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RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9 PDF Download

RS Aggarwal Solutions: Exercise 2B - Polynomials

Q.1. If p(x) = 5 − 4x + 2x2, find (i) p(0), (ii) p(3), (iii) p(−2)
Ans.
(i) p(x)=5−4x+2x2
⇒ p(0)=(5−4×0+2×02)
=(5 − 0 + 0) = 5
= 5
(ii) p(x)=5−4x+2x2
⇒ p(3)=(5−4×3+2×32)
= (5−12+18)
= 11
(iii) p(x)=5−4x+2x2
⇒ p(−2)=[5−4×(−2)+2×(−2)2]            
=(5+8+8)
= 21

Q.2. If p(y) = 4 + 3y − y2 + 5y3, find (i) p(0), (ii) p(2), (iii) p(−1).
Ans. 
(i) p(y)=4+3y−y+ 5y3
⇒p(0) = (4 + 3 × 0− 02 + 5 × 03)
=(4 + 0 − 0 + 0)
= 4
(ii) p(y) = 4 + 3y − y+ 5y3
⇒p(2) = (4 + 3 × 2 − 2+ 5 × 23)
=(4 + 6 − 4 + 40)
= 46
(iii) p(y) = 4 + 3y − y+ 5y3
⇒ p(−1) = [4 + 3 × (−1) − (−1)+ 5 × (−1)3]
=(4 − 3 − 1 − 5)
= − 5

Q.3. If f(t) = 4t2 − 3t + 6, find (i) f(0), (ii) f(4), (iii) f(−5).
Ans.
(i) f(t) = 4t− 3t + 6
⇒ f(0) = (4 × 0− 3 × 0 + 6)
= (0 − 0 + 6)
= 6
(ii) f(t) = 4t− 3t + 6
⇒ f(4) = (4 × 4− 3 × 4 + 6)
= (64 − 12 + 6)
= 58
(iii) f(t) = 4t− 3t + 6
⇒ f(−5) = [4×(−5)− 3× (−5)+6]
=(100+15+6)
=121

Q.4. If p(x)=x− 3x+ 2x, find p(0), p(1), p(2). What do you conclude?
Ans.
p(x) = x− 3x+ 2x       .....(1)
Putting x = 0 in (1), we get
p(0) = 0− 3 × 0+ 2 × 0 = 0
Thus, x = 0 is a zero of p(x).
Putting x = 1 in (1), we get
p(1)=13−3×12+2×1=1−3+2=0
Thus, x = 1 is a zero of p(x).
Putting x = 2 in (1), we get
p(2) = 23−3 × 2+ 2 × 2 = 8 − 3 × 4 + 4 = 8 − 12 + 4 = 0
Thus, x = 2 is a zero of p(x).

Q.5. If p(x) = x3 + x2 – 9x – 9, find p(0), p(3), p(–3) and p(–1). What do you conclude about the zero of p(x)? Is 0 a zero of p(x)?
Ans.
p(x) = x3 + x2 – 9x – 9         .....(1)
Putting x = 0 in (1), we get
p(0) = 03 + 02 – 9 × 0 – 9 = 0 + 0 – 0 – 9 = –9 ≠ 0
Thus, x = 0 is not a zero of p(x).
Putting x = 3 in (1), we get
p(3) = 33 + 32 – 9 × 3 – 9 = 27 + 9 – 27 – 9 = 0

Thus, x = 3 is a zero of p(x).
Putting x = –3 in (1), we get
p(–3) = (–3)3 + (–3)2 – 9 × (–3) – 9 = –27 + 9 + 27 – 9 = 0
Thus, x = –3 is a zero of p(x).
Putting x = –1 in (1), we get
p(–1) = (–1)3 + (–1)2 – 9 × (–1) – 9 = –1 + 1 + 9 – 9 = 0
Thus, x = –1 is a zero of p(x).

Q.6. Verify that:
(i) 4 is a zero of the polynomial p(x) = x − 4.
(ii) −3 is a zero of the polynomial q(x) = x + 3.
(iii) 2/5 is a zero of the polynomial, f(x) = 2 − 5x.
(iv) RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9 is a zero of the polynomial g(y) = 2y + 1.
Ans.
(i) p(x)=x−4
⇒ p(4)=4−4
= 0
Hence, 4 is the zero of the given polynomial.
(ii) p(x) =  (−3) + 3
⇒ p(3)=0
Hence, 3 is the zero of the given polynomial.
(iii) p(x) = 2 − 5x
RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9
= 2−2
= 0
Hence, 2/5  is the zero of the given polynomial.
(iv) p(y) = 2y + 1
RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9
= − 1 + 1
= 0
Hence,RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9 is the zero of the given polynomial.

Q.7. Verify that
(i) 1 and 2 are the zeros of the polynomial p(x) = x2 − 3x + 2.
(ii) 2 and −3 are the zeros of the polynomial q(x) = x2 + x − 6.
(iii) 0 and 3 are the zeros of the polynomial r(x) = x2 − 3x.
Ans.
(i) p(x)=x− 3x + 2 = (x−1) (x−2)
⇒ p(1)=(1−1)×(1−2)
=0×(−1)
=0
Also,
p(2) = (2−1)(2−2)
= (−1) × 0
= 0
Hence, 1 and 2 are the zeroes of the given polynomial.
(ii) p(x) = x+ x − 6
⇒ p(2) = 2+ 2 − 6
=4 − 4
= 0
Also,
p(−3) = (−3)+ (−3) − 6
=9 − 9
= 0
Hence, 2 and −3 are the zeroes of the given polynomial.
(iii) p(x) = x− 3x
⇒ p (0) = 0− 3 × 0
Also,
p(3) = 3− 3 × 3
= 9 − 9
= 0
Hence, 0 and 3 are the zeroes of the given polynomial.

Q.8. Find the zero of the polynomial:
(i) p(x) = x − 5
(ii) q(x) = x + 4
(iii) r(x) = 2x + 5
(iv) f(x) = 3x + 1
(v) g(x) = 5 − 4x
(vi) h(x) = 6x − 2
(vii) p(x) = ax, a ≠ 0
(viii) q(x) = 4x
Ans.
(i) p(x) = 0⇒ x − 5 = 0
⇒ x = 5
Hence, 5 is the zero of the polynomial p(x).
(ii) q(x) = 0 ⇒ x + 4 = 0
⇒x= − 4
Hence, − 4 is the zero of the polynomial q(x).
(iii) r(x) = 0 ⇒ 2x + 5 = 0
RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9
Hence,RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9is the zero of the polynomial p(t).
(iv) f(x) = 0 ⇒ 3x + 1 = 0
RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9
Hence,RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9is the zero of the polynomial f(x).
(v) g(x) = 0 ⇒ 5 − 4x = 0

RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9
Hence, 5/4 is the zero of the polynomial g(x).
(vi) h (x) = 0 ⇒ 6x − 2 = 0
RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9
Hence, 1/3 is the zero of the polynomial h(x).
(vii) p(x) = 0 ⇒ ax = 0
⇒ x = 0
Hence, 0 is the zero of the polynomial p(x).
(viii) q(x) = 0 ⇒ 4x = 0
⇒x = 0
Hence, 0 is the zero of the polynomial q(x).

Q.9. If 2 and 0 are the zeros of the polynomial f(x)=2x3−5x+ ax + b then find the values of a and b.
Hint f(2) = 0 and f(0) = 0.
Ans.
It is given that 2 and 0 are the zeroes of the polynomial f(x)=2x− 5x+ ax + b.
∴ f(2) = 0
⇒2 × 2− 5 × 2+ a × 2 + b = 0
⇒16 − 20 + 2a + b = 0
⇒ −4 + 2a + b = 0
⇒ 2a + b = 4    .....(1)
Also,
f(0) = 0
⇒ 2 × 03− 5 × 0+ a × 0 + b = 0
⇒ 0 − 0 + 0 + b = 0
⇒ b = 0
Putting b = 0 in (1), we get
2a + 0=4
⇒ 2a = 4
⇒ a = 2

Thus, the values of a and b are 2 and 0, respectively.


RS Aggarwal Solutions: Exercise 2C - Polynomials

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.
RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9
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)=2x− 6x+ 2x− x + 2 and g(x) = x + 2.
Ans.
p(x)=2x− 6 x+ 2x−x + 2 and g(x) = x + 2
RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9
Quotient = 2x3−10x+ 22x − 45
Remainder = 92
Verification:
Divisor × Quotient + Remainder
=(x+2)×(2x3−10x+ 22x − 45) + 92
=x(2x− 10x+ 22x − 45) + 2(2x3−10x2+22x − 45) + 92
=2x− 10x+ 22x− 45x + 4x− 20x+ 44x − 90 + 92
=2x− 6x+ 2x− 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)=x3−6x+ 9x + 3, g(x) = x−1.
Ans. p(x)=x− 6x+ 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)=1− 6 × 1+ 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)=2x− 7x+ 9x − 13, g(x) = x − 3.
Ans.
p(x)=2x− 7x+ 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) = 2×3− 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)=2x− 9x+ x + 15, g(x) = 2x − 3.
Ans.
p(x)=2x− 9x2 + x + 15
g(x) = 2x − 3 = 2 RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9
By remainder theorem, when p(x) is divided by (2x − 3), then the remainder = RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9
Putting RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9in p(x), we get
RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9
∴ 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)=x− 2x2− 8x − 1, g(x) = x + 1.
Ans. p(x)=x3−2x− 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) = 2x+ x2− 15x − 12, g(x) = x + 2.
Ans.
p(x) = 2x+ x−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)+ (−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)=6x+ 13x+ 3, g(x) = 3x + 2.
Ans.
p(x) = 6x+ 13x+ 3
RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9
By remainder theorem, when p(x) is divided by (3x + 2), then the remainder = p RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9
Putting x = RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9in p(x), we get
RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9
∴ 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) = x− 6x+ 2x − 4, g (x) = 1 - RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9
Ans.
p(x)=x− 6x+ 2x − 4
RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9
By remainder theorem, when p(x) is divided by RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9then the remainder = 

RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9
Putting x RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9in p(x), we get
RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9
RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9
∴ Remainder = RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9
Thus, the remainder when p(x) is divided by g(x) is RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9

Q.11. Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
p(x) = 2x+ 3x− 11x − 3, g(x) = RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9
Ans.
p(x) = 2x+ 3x− 11x − 3
RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9
By remainder theorem, when p(x) is divided by RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9then the remainder =RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9
Putting x = RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9 in p(x), we get
RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9
RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9
∴ 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)=x−ax+ 6x − a, g(x) = x − a.
Ans.
p(x)=x− ax+ 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) = a− a × a+ 6 × a − a = a− a+ 6a − a = 5a
∴ Remainder = 5a
Thus, the remainder when p(x) is divided by g(x) is 5a.

Q.13. The polynomials (2x+ x− ax + 2) and (2x− 3x−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)=2x− 3x− 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) = 2×2− 3 × 2− 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 + 3x− 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) = (1− 2 × 1+ 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) = x−2x3 + 3x2 − 5x + 8
Also,
When p(x) is divided by (x−2), the remainder is p(2).
Thus, we have:
p(2)=(2− 2 × 23+ 3 × 2− 5 × 2 + 8)   [a=5 and b=8]
=(16 − 16 + 12 − 10 + 8)
=10

Q.15. If p(x)=x− 5x+ 4x − 3 and g(x)= x − 2, show that p(x) is not a multiple of g(x).
Ans.
p(x) = x− 5x+ 4x−3
g(x) = x − 2
Putting x = 2 in p(x), we get
p(2) = 2− 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) = 2x3−11x− 4x + 5
RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9 
Putting RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9in p(x), we get
RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9
RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9
Therefore, by factor theorem, (2x + 1) is not a factor of p(x).
Hence, g(x) is not a factor of p(x).


RS Aggarwal Solutions: Exercise 2D - Polynomials

Q.1. Using factor theorem, show that g(x) is a factor of p(x), when
p(x) = x3 – 8, g(x) = x – 2
Ans.
Let:
p(x) = x3 – 8
Now,
g(x) = 0 ⇒ x −2 = 0 ⇒ x =2
By the factor theorem, (x – 2) is a factor of the given polynomial if p(2) = 0.
Thus, we have:
p(2)=(2− 8)=0
Hence, (x − 2) is a factor of the given polynomial.

Q.2. Using factor theorem, show that g(x) is a factor of p(x), when
p(x) = 2x+ 7x2 – 24x – 45, g(x) = x – 3
Ans.
Let:
p(x) = 2x3 + 7x2 – 24x – 45
Now,
x − 3 = 0 ⇒ x = 3
By the factor theorem, (x − 3) is a factor of the given polynomial if p(3) = 0.
Thus, we have:
p(3) = (2 × 3− 7 × 3− 24 × 3 − 45)
= (54 + 63 − 72 − 45)
= 0
Hence, (x − 3) is a factor of the given polynomial

Q.3. Using factor theorem, show that g(x) is a factor of p(x), when
p(x) = 2x4 + 9x3 + 6x2 – 11x – 6, g(x) = x – 1
Ans.
Let:
p(x) = 2x4 + 9x3 + 6x2 – 11x – 6
Here,
x − 1 = 0 ⇒ x=1
By the factor theorem, (x − 1) is a factor of the given polynomial if p(1) = 0.
Thus, we have:
p(1)=(2×1+ 9 × 1+ 6 × 1− 11 × 1 − 6)
= (2 + 9 + 6 − 11 −6)
= 0
Hence, (x − 1) is a factor of the given polynomial.

Q.4. Using factor theorem, show that g(x) is a factor of p(x), when
p(x) = x4 – x2 – 12, g(x) = x + 2
Ans.
Let:
p(x) = x4 – x2 – 12
Here,
x+2 = 0 ⇒ x= −2
By the factor theorem, (x + 2) is a factor of the given polynomial if p (−2) = 0.
Thus, we have:
p(−2) = [(−2)− (−2)− 12]
= (16 − 4 − 12)
=0
Hence, (x + 2) is a factor of the given polynomial.

Q.5. Using factor theorem, show that g(x) is a factor of p(x), when
p(x) = 69 + 11x – x2 + x3, g(x) = x + 3
Ans.
p(x)=69+11x−x2+x3
g(x) = x + 3
Putting x = −3 in p(x), we get
p(−3) = 69 + 11 × (−3) − (−3)+ (−3)= 69 − 33 − 9 − 27 = 0
Therefore, by factor theorem, (x + 3) is a factor of p(x).
Hence, g(x) is a factor of p(x).

Q.6. Using factor theorem, show that g(x) is a factor of p(x), when
p(x) = 2x3 + 9x2 – 11x – 30, g(x) = x + 5
Ans.
Let:
p(x) = 2x+ 9x− 11x − 30
Here,
x + 5 = 0 ⇒ x = −5
By the factor theorem, (x + 5) is a factor of the given polynomial if p (−5) = 0.
Thus, we have:
p(−5) = [2 × (−5)+9 × (−5)−11 × (−5)−30]
= (−250 + 225 + 55 − 30)
= 0
Hence, (x + 5) is a factor of the given polynomial.

Q.7. Using factor theorem, show that g(x) is a factor of p(x), when

p(x) = 2x4 + x3 – 8x2 – x + 6, g(x) = 2x – 3

Ans.
Let:
p(x)=2x+ x− 8x− x + 6
Here,
2x − 3 = 0 ⇒ x = RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9
By the factor theorem, (2x − 3) is a factor of the given polynomial if pRS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9= 0
Thus, we have:
RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9
RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9
= 0
Hence, (2x − 3) is a factor of the given polynomial.


Q.8. Using factor theorem, show that g(x) is a factor of p(x), when
p(x) = 3x3 + x2 – 20x + 12, g(x) = 3x – 2
Ans.
p(x) = 3x+ x− 20x + 12
g(x) = 3x − 2 = 3RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9
Putting RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9in p(x), we get
RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9
RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9
Therefore, by factor theorem, (3x − 2) is a factor of p(x).
Hence, g(x) is a factor of p(x).

Q.9. Using factor theorem, show that g(x) is a factor of p(x), when

p(x) = 7x−4√2x − 6, g(x) = x − √2

Ans.
Let:
p(x)=7x2−4√2x − 6
Here,
x−√2 = 0 ⇒ x = √2
By the factor theorem, (x - √2) is a factor of the given polynomial if p(√2)=0
Thus, we have:
p(√2) = [7×(√2)−4√2 × √2 −6]
=(14 − 8 − 6)
= 0
Hence, (x − √2) is a factor of the given polynomial.

Q.10. Using factor theorem, show that g(x) is a factor of p(x), when
p(x)=2√2x+ 5x + √2, g(x) =x + √2

Ans.
Let:
p(x)=2√2x+ 5x + √2

Here,
x + √2 = 0 ⇒ x = −√2
By the factor theorem, (x+2) will be a factor of the given polynomial if p(−2) = 0.
Thus, we have:
p(−√2) = [2√2 × (−√2)+ 5 × (−√2)+√2]
=(4√2−5√2+√2)

= 0
Hence, (x + √2) is a factor of the given polynomial.

Q.11. Show that (p – 1) is a factor of (p10 – 1) and also of (p11 – 1).
Ans.
Let f(p) = p10 – 1 and g(p) = p11 – 1.
Putting p = 1 in f(p), we get
f(1) = 110 − 1 = 1 − 1 = 0
Therefore, by factor theorem, (p – 1) is a factor of (p10 – 1).
Now, putting p = 1 in g(p), we get
g(1) = 111 − 1 = 1 − 1 = 0
Therefore, by factor theorem, (p – 1) is a factor of (p11 – 1).

Q.12. Find the value of k for which (x − 1) is a factor of (2x3 + 9x2 + x + k).
Ans.

Let:

f(x)=2x+ 9x+ x + k

(x−1) is a factor of f(x) = 2x+ 9x+ x + k.
⇒ f(1) = 0
⇒2 × 1+ 9 × 1+ 1 + k = 0
⇒12 + k = 0
⇒k= −12
Hence, the required value of k is −12.


Q.13. Find the value of a for which (x − 4) is a factor of (2x3 − 3x2 − 18x + a).
Ans.
Let:

f(x)=2x− 3x− 18x + a
(x−4) is a factor of f(x)=2x− 3x− 18x + a.
⇒ f(4) = 0
⇒2 × 4− 3 × 4− 18 × 4 + a = 0
⇒8 + a = 0 ⇒ a = −8
Hence, the required value of a is −8.

Q.14. Find the value of a for which (x + 1) is a factor of (ax3 + x2 – 2x + 4a – 9).
Ans.
Let f(x) = ax3 + x2 – 2x + 4a – 9
It is given that (x + 1) is a factor of f(x).
Using factor theorem,  we have
f(−1) = 0
⇒a×(−1)3+(−1)2−2×(−1) + 4a − 9 = 0
⇒− a + 1 + 2+ 4a −9 = 0
⇒ 3a −6 = 0
⇒ 3a = 6
⇒a = 2
Thus, the value of a is 2.

Q.15. Find the value of a for which (x + 2a) is a factor of (x5 – 4a2 x3 + 2x + 2a +3).
Ans.

Let f(x) = x5 – 4a2 x3 + 2x + 2a +3
It is given that (x + 2a) is a factor of f(x).
Using factor theorem, we have
f(−2a) = 0
⇒(−2a)5−4a2×(−2a)3+2×(−2a)+2a+3=0
⇒−32a5−4a2×(−8a3)+2×(−2a)+2a+3=0
⇒−32a5+32a5−4a+2a+3=0
⇒ −2a + 3 = 0
⇒ 2a = 3
RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9
Thus, the value of a is 3/2.


Q.16. Find the value of m for which (2x – 1) is a factor of (8x4 + 4x3 – 16x2 + 10x + m).
Ans.
Let f(x) = 8x4 + 4x3 – 16x2 + 10x + m
It is given that (2x−1) = 2 RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9is a factor of f(x).
Using factor theorem,  we have
RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9
RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9
RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9
⇒2 + m = 0
⇒m = −2

Thus, the value of m is −2.


Q.17. Find the value of a for which the polynomial (x4 − x3 − 11x2 − x + a) is divisible by (x + 3).
Ans.
Let:
f(x) = x− x− 11x− x + a
Now,
x + 3 = 0 ⇒ x = −3
By the factor theorem, f(x) is exactly divisible by (x+3) if f (−3) = 0.
Thus, we have:
f(−3) = [(−3)4−(−3)3−11×(−3)2−(−3)+a]
=(81 + 27 − 99 + 3 + a)
=12 + a
Also,
f( − 3) = 0
⇒ 12 + a = 0
⇒ a = −12

Hence, f(x) is exactly divisible by (x+3) when a is −12.

Q.18. Without actual division, show that (x3 − 3x2 − 13x + 15) is exactly divisible by (x2 + 2x − 3).
Let:
f(x)=x− 3x−13 x + 15
And,
g(x) = x+ 2x − 3
=x2 + x − 3x −3
=x(x − 1)+3(x − 1)
=(x − 1)(x + 3)
Now, f(x) will be exactly divisible by g(x) if it is exactly divisible by (x−1) as well as (x+3).
For this, we must have:
f(1) = 0 and f(−3) = 0
Thus, we have:
f(1) = (1− 3 × 12 − 13 × 1 + 15)
=(1 − 3 − 13 + 15)
= 0
And,
f(−3)=[(−3)3−3×(−3)2−13×(−3)+15]
=(−27−27+39+15)
= 0
f(x) is exactly divisible by (x−1) as well as (x+3) . So, f(x) is exactly divisible by (x−1)(x+3).
Hence, f(x) is exactly divisible by x+ 2x − 3.

Q.19. If (x+ ax2 + bx + 6) has (x − 2) as a factor and leaves a remainder 3 when divided by (x − 3), find the values of a and b.
Ans.
Let:
f(x)=x+ ax+ bx + 6
(x−2) is a factor of f(x)=x+ ax2 + bx + 6.
⇒f(2) = 0
⇒2+ a × 2+ b × 2 + 6 = 0
⇒14 + 4a + 2b =0
⇒4a + 2b = −14
⇒ 2a + b= −7         ...(1)
Now,
x − 3 = 0 ⇒ x = 3
By the factor theorem, we can say:
When f(x) will be divided by (x−3), 3 will be its remainder.
⇒ f(3) = 3
Now,
f(3)=3+ a × 32 + b × 3 + 6
=(27 + 9a + 3b + 6)
= 33 + 9a + 3b
Thus, we have:
f (3) = 3
⇒33 + 9a + 3b = 3
⇒9a + 3b = −30
⇒3a + b = −10      ...(2)
Subtracting (1) from (2), we get:
a = −3
By putting the value of a in (1), we get the value of b, i.e., −1.
∴ a = −3 and b = −1

Q.20. Find the values of a and b so that the polynomial (x3 − 10x2 + ax + b) is exactly divisible by (x −1) as well as (x − 2).
Ans.
Let:
f(x) =x− 10 x2 + ax + b
Now,
x − 1 = 0 ⇒ x = 1
By the factor theorem, we can say:
f(x) will be exactly divisible by (x − 1) if f(1) = 0
Thus, we have:
f(1)=1− 10 × 12 + a × 1 + b
=(1 − 10 + a + b)
= −9 + a + b
∴ f(1) = 0 ⇒ a + b = 9 ...(1)
Also,
x − 2 = 0 ⇒ x = 2
By the factor theorem, we can say:
f(x) will be exactly divisible by (x−2) if f(2)=0.
Thus, we have:
f(2)=2− 10 × 2+ a × 2 + b
=(8 − 40 + 2a + b)
= − 32 + 2a + b
∴ f(2) = 0 ⇒ 2a+b=32       ...(2)
Subtracting (1) from (2), we get:
a = 23
Putting the value of a, we get the value of b, i.e., −14.
∴ a = 23 and b = −14

Q.21. Find the values of a and b so that the polynomial (x4 + ax3 − 7x2 − 8x + b) is exactly divisible by (x + 2) as well as (x + 3).
Ans.
Let:
f(x)=x+ ax− 7x− 8x + b
Now,
x + 2 = 0 ⇒ x = −2
By the factor theorem, we can say:
f(x) will be exactly divisible by (x+2) if f(−2)=0 .
Thus, we have:
f(−2)=[(−2)4+a×(−2)3−7×(−2)2−8×(−2)+b]
=(16 − 8a − 28 +16+ b)
=(4 − 8a + b)

∴ f(−2) = 0 ⇒ 8a − b = 4      ...(1)
Also,
x + 3 = 0 ⇒ x = −3

By the factor theorem, we can say:
f(x) will be exactly divisible by (x+3) if f(−3)=0 .Thus, we have:
f(−3) = [(−3)4+a×(−3)3−7×(−3)2−8×(−3)+b
=(81 −27a − 63 + 24 + b)
= (42 − 27 a + b)
∴ f(−3) = 0 ⇒ 27a − b = 42   ...(2)
Subtracting (1) from (2), we get:
⇒19a = 38
⇒ a = 2
Putting the value of a, we get the value of b, i.e., 12.
∴ a = 2 and b = 12

Q.22. If both (x – 2) and RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9are factors of px2 + 5x + r, prove that p = r.
Ans.
Let f(x) = px2 + 5x + r
It is given that (x – 2) is a factor of f(x).
Using factor theorem,  we have
f(2) = 0
⇒p × 2+ 5 × 2 + r = 0
⇒ 4p + r= −10 .....(1)
Also,RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9is a factor of f(x).
Using factor theorem,  we have
RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9
RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9
From (1) and (2), we have
4p + r = p + 4r
⇒ 4p − p= 4r − r
⇒3p = 3r
⇒ p = r

Q.23. Without actual division, prove that 2x4 – 5x3 + 2x2 – x + 2 is divisible by x2 – 3x + 2.
Ans.
Let f(x) = 2x4 – 5x3 + 2x2 – x + 2 and g(x) = x2 – 3x + 2.
x2 − 3x + 2
=x− 2x − x + 2
= x(x − 2) − 1(x − 2)
=(x−1)(x − 2)
Now, f(x) will be divisible by g(x) if f(x) is exactly divisible by both (x − 1) and (x − 2).
Putting x = 1 in f(x), we get
f(1) = 2 × 14 – 5 × 13 + 2 × 12 – 1 + 2 = 2 – 5 + 2 – 1 + 2 = 0
By factor theorem, (x − 1) is a factor of f(x). So, f(x) is exactly divisible by (x − 1).
Putting x = 2 in f(x), we get
f(2) = 2 × 24 – 5 × 23 + 2 × 22 – 2 + 2 = 32 – 40 + 8 – 2 + 2 = 0
By factor theorem, (x − 2) is a factor of f(x). So, f(x) is exactly divisible by (x − 2).
Thus, f(x) is exactly divisible by both (x − 1) and (x − 2).

Hence, f(x) = 2x4 – 5x+ 2x– x + 2 is exactly divisible by (x − 1)(x − 2) = x2 – 3x + 2.

Q.24. What must be added to 2x4 – 5x3 + 2x2 – x – 3 so that the result is exactly divisible by (x – 2)?
Ans.

Let k be added to 2x4 – 5x3 + 2x2 – x – 3 so that the result is exactly divisible by (x – 2). Here, k is a constant.
∴ f(x) = 2x4 – 5x3 + 2x2 – x – 3 + k is exactly divisible by (x – 2).
Using factor theorem, we have
f(2)=0
⇒2×24−5×23+2×22−2−3+k=0
⇒32 − 40 + 8 − 5 + k = 0

⇒ −5 + k = 0
⇒k = 5
Thus, 5 must be added to 2x– 5x3 + 2x2 – x – 3 so that the result is exactly divisible by (x – 2).

Q.25. What must be subtracted from (x4 + 2x3 – 2x2 + 4x + 6) so that the result is exactly divisible by (x+ 2x – 3)?
Ans.
Dividing (x4 + 2x3 – 2x2 + 4x + 6) by (x2 + 2x – 3) using long division method, we have
RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9
Here, the remainder obtained is (2x + 9).
Thus, the remainder (2x + 9) must be subtracted from (x4 + 2x3 – 2x2 + 4x + 6) so that the
result is exactly divisible by (x2 + 2x – 3).

Q.26. Use factor theorem to prove that (x + a) is a factor of (xn + an) for any odd positive integer.
Ans.
Let f(x) = xn + an
Putting x = −a in f(x), we get
f(−a) = (−a)n + an
If n is any odd positive integer, then
f(−a) = (−a)n + an = −an + an = 0
Therefore, by factor theorem, (x + a) is a factor of (xn + an) for any odd positive integer.

The document RS Aggarwal Solutions: Polynomials- 2 | Mathematics (Maths) Class 9 is a part of the Class 9 Course Mathematics (Maths) Class 9.
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FAQs on RS Aggarwal Solutions: Polynomials- 2 - Mathematics (Maths) Class 9

1. What are the RS Aggarwal Solutions?
Ans. RS Aggarwal Solutions are the detailed answers to the questions and problems given in the RS Aggarwal textbook. These solutions are provided to help students understand and solve the exercises and problems effectively.
2. What is Exercise 2B - Polynomials in RS Aggarwal Solutions?
Ans. Exercise 2B - Polynomials is a section in the RS Aggarwal textbook that focuses on the topic of polynomials. It consists of various problems and exercises related to polynomials, which help students practice and strengthen their understanding of the topic.
3. What is the purpose of Exercise 2C - Polynomials in RS Aggarwal Solutions?
Ans. The purpose of Exercise 2C - Polynomials in RS Aggarwal Solutions is to further enhance students' knowledge and skills in dealing with polynomials. This exercise includes more challenging problems and exercises to test their understanding and problem-solving abilities.
4. What does Exercise 2D - Polynomials in RS Aggarwal Solutions cover?
Ans. Exercise 2D - Polynomials in RS Aggarwal Solutions covers advanced topics and concepts related to polynomials. It includes problems involving factorization, division of polynomials, finding roots, and other higher-level applications of polynomials.
5. Are the RS Aggarwal Solutions for Polynomials-2 suitable for competitive exams?
Ans. Yes, the RS Aggarwal Solutions for Polynomials-2 are suitable for competitive exams. These solutions provide a comprehensive understanding of polynomials and help students prepare for various competitive exams that include questions on polynomials. By practicing these solutions, students can improve their problem-solving skills and increase their chances of performing well in such exams.
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