Class 10 R S AGGARWAL AND V AGGARWAL Solutions Maths Chapter 2 - Polynomials
Polynomials Exercise Ex. 2A
Solution 1
Solution 2
Solution 3
Solution 5
We have
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 12
Let
Solution 14
Solution 15
Solution 16
Solution 18
Solution 20
Solution 21
Solution 4
2x2 - x - 6 = 0
⇒ 2x2 - 4x + 3x - 6 = 0
⇒ 2x(x - 2) + 3(x - 2) = 0
⇒ (x - 2)(2x + 3) = 0
⇒ (x - 2) = 0 or (2x + 3) = 0
So, the zeroes of
the given quadratic polynomial are 2 and 
Sum of the zeros 
Product of the
zeros 
Solution 11
5x2 + 10x = 0
⇒ 5x(x + 2) = 0
⇒ 5x = 0 or (x + 2) = 0
⇒ x = 0 or x = -2
So, the zeroes of the given quadratic polynomial are 0 and -2.
Sum of the zeros 
Product of the
zeros 
Solution 13
p(x) = 2x2 + 5x + k
Here, a = 2, b = 5 and c = k
α + β = 
and
αβ = 
α2 + β2
+ αβ = 
…. Given
Solution 17
Let α = 2 and β = -6
Sum of zeros = α + β = 2 + (-6) = -4
Product of zeros = αβ = 2(-6) = -12
So, the required quadratic polynomial is
x2 - (α + β)x + αβ = x2 + 4x - 12
Here, a = 1, b = 4 and c = -12
Sum of the zeros 
Product of the zeros 
Solution 19
Polynomials Exercise Ex. 2B
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Here, remainder = 2x + 3 = px + q
⇒ p = 2 and q = 3
Solution 11
Solution 12
Solution 13
1
Solution 14
Solution 15
Solution 16
Solution 17
Solution 21
f(x) = 3x3 + 16x3 + 15x - 18
is one zero
of f(x).
is a factor
of f(x).
⇒ (3x - 2) is also a factor of f(x).
Solution 22
Here, f(x) = 2x4 - 3x3 - 3x2 + 6x - 2
1 and 
are
two zeros of f(x).
⇒ 
is a factor
of f(x).
⇒
is a factor
of f(x).
Solution 18
Solution 19
Here, f(x) = x4 + x3 - 14x2 - 2x + 24
Since, √2 and -√2 are two zeros of f(x)
⇒ (x - √2) and (x + √2) is a factor of f(x).
⇒ (x - √2) and (x + √2) = x - 2 is a factor of f(x).
Solution 20
Here, f(x) = 2x4 - 13x3 + 19x2 + 7x - 3
Let α = (2 + √3) and β = (2 - √3). Then,
α + β = (2 + √3).+ (2 - √3) = 4
αβ = (2 + √3).(2 - √3) =4 - 3 = 1
So, the quadratic polynomial whose roots are α and β is given by
x2 - (α + β)x + αβ = x2 - 4x + 1
⇒ x2 - 4x + 1 is a factor of f(x).
Solution 23
Polynomials Exercise Ex. 2C
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
Solution 17
If f(x) and g(x) are any two polynomials with g(x) ≠ 0, then we can always find polynomials q(x) and r(x) such that f(x) = q(x)g(x) + r(x),
where r(x) = 0 or degree r(x) < degree g(x).
Solution 18
Solution 19
Solution 20
Solution 21
Solution 22
Solution 23
Solution 24
Solution 25
Polynomials Exercise MCQ
Solution 1
Correct answer: (d)
An expression of the form p(x) = a0 + a1x + a2x2 + ….. + anxn, where an ≠ 0, is called a polynomial in x of degree n.
Here, a0, a1, a2, ……, an are real numbers and each power of x is a non-negative integer.
Solution 2
Correct answer: (d)
An expression of the form p(x) = a0 + a1x + a2x2 + ….. + anxn, where an ≠ 0, is called a polynomial in x of degree n.
Here, a0, a1, a2, ……, an are real numbers and each power of x is a non-negative integer.
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
Solution 19
Solution 20
Solution 21
Solution 22
Solution 23
Solution 24
Solution 25
Solution 26
Solution 27
Solution 28
Polynomials Exercise FA
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
Solution 19
Solution 20
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