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Given below are the
Class 10 Maths Important Questions for Polynomials
a) Concepts questions
b) Calculation problems
c) Multiple choice questions
d) Long answer questions
e) Fill in the blank's
f) Match the column
Match the column
>
A)
Degree of polynomial

Polynomial

1

x^{5} 3x^{2} +1

2

x1

3

x^{4}3x^{2}+2+ 3x^{3}

4

x^{2} 2x1

5

13x^{3}

B)
type of polynomial

Polynomial

monomial

X^{3} 4x^{2} +1

binomial

x1

trinomial

x^{4}3x^{2}+2+ 3x^{3}

No appropriate match

x^{2} 2


3x^{3}

C)
P(x)=5x
^{3} 3x
^{2}+7x+2
P(0)

P(1)

P(5)

P(1)

P(2)






Multiple choice Questions
Question 1 Find the remainder when x
^{4}+x
^{3}2x
^{2}+x+1 is divided by x1
a)1
b)5
c)2
d)3
Solution ( c)
Question 2Which of these identities is not true?
Solution (d)
Question 3True or False statement
 P(x) =x1 and g(x) =x^{2}2x +1 . p(x) is a factor of g(x)
 The factor of 3x^{2} –x4 are (x+1)(3x4)
 Every linear polynomial has only one zero
 Every real number is the zero’s of zero polynomial
 A binomial may have degree 6
 1,2 are the zeroes of x^{2}3x+2
 The degree of zero polynomial is not defined
 Graph of polynomial (x^{2}1) meets the xaxis at one point
 Graph of constant polynomial never meets x axis
Solution
 True, as g(1)=0
 True, we can get this by split method
 True
 True
 True , example x^{6} +1
 True
 True
 False as it meets at two points
 True
Question 4Factorize following
 x^{2} +9x+18
 3x^{3} –x^{2}3x+1
 x^{3}23x^{2}+142x120
 1+8x^{3}
Solution
a) (x+6)(x+3)
b) )(3x1)(x1)(x+1)
c)(x1)(x10)(x12)
d) (2x+1)(4x
^{2}2x+1)
Match the column
Graph of polynomial

Number of Zeros


0


1


2


3


4


5


6


7

Solution
a) it cuts the xaxis at two points ,so 2 zeroes
b) it cuts the xaxis at four points ,so 4 zeroes
c) Since it does not cut the axis, so 0 zeroes
d) it cuts the xaxis at 1 points ,so 1 zero’s
e) it cuts the xaxis at 1 points ,so 1 zero’s
f) Since it does not cut the axis, so 0 zeroes
g) Since it does not cut the axis, so 0 zeroes
h) it cuts the xaxis at two points ,so 2 zeroes
Graph of polynomial

Type of polynomial


Linear polynomial


Quadratic polynomial


Cubic polynomial


Constant polynomial









Solution
a) Quadratic as parabola
b) Three zeroes,So cubic polynomial
c) Contant value polynomial
d) Linear polynomial
e) One zeroes but not straight line. So no appropriate match found
f) Quadratic as parabola
g) Quadratic as parabola
e) Cubic as has three zeroes ,two of them same
Multiple Choice Questions
Question 1 If a and b are the zeroes of the polynomial x
^{2}11x +30, Find the value of a
^{3} + b
^{3}
a)134
b)412
c)256
d)341
Solution
a
^{3} + b
^{3}= (a+b) (a
^{2}+b
^{2}ab)=(a+b) {(a+b)
^{2} 3ab}
Now a+b=(11)/1=11
ab=30
So a
^{3}+b
^{3}=11( 121 90)=341
Question 2 S(x) = px
^{2}+(p2)x +2. If 2 is the zero of this polynomial,what is the value of p
a)1
b)1/2
c) 1/2
d)+1
Solution
S(2)=4p+0+2=0 => p=1/2
Question 3if the zeroes of the quadratic equation are 11 and 2 ,what is expression for quadratic
a) x
^{2}13x+22
b) x
^{2}11x+22
c) x
^{2}13x22
d) x
^{2}+13x22
Solution (a)
P(x) =(x11)(x2)
Question 4 p(x) = x
^{4} 6x
^{3} +16x
^{2} 25x +10
q(x) = x
^{2}2x+k
It is given
p(x) = r(x) q(x) + (x+a)
Find the value of k and a
a) 2,2
b) 5 ,5
c) 7,3
d) 3,1
Solution (b)
Dividing p(x) by q(x) ,we get the remainder
(2k9)x –(8k)k +10
Comparing this with (x+a)
We get
K=5 and a=5
Question 5 A cubic polynomial is given below
S(x) =x
^{3} 3x
^{2}+x+1
The zeroes of the polynomial are given as (pq) ,p and (p+q). What is the value p and q
a) 1 ,
b) 1,2
c) 1,2
d) None of these
Solution (a)
Division of polynomial
s(x) =r(x) s(x) + w(x)
Find the value of r(x) and w(x) in each case
a) p(x) =x
^{4}+x
^{3}+2x
^{2}+3x+4
s(x) =x+2
b) p(x) =x
^{4}+4
s(x)=x
^{2}+x+1
Solution
a) r(x)=x
^{3}x
^{2} +4x5 w(x)=14
b) r(x)=x
^{2}x w(x) =x+4
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