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RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Test Yourself

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Test Yourself

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 2 Polynomials Ex 2A
• RS Aggarwal Solutions Class 10 Chapter 2 Polynomials Ex 2B
• RS Aggarwal Solutions Class 10 Chapter 2 Polynomials Ex 2C
• RS Aggarwal Solutions Class 10 Chapter 2 Polynomials MCQS
• RS Aggarwal Solutions Class 10 Chapter 2 Polynomials Test Yourself

MCQ
Question 1.
Solution:
(b) Polynomial is x² – 2x – 3
=> x² – 3x + x – 3
= x(x – 3) + 1(x – 3)
= (x – 3) (x + 1)
Either x – 3 = 0, then x = 3
or x + 1 = 0, then x = -1
Zeros are 3, -1

Question 2.
Solution:
(a) α, β, γ are the zeros of Polynomial is x³ – 6x² – x + 30
Here, a = 1, b = -6, c = -1, d = 30
αβ + βγ + γα = \(\frac { c }{ a }\) = \(\frac { -1 }{ 1 }\) = -1

Question 3.
Solution:
(c)

Question 4.
Solution:
(c) Let α and β be the zeros of the polynomial 4x² – 8kx + 9

Short-Answer Questions
Question 5.
Solution:

Either x + 15 = 0, then x = -15
or x – 13 = 0, then x = 13
Zeros are 13, -15

Question 6.
Solution:
The polynomial is (a² + 9) x² + 13x + 6a

Question 7.
Solution:
Zeros are 2 and -5
Sum of zeros = 2 + (-5) = 2 – 5 = -3
and product of zeros = 2 x (-5) = -10
Now polynomial will be
x² – (Sum of zeros) x + Product of zeros
= x² – (-3)x + (-10)
= x² + 3x – 10

Question 8.
Solution:
(a – b), a, (a + b) are the zeros of the polynomial x³ – 3x² + x + 1
Here, a = 1, b = -3, c = 1, d = 1
Now, sum of zeros = \(\frac { -b }{ a }\) = \(\frac { -(-3) }{ 1 }\) = 3
a – b + a + a + b = 3

Question 9.
Solution:
Let f(x) = x³ + 4x² – 3x – 18
If 2 is its zero, then it will satisfy it
Now, (x – 2) is a factor Dividing by (x – 2)
Hence, x = 2 is a zero of f(x)

Question 10.
Solution:
Sum of zeros = -5
and product of zeros = 6
Quadratic polynomial will be
x² – (Sum of zeros) x + Product of zeros
= x² – (-5) x + 6
= x² + 5x + 6

Question 11.
Solution:

Question 12.
Solution:
p(x) = x³ + 3x² – 5x + 4
g(x) = x – 2
Let x – 2 = 0, then x = 2
Remainder = p(2) = (2)³ + 3(2)² – 5(2) + 4 = 8 + 12 – 10 + 4 = 14

Question 13.
Solution:
f(x) = x³ + 4x² + x – 6
and g(x) = x + 2
Let x + 2 = 0, then x = -2
f(-2) = (-2)³ + 4(-2)² + (-2) – 6 = -8 + 16 – 2 – 6 = 0
Remainder is zero, x + 2 is a factor of f(x)

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:
f(x) = x⁴ + 4x² + 6
=> (x²)² + 4(x²) + 6 = y² + 4y + 6 (Let x² = y)
Let α, β be the zeros of y² + 4y + 6
Sum of zeros = -4
and product of zeros = 6
But there are no factors of 6 whose sum is -4 {Factors of 6 = 1 x 6 and 2 x 3}
Hence, f(x) Has no zero (real).

Long-Answer Questions
Question 17.
Solution:
3 is one zero of p(x) = x³ – 6x² + 11x – 6
(x – 3) is a factor of p(x)
Dividing, we get

Question 18.
Solution:

Question 19.
Solution:
p(x) = 3x⁴ + 5x³ – 7x² + 2x + 2
Dividing by x² + 3x + 1,
we get,
Quotient = 3x² – 4x + 2

Question 20.
Solution:
Let p(x) = x³ + 2x² + kx + 3
g(x) = x – 3
and r(x) = 21
Dividing p(x) by g(x), we get

But remainder = 21
3 + 3k + 45 = 21
3k = 21 – 45 – 3
=> 3k = 21 – 48 = -27
k = -9
Second method:
x – 3 is a factor of p(x) : x = 3
Substituting the value of x in p(x)
p(3) = 3³ + 2 x 3² + k x 3 + 3
= 27 + 18 + 3k + 3
48 + 3k = 21
=> 3k = -48 + 21 = -27
k = -9
Hence, k = -9

Hope given RS Aggarwal Solutions Class 10 Chapter 2 Polynomials Test Yourself are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2I

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2I.

Other Exercises

• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2A
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2B
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2C
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2D
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2E
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2F
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2G
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2H
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2I
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2J
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2K

Question 1.
Solution:
(i)(3x+2)³

Question 2.
Solution:

Question 3.
Solution:
(i)(95)³ = (100 – 5)³

Hope given RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2I are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2H

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2H.

Other Exercises

• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2A
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2B
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2C
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2D
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2E
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2F
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2G
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2H
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2I
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2J
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2K

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:
9x² + 16y² + 4z² – 24xy + 16yz – 12xz

Question 5.
Solution:
25x² + 4y² + 9z² – 20xy – 12yz + 30xz

Question 6.
Solution:

Hope given RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2H are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.