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Class X
Chapter 2 – Polynomials
Maths
Exercise 2
...
Find the number of zeroes of p(x), in each case
...
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Class X
Chapter 2 – Polynomials
Maths
(iv)
(v)
(v)
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(ii) The number of zeroes is 1 as the graph intersects the x-axis at
only 1 point
...
(iv) The number of zeroes is 2 as the graph intersects the x-axis at 2
points
...
(vi) The number of zeroes is 3 as the graph intersects the x-axis at 3
points
...
vidhyarjan
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com
Mobile: 9999 249717
Head Office: 1/3-H-A-2, Street # 6, East Azad Nagar, Delhi-110051
(One Km from ‘Welcome Metro Station)
Class X
Chapter 2 – Polynomials
Maths
Exercise 2
...
Answer:
The value of
is zero when x − 4 = 0 or x + 2 = 0, i
...
, when x
= 4 or x = −2
Therefore, the zeroes of
are 4 and −2
...
e
...
Sum of zeroes =
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e
...
Sum of zeroes =
Product of zeroes =
The value of 4u2 + 8u is zero when 4u = 0 or u + 2 = 0, i
...
, u = 0 or
u = −2
Therefore, the zeroes of 4u2 + 8u are 0 and −2
...
vidhyarjan
...
com
Mobile: 9999 249717
Head Office: 1/3-H-A-2, Street # 6, East Azad Nagar, Delhi-110051
(One Km from ‘Welcome Metro Station)
Class X
Chapter 2 – Polynomials
The value of t2 − 15 is zero when
Therefore, the zeroes of t2 − 15 are
Maths
or
and
, i
...
, when
...
e
...
Sum of zeroes =
Product of zeroes
Question 2:
Find a quadratic polynomial each with the given numbers as the sum
and product of its zeroes respectively
...
vidhyarjan
...
com
Mobile: 9999 249717
Head Office: 1/3-H-A-2, Street # 6, East Azad Nagar, Delhi-110051
(One Km from ‘Welcome Metro Station)
Class X
Chapter 2 – Polynomials
Maths
Answer:
Let the polynomial be
, and its zeroes be
and
...
Therefore, the quadratic polynomial is 4x2 − x − 4
...
, and its zeroes be
and
...
vidhyarjan
...
com
Mobile: 9999 249717
Head Office: 1/3-H-A-2, Street # 6, East Azad Nagar, Delhi-110051
(One Km from ‘Welcome Metro Station)
Class X
Chapter 2 – Polynomials
Therefore, the quadratic polynomial is
Let the polynomial be
...
and
...
, and its zeroes be
Therefore, the quadratic polynomial is
Let the polynomial be
Maths
...
Therefore, the quadratic polynomial is
...
vidhyarjan
...
com
Mobile: 9999 249717
Head Office: 1/3-H-A-2, Street # 6, East Azad Nagar, Delhi-110051
(One Km from ‘Welcome Metro Station)
Class X
Chapter 2 – Polynomials
Maths
Exercise 2
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Class X
Chapter 2 – Polynomials
Maths
Quotient = x2 + x − 3
Remainder = 8
Quotient = −x2 − 2
Remainder = −5x +10
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,
is not a factor of
...
vidhyarjan
...
com
Mobile: 9999 249717
Head Office: 1/3-H-A-2, Street # 6, East Azad Nagar, Delhi-110051
(One Km from ‘Welcome Metro Station)
Class X
Chapter 2 – Polynomials
Maths
Question 3:
Obtain all other zeroes of
, if two of its zeroes are
...
...
vidhyarjan
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Mobile: 9999 249717
Head Office: 1/3-H-A-2, Street # 6, East Azad Nagar, Delhi-110051
(One Km from ‘Welcome Metro Station)
Class X
Chapter 2 – Polynomials
Maths
We factorize
Therefore, its zero is given by x + 1 = 0
x = −1
As it has the term
, therefore, there will be 2 zeroes at x = −1
...
Question 4:
On dividing
by a polynomial g(x), the quotient and
remainder were x − 2 and − 2x + 4, respectively
...
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vidhyarjan
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com
Mobile: 9999 249717
Head Office: 1/3-H-A-2, Street # 6, East Azad Nagar, Delhi-110051
(One Km from ‘Welcome Metro Station)
Class X
Chapter 2 – Polynomials
Maths
(i) deg p(x) = deg q(x)
(ii) deg q(x) = deg r(x)
(iii) deg r(x) = 0
Answer:
According to the division algorithm, if p(x) and g(x) are two
polynomials with
g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that
p(x) = g(x) × q(x) + r(x),
where r(x) = 0 or degree of r(x) < degree of g(x)
Degree of a polynomial is the highest power of the variable in the
polynomial
...
e
...
by 2
...
e
...
Checking for division algorithm,
p(x) = g(x) × q(x) + r(x)
= 2(
)
=
Thus, the division algorithm is satisfied
...
vidhyarjan
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Mobile: 9999 249717
Head Office: 1/3-H-A-2, Street # 6, East Azad Nagar, Delhi-110051
(One Km from ‘Welcome Metro Station)
Class X
Chapter 2 – Polynomials
Maths
(ii) deg q(x) = deg r(x)
Let us assume the division of x3 + x by x2,
Here, p(x) = x3 + x
g(x) = x2
q(x) = x and r(x) = x
Clearly, the degree of q(x) and r(x) is the same i
...
, 1
...
(iii)deg r(x) = 0
Degree of remainder will be 0 when remainder comes to a constant
...
Here, p(x) = x3 + 1
g(x) = x2
q(x) = x and r(x) = 1
Clearly, the degree of r(x) is 0
...
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4
Question 1:
Verify that the numbers given alongside of the cubic polynomials
below are their zeroes
...
Comparing the given polynomial with
, we obtain a = 2,
b = 1, c = −5, d = 2
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(ii)
Therefore, 2, 1, 1 are the zeroes of the given polynomial
...
Verification of the relationship between zeroes and coefficient of the
given polynomial
Multiplication of zeroes taking two at a time = (2)(1) + (1)(1) + (2)(1)
=2 + 1 + 2 = 5
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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 2, − 7, − 14
respectively
...
It is given that
If a = 1, then b = −2, c = −7, d = 14
...
Answer:
Zeroes are a − b, a + a + b
Comparing the given polynomial with
, we obtain
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Hence, a = 1 and b =
or
...
Answer:
Given that 2 +
and 2
are zeroes of the given polynomial
...
vidhyarjan
...
com
Mobile: 9999 249717
Head Office: 1/3-H-A-2, Street # 6, East Azad Nagar, Delhi-110051
(One Km from ‘Welcome Metro Station)
Class X
Chapter 2 – Polynomials
Maths
For finding the remaining zeroes of the given polynomial, we will find
by x2 − 4x + 1
...
And
=
Therefore, the value of the polynomial is also zero when
or
Or x = 7 or −5
Hence, 7 and −5 are also zeroes of this polynomial
...
vidhyarjan
...
com
Mobile: 9999 249717
Head Office: 1/3-H-A-2, Street # 6, East Azad Nagar, Delhi-110051
(One Km from ‘Welcome Metro Station)
Class X
Chapter 2 – Polynomials
Maths
Question 5:
If the polynomial
polynomial
is divided by another
, the remainder comes out to be x + a, find k and
a
...
Let us divide
by
It can be observed that
will be 0
...
vidhyarjan
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Mobile: 9999 249717
Head Office: 1/3-H-A-2, Street # 6, East Azad Nagar, Delhi-110051
(One Km from ‘Welcome Metro Station)
Class X
Chapter 2 – Polynomials
Therefore,
= 0 and
For
Maths
=0
= 0,
2 k =10
And thus, k = 5
=0
For
10 − a − 8 × 5 + 25 = 0
10 − a − 40 + 25 = 0
−5−a=0
Therefore, a = −5
Hence, k = 5 and a = −5
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