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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 blanks
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
Answer is ( c)
Let P(x) =x^{4}+x^{3}2x^{2}+x+1
Remainder when divide by x1
P(1) = 1+ 12+1+1=2
Question 2 Which of these identities is not true?
Solution
Answer is (d)
Question 3 True 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
A)
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
B.
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. Constant 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
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
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. p=1,q=√2 or √2
b. p=1,q=2 or 2
c. p=1,q=1 or 1
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
More Practice Questions
Question 1. If the polynomial f(x) = x
^{4} 6x
^{3} + 16x
^{2} – 25x + 10 is divided by another polynomial x
^{2} 2x + k, the remainder comes out to be x + a, find k and a.
Answer
Dividend = Divisor X Quotient + Remainder
Dividend  Remainder = Divisor X Quotient
Dividend  Remainder is always divisible by the divisor.
Now, it is given that f(x) when divided by x
^{2} 2x + k leaves (x + a) as remainder.
So f(x) (x+a) is divided by x
^{2} 2x + k
x
^{4} 6x
^{3} + 16x
^{2} – 25x + 10 (x+a) =x
^{4} 6x
^{3} + 16x
^{2} – 26x + 10 a
Now doing the division
Now remainder should be zero
(10 + 2k)x + (10  a  8k + k
^{2}) = 0
10 + 2k = 0 or k=5
Now (10 a  8k + k
^{2}) = 0
or 10  a  8 (5) + 5
^{2} = 0
or  a  5 = 0
a =5
Question 2. Find all the zeroes of the polynomial x
^{4} – 3x
^{3 }+ 6x – 4, if two of its zeroes are √2 and √2
Question 3. If p and q are he zeroes of the quadratic polynomial f(x) = x
^{2} – 2x + 3, find a polynomial whose roots are:
 p + 2, q + 2
 (p1)/(p+1) , (q1)/(q+1)
Question 4. For what value of k, 7 is the zero of the polynomial 2x
^{2} + 11x + (6k – 3)? Also find the other zero of the polynomial
Question 5. What must be added to f(x) = 4x
^{4 }+ 2x
^{3} – 2x
^{2} + x – 1 so that the resulting polynomial is divisible by g(x) = x
^{2} + 2x 3?
Question 6. Find k so that x
^{2} + 2x + k is a factor of 2x
^{4} + x
^{3} – 14 x
^{2} + 5x + 6. Also find all the zeroes of the two polynomials.
Question 7. If the zeroes of the quadratic polynomial ax
^{2} + bx + c, c ≠ 0 are equal, then
(A) c and a have opposite signs
(B) c and b have opposite signs
(C) c and a have the same sign
(D) c and b have the same sign
Answer
Given that, the zeroes of the quadratic polynomial ax^{2} + bx + c , c ≠ 0 are equal
Value of discriminant (D) has to be zero
b^{2}– 4ac = 0
b^{2} = 4ac
Since L.H.S. (b^{2}) can never be negative
R.H.S. also can never be negative.
a and c must have same sign
Question 8. Find the zeroes of 2x
^{3} – 11x
^{2} + 17x – 6.
Question 9. If (x  2) and [x – ½ ] are the factors of the polynomials qx
^{2} + 5x + r prove that q = r
Answer
4q+10+r=0 (1)
q/4 +5/2 +r=0 or q+10+4r=0 (2)
Subtracting 1 from 2
3q3r=0
q=r
Question 10. Find k so that the polynomial x
^{2} + 2x + k is a factor of polynomial 2x
^{4} + x
^{3} – 14x
^{2} + 5x + 6. Also, find all the zeroes of the two polynomials.
Answer
For x
^{2} + 2x + k is a factor of polynomial 2x
^{4} + x
^{3} – 14x
^{2} + 5x + 6, it should be able to divide the polynomial without any remainder
Comparing the coefficient of x we get.
21+7k=0 or k=3
So x
^{2} + 2x + k becomes x
^{2} + 2x 3 = (x1)(x+3)
Now
2x
^{4} + x
^{3} – 14x
^{2} + 5x + 6= (x
^{2} + 2x 3)(2x
^{2}3x8+2k)
=(x
^{2} + 2x 3)(2x
^{2}3x2)
=(x1)(x+3)(x2)(2x+1)
or x= 1,3,2,=1/2
Question 11. On dividing p(x) = x
^{3} – 3x
^{2} + x + 2 by a polynomial q(x), the quotient and remainder were x – 2 and 2x + 4, respectively. Find g(x).
Question 12. a, b, c are zeroes of cubic polynomial x
^{3} – 2x
^{2 }+ qx – r. If a + b = 0 then show that 2q = r.
Question 13. Find the remainder when x
^{51} +51 is divided by (x+1).
Question 14. a,b and c are zeroes of polynomial x
^{3 }+ px
^{2} + qx + 2 such that a b + 1 = 0. Find the value of 2p + q + 5.
Answer
By cubic polynomial equation, we have
a+b+c=p
ab+bc+ac=q
abc=2
Now given ab+1=0 or ab=1
So abc=2
(1)c=2 or c=2
2p + q + 5
=2(a+b+c) +(ab+bc+ac) +5
=2(a+b+2) +[1+2(a+b)] +5
=4 1+5=0
Question 15. Find the quadratic polynomial, the sum and product of whose zeroes are 4 and 1, respectively
Other Answer
2. √2 , √2, 2,1
4. 3, 3/2, 7
5. 61x – 65
11. x^{2} – x +1
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Class 10 Maths
Class 10 Science