RS Aggarwal Solutions: Exercise 2C - Polynomials | Extra Documents & Tests for Class 9 PDF Download

Q.1. By actual division, find the quotient and the remainder when (x⁴ + 1) is divided by (x – 1).
Verify that remainder = f(1).
Ans. Let f(x) = x⁴ + 1 and g(x) = x – 1.

Quotient = x³ + x² + x + 1

Remainder = 2
Verification:

Putting x = 1 in f(x), we get
f(1) = 1⁴ + 1 = 1 + 1 = 2 = Remainder, when f(x) = x⁴ + 1 is divided by g(x) = x – 1

Q.2. Verify the division algorithm for the polynomials
p(x)=2x⁴ − 6x³ + 2x² − x + 2 and g(x) = x + 2.
Ans.
p(x)=2x⁴ − 6 x³ + 2x² −x + 2 and g(x) = x + 2

Quotient = 2x³−10x² + 22x − 45
Remainder = 92
Verification:
Divisor × Quotient + Remainder
=(x+2)×(2x³−10x² + 22x − 45) + 92
=x(2x³ − 10x² + 22x − 45) + 2(2x³−10x²+22x − 45) + 92
=2x⁴ − 10x³ + 22x² − 45x + 4x³ − 20x² + 44x − 90 + 92
=2x⁴ − 6x³ + 2x² − x + 2
= Dividend
Hence verified.

Q.3. Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
p(x)=x³−6x² + 9x + 3, g(x) = x−1.
Ans. p(x)=x³ − 6x² + 9x + 3
g(x)=x−1

By remainder theorem, when p(x) is divided by (x − 1), then the remainder = p(1).
Putting x = 1 in p(x), we get
p(1)=1³ − 6 × 1² + 9 × 1 + 3 = 1 − 6 + 9 + 3 = 7
∴ Remainder = 7
Thus, the remainder when p(x) is divided by g(x) is 7.

Q.4. Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
p(x)=2x³ − 7x² + 9x − 13, g(x) = x − 3.
Ans.
p(x)=2x³ − 7x² + 9x − 13
g(x) = x − 3
By remainder theorem, when p(x) is divided by (x − 3), then the remainder = p(3).
Putting x = 3 in p(x), we get
p(3) = 2×3³ − 7×3²+9×3−13=54−63+27−13=5
∴ Remainder = 5
Thus, the remainder when p(x) is divided by g(x) is 5.

Q.5. Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where

p(x)=3x⁴−6x²−8x−2, g(x)=x−2.
Ans.
p(x) = 3x4 − 6x2 − 8x − 2
g(x) = x − 2
By remainder theorem, when p(x) is divided by (x − 2), then the remainder = p(2).
Putting x = 2 in p(x), we get

p(2)=3 × 2⁴ −6 × 2² − 8 × 2 − 2 = 48 − 24− 16 − 2=6
∴ Remainder = 6
Thus, the remainder when p(x) is divided by g(x) is 6.

Q.6. Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
p(x)=2x³ − 9x² + x + 15, g(x) = 2x − 3.
Ans.
p(x)=2x³ − 9x² + x + 15
g(x) = 2x − 3 = 2
By remainder theorem, when p(x) is divided by (2x − 3), then the remainder =
Putting in p(x), we get

∴ Remainder = 3
Thus, the remainder when p(x) is divided by g(x) is 3.

Q.7. Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where

p(x)=x³ − 2x²− 8x − 1, g(x) = x + 1.
Ans. p(x)=x³−2x² − 8x − 1
g(x)=x+1
By remainder theorem, when p(x) is divided by (x + 1), then the remainder = p(−1).
Putting x = −1 in p(x), we get
p(−1)=(−1)³−2×(−1)²−8×(−1)−1=−1−2+8−1=4
∴ Remainder = 4
Thus, the remainder when p(x) is divided by g(x) is 4.

Q.8. Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
p(x) = 2x³ + x²− 15x − 12, g(x) = x + 2.
Ans.
p(x) = 2x³ + x² −15x − 12
g(x) = x + 2
By remainder theorem, when p(x) is divided by (x + 2), then the remainder = p(−2).
Putting x = −2 in p(x), we get
p(−2) = 2 × (−2)³ + (−2)2 − 15 × (−2) −12 = −16 + 4 + 30 − 12 = 6
∴ Remainder = 6
Thus, the remainder when p(x) is divided by g(x) is 6.

Q.9. Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
p(x)=6x³ + 13x² + 3, g(x) = 3x + 2.
Ans.
p(x) = 6x³ + 13x² + 3

By remainder theorem, when p(x) is divided by (3x + 2), then the remainder = p
Putting x = in p(x), we get

∴ Remainder = 7
Thus, the remainder when p(x) is divided by g(x) is 7.

Q.10. Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
p(x) = x³ − 6x² + 2x − 4, g (x) = 1 -
Ans.
p(x)=x³ − 6x² + 2x − 4

By remainder theorem, when p(x) is divided by then the remainder = 

Putting x in p(x), we get


∴ Remainder =
Thus, the remainder when p(x) is divided by g(x) is

Q.11. Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
p(x) = 2x³ + 3x² − 11x − 3, g(x) =
Ans.
p(x) = 2x³ + 3x² − 11x − 3

By remainder theorem, when p(x) is divided by then the remainder =
Putting x =  in p(x), we get


∴ Remainder = 3
Thus, the remainder when p(x) is divided by g(x) is 3.

Q.12. Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where

p(x)=x³ −ax² + 6x − a, g(x) = x − a.
Ans.
p(x)=x³ − ax² + 6x − a
g(x) = x − a
By remainder theorem, when p(x) is divided by (x − a), then the remainder = p(a).
Putting x = a in p(x), we get
p(a) = a³ − a × a² + 6 × a − a = a³ − a³ + 6a − a = 5a
∴ Remainder = 5a
Thus, the remainder when p(x) is divided by g(x) is 5a.

Q.13. The polynomials (2x³ + x² − ax + 2) and (2x³ − 3x² −3x + a) when divided by (x – 2) leave the same remainder. Find the value of a.
Ans.
Let f(x) = 2x³ + x²− ax + 2 and g(x)=2x³ − 3x² − 3x + a.
By remainder theorem, when f(x) is divided by (x – 2), then the remainder = f(2).
Putting x = 2 in f(x), we get
f(2) = 2× 2³ + 2² − a × 2 + 2 = 16 + 4 − 2a + 2= −2a + 22
By remainder theorem, when g(x) is divided by (x – 2), then the remainder = g(2).
Putting x = 2 in g(x), we get
g(2) = 2×2³ − 3 × 2² − 3 × 2 + a = 16 − 12 − 6 + a = −2 + a
It is given that,
f(2)=g(2)
⇒ −2a + 22 = −2 + a
⇒ −3a = −24
⇒ a = 8
Thus, the value of a is 8.

Q.14. The polynomial p(x) = x⁴ − 2x³ + 3x² − ax + b when divided by (x − 1) and (x + 1) leaves the remainders 5 and 19 respectively. Find the values of a and b. Hence, find the remainder when p(x) is divided by (x − 2).
Ans.
Let:
p(x) = x4 − 2x³ + 3x² − ax + b
Now,
When p(x) is divided by (x−1), the remainder is p(1).
When p(x) is divided by (x+1), the remainder is p(−1). 

Thus, we have:
p(1) = (1⁴ − 2 × 1³ + 3 × 1² − a × 1 + b)
=(1 − 2 + 3 − a + b)
= 2 − a + b
And,
p(−1)=[(−1)⁴−2×(−1)³+3×(−1)²−a×(−1)+b]
= (1 + 2 + 3 + a + b)
= 6 + a + b
Now,
2 − a + b = 5    ...(1)
6 + a + b = 19  ...(2)
Adding (1) and (2), we get:
8 + 2b = 24
⇒ 2b = 16
⇒ b = 8
By putting the value of b, we get the value of a, i.e., 5.
∴ a = 5 and b = 8
Now,
f(x) = x⁴ −2x³ + 3x² − 5x + 8
Also,
When p(x) is divided by (x−2), the remainder is p(2).
Thus, we have:
p(2)=(2⁴ − 2 × 2³+ 3 × 2² − 5 × 2 + 8)   [a=5 and b=8]
=(16 − 16 + 12 − 10 + 8)
=10

Q.15. If p(x)=x³ − 5x² + 4x − 3 and g(x)= x − 2, show that p(x) is not a multiple of g(x).
Ans.
p(x) = x³ − 5x² + 4x−3
g(x) = x − 2
Putting x = 2 in p(x), we get
p(2) = 2³ − 5 × 2² + 4 × 2 − 3 = 8 − 20 + 8 − 3 = − 7 ≠ 0
Therefore, by factor theorem, (x − 2) is not a factor of p(x).
Hence, p(x) is not a multiple of g(x).

Q.16. If p(x) = 2x³ − 11x² − 4x+5 and g(x) = 2x + 1, show that g(x) is not a factor of p(x).
Ans.
p(x) = 2x³−11x² − 4x + 5
 
Putting in p(x), we get


Therefore, by factor theorem, (2x + 1) is not a factor of p(x).
Hence, g(x) is not a factor of p(x).