question: 2
Check whether the first polynomial is a factor of
the second polynomial by dividing the second
polynomial by the first polynomial:
(i)
2
t 3,
4 3 2
2t 3t 2t 9t 12
(ii)
2
x 3x 1,
4 3 2
3x 5x 7x 2x 2
(iii)
3
x 3x 1,
5 3 2
x 4x x 3x 1
Solution:
(i)
2
t 3,
4 3 2
2t 3t 2t 9t 12
By long division method,
The remainder is 0.
Therefore, is a factor of
2
t 3
.
4 3 2
2t 3t 2t 9t 12
(ii)
2
x 3x 1,
4 3 2
3x 5x 7x 2x 2
By long division method,
The remainder is 0.
Therefore, is a factor of
2
x 3x 1
.
4 3 2
3x 5x 7x 2x 2
(iii)
3
x 3x 1,
5 3 2
x 4x x 3x 1
By long division method,
The remainder is 2.
Therefore, is not a factor of
3
x 3x 1
.
5 3 2
x 4x x 3x 1
Question: 3
Obtain all other zeroes of ,
4 3 2
3x +6x 2x 10x 5
if two of its zeroes are and .
5
3
5
3
Solution:
Let
p x
4 3 2
3x +6x 2x 10x 5
The two zeroes are and .
5
3
5
3
So,
2
5 5 5
x x x
3 3 3
Thus, is a factor of
2
5
x
3
.
4 3 2
3x +6x 2x 10x 5
So, divide the given polynomial by .
2
5
x
3
2 2
5
x 3x 6x 3
3
4 3 2
3x +6x 2x 10x 5
2 2
5
3 x x 2x 1
3
Now, factorize .
2
2
x 2x 1 x 1
Thus, its zeroes is
2
x 1 0
x 1 0
x 1
Therefore, the two zeroes are and .
1
1
Hence, the zeroes of the given polynomial are
5
,
3
and .
5
,
3
1
1
Question: 4
On dividing by a polynomial ,
3 2
x 3x x 2
g x
the quotient and remainder were and
x 2
, respectively. Find .
2x 4
g x
Solution:
Let
3 2
p x x 3x x 2
g x ?
Quotient
x 2
Remainder
2x 4
We know that,
Dividend Divisor Quotient Remainder
3 2
x 3x x 2 g x x 2 2x 4
3 2
x 3x x 2 2x 4 g x x 2
3 2
x 3x 3x 2 g x x 2
is the quotient when we divide
g x
by .
3 2
x 3x 3x 2
x 2
Thus,
2
g x x x 1
Question: 5
Give examples of polynomials and
p x ,
g x ,
q x
, which satisfy the division algorithm and
r x
(i)
deg p x deg q x
(ii)
deg q x deg r x
(iii)
degr x 0
Solution:
(i) Let
2
p x 8x 4x 4
g x 4
So,
2
q x 2x x 1
r x 0
Here, the degree of and is same.
q x
p x
Now, check for division algorithm,
p x g x q x r x
2 2
8x 4x 4 4 2x x 1 0
2 2
8x 4x 4 4 2x x 1
2 2
8x 4x 4 8x 4x 4
Therefore, the division algorithm is satisfied.
(ii) Let
4
p x x x
3
g x x
So,
q x x
r x x
Here, the degree of and is same.
q x
r x
Now, check for division algorithm,
p x g x q x r x
4 3
x x x x x
4 4
x x x x
Therefore, the division algorithm is satisfied.
(iii) Let
2
p x x 1
g x x
So,
q x x
r x 1
Here, the degree of is 0.
r x
Now, check for division algorithm,
p x g x q x r x
2
x 1 x x 1
2 2
x 1 x 1
Therefore, the division algorithm is satisfied.
Exercise 2.4 (5)
Question: 1
Verify that the numbers given alongside of the
cubic polynomials below are their zeroes. Also
verify the relationship between the zeroes and the
coefficients in each case:
(i)
3 2
2x x 5x 2;
1
,
2
1,
2
(ii)
3 2
x 4x 5x 2;
2,
1,
1
Solution:
(i) Let
3 2
p x 2x x 5x 2
The zeroes of the polynomials are .
1
,
2
1,
2
Now, check the zeroes of the polynomial.
For ,
1
x
2
3 2
1 1 1 1
p 2 5 2
2 2 2 2
 
1 1 5
2
4 4 2
1 1 10 8
4
0
For ,
x 1
2 1 5 2
5 5
0
For ,
x 2
 
3 2
p 2 2 2 2 5 2 2
16 4 10 2
16 16
0
Hence, and are the zeroes of the given
1
,
2
1
2
polynomial.
Now, compare the given polynomial with
.
3 2
ax bx cx d
So, .
a 2,
b 1,
c 5,
d 2
Now, verify the relationship between zeroes
and coefficient of the given polynomial.
Also, and .
1
,
2
1
2
Now,
1
1 2
2
1
2
b
a
   
1 1
1 1 2 2
2 2
1
2 1
2
1 4 2
2
5
2
c
a
1
1 2
2
  
1
1
2
2
d
a
Thus, the relationship between the zeroes and
the coefficients is verified.
(ii) Let
3 2
p x x 4x 5x 2
The zeroes of the polynomial are and .
2,
1
1
Now, check the zeroes of the polynomial.
For ,
x 2
3 2
p 2 2 4 2 5 2 2
8 16 10 2
0
For ,
x 1
1 4 5 2
0
Hence, and are the zeroes of the given
2,
1
1
polynomial.
Now, compare the given polynomial with
.
3 2
ax bx cx d
So, .
a 1,
b 4,
c 5,
d 2
Now, verify the relationship between zeroes
and coefficients of the given polynomial.
Also, and .
2,
1
1
Now,
2 1 1
4
4
1
b
a
  2 1 1 1 2 1
2 1 2
5
5
1
c
a
2 1 1
2
2
1
d
a
Thus, the relationship between the zeroes and
the coefficients is verified.
Question: 2
Find a cubic polynomial with the sum, sum of the
product of its zeroes taken two at a time, and the
product of its zeroes as respectively.
2,
7,
14
Solution:
Let the polynomial be and the
3 2
ax bx cx d
zeroes are and .
,
Now,
2
1
b
a
7
1
 
c
a
14
1

d
a
So,
a 1,
b 2,
c 7,
d 14
Therefore, the polynomials is .
3 2
x 2x 7x 14
Question: 3
If the zeroes of the polynomial are
3 2
x 3x x 1
find and .
a b,
a,
a b
a
b
Solution:
Let
3 2
p x x 3x x 1
The zeroes of the polynomial are .
a b,
a,
a b
Now, compare the given polynomial with
.
3 2
px qx rx s
So,
p 1,
q 3,
r 1,
t 1
Sum of zeroes
a b a a b
q
3a
p
3 3a
a 1
The zeroes are .
1 b,
1,
1 b
Multiplication of zeroes
1 1 1 b b
2
t
1 b
p
2
1
1 b
1
2
1 b 1
2
b 1 1
2
b 2
b 2
Thus, and or .
a 1
b 2
2
Question: 4
If two zeroes of the polynomial
are find other
4 3 2
x 6x 26x 138x 35
2 3
zeroes.
Solution:
Given and are zeroes of the given
2 3
2 3
polynomial.
So, is a factor of polynomial.
2 3 2 3
Thus,
x 2 3 x 2 3 x 2 3 x 2 3
2
2
x 2 3
2
x 4 4x 3
2
x 4x 1
So, is a factor of the given polynomial.
2
x 4x 1
Now, divide the polynomial
by for finding
4 3 2
x 6x 26x 138x 35
2
x 4x 1
the remaining zeroes.
Clearly,
4 3 2 2 2
x 6x 26x 138x 35 x 4x 1 x 2x 35
Now,
2 2
x 2x 35 x 7x 5x 35
x x 7 5 x 7
x 7 x 5
The value of the polynomial is also zero when
or
x 7 0
x 5 0
or
x 7
x 5
Therefore, 7 and -5 are also the zeroes of given
polynomial.
Question: 5
If the polynomial is divided
4 3 2
x 6x 16x 25x 10
by another polynomial
,
the remainder
2
x 2x k
comes out to be find and .
x a
k
a
Solution:
By the division algorithm,
Dividend Divisor Quotient Remainder
Dividend Remainder Divisor Quotient
4 3 2 4 3 2
x 6x 16x 25x 10 x a x 6x 16x 26x 10 a
is divisible by .
2
x 2x k
Now, divide by
4 3 2
x 6x 16x 26x 10 a
.
2
x 2x k
Observe that will be
2
10 2k x 10 a 8k k
zero.
So,
and
10 2k 0
2
10 a 8k k 0
10 2k 0
10 2k 0
2k 10
k 5
Now,
2
10 a 8k k 0
2
10 a 8 5 5 0
10 a 40 25 0
5 a 0
a 5
Therefore,
and
k 5
a 5