Class 10 Maths Important Questions for Polynomials
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
x-1
3
x^{4}-3x^{2}+2+ 3x^{3}
4
x^{2} -2x-1
5
1-3x^{3}
B.
type of polynomial
Polynomial
monomial
X^{3} -4x^{2} +1
binomial
x-1
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 x-1
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 x-1
P(1) = 1+ 1-2+1+1=2
a. it cuts the x-axis at two points ,so 2 zeroes
b. it cuts the x-axis at four points ,so 4 zeroes
c. Since it does not cut the axis, so 0 zeroes
d. it cuts the x-axis at 1 points ,so 1 zero’s
e. it cuts the x-axis 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 x-axis at two points ,so 2 zeroes
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}+(p-2)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}-13x-22
d. x^{2}+13x-22 Solution
P(x) =(x-11)(x-2)
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
(2k-9)x –(8-k)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 (p-q) ,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} +4x-5 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
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
(p-1)/(p+1) , (q-1)/(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
3q-3r=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 = (x-1)(x+3)
Now
2x^{4} + x^{3} – 14x^{2} + 5x + 6= (x^{2} + 2x -3)(2x^{2}-3x-8+2k)
=(x^{2} + 2x -3)(2x^{2}-3x-2)
=(x-1)(x+3)(x-2)(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
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