RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2D

RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2D

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

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

Using factor theorem, show that :

Question 1.
Solution:
By factor theorem, x – 2 will be a factor of f(x) = x³ – 8 if f(2) = 0
(∴ x-2 = 0=>x = 2)
Now f(2) = (2)³ – 8 = 8- 8 = 0
Hence (x – 2) is a factor of f(x) Ans.

Question 2.
Solution:
By factor theorem,

Question 3.
Solution:
By factor theorem,
(x – 1) is a factor of f(x)=(2x⁴ + 9x³ + 6x²– 11x – 6)

Question 4.
Solution:
By factor theorem, (x + 2) will
a factor of f (x) = x⁴ – x⁴ + 2

Question 5.
Solution:
By factor theorem, (x + 5) will be a factor of f(x) = 2x³ + 9x² – 11x – 30 if f(-5) = 0

Question 6.
Solution:
By factor theorem, (2x – 3) is a factor of f(x) = 2x⁴ + x³ – 8x² – x + 6

Question 7.
Solution:
By factor theorem, (x – √2 ) will be a factor of f(x) = 7x² – 4√2x – 6

Question 8.
Solution:
By factor theorem, (x + √2) will be a factor of f(x) = 2√2 x³ + 5x + √2

Question 9.
Solution:
Let f(x) = 2x³ + 9x² + x + k and x – 1 is a factor of f(x)

Question 10.
Solution:
Let f(x) = 2x³ – 3x² – 18x + a and x – 4 is its factor

Question 11.
Solution:
Let f(x) – x⁴ – x³ – 11x² – x + a
f(x) is divisible by (x + 3)

Question 12.
Solution:
Let f(x) = 2x³ + ax² + 11x + a + 3
and (2x – 1) is its factor
Let 2x – 1 = 0 then 2x = 1

Question 13.
Solution:
Let f(x) = x³ – 10x² + ax + b and (x – 1) and (x – 2) are its factors
∴ x – 1 = 0 =>x=1
and x – 2 = 2 =>x=2

Question 14.
Solution:
Let f(x) = x⁴ + ax³ – 7x² – 8x + b ,
and (x + 2) and (x + 3) are its factors
∴x + 2 = 0 => x = -2
and x + 3= 0 => x = -3

Question 15.
Solution:
Let f(x) = x³ – 3x² – 13x + 15
Now x² + 2x – 3 = x² + 3x – x – 3

Question 16.
Solution:
Let f(x) = x³ + ax² + bx + 6 and (x – 2) is its factor
Let x – 2 = 0 then x = 2

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

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