RD Sharma Class 9 Solutions: Complete Guide [2026]

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RD Sharma Class 9 Solutions Chapter 6 MCQs: Class 9 mein maths ka topper banna hai? Perfect study material choose karna bohot zaroori hai, aur iske liye aap hamesha RD Sharma Solutions Class 9 Maths par bharosa kar sakte hain. Chahe aapke class tests hon ya assignments, humara detailed solutions guide hamesha aapke saath hai. RD Sharma Class 9 Solutions Chapter 6 MCQs ke baare mein complete information ke liye, poora blog padhiye.

Complete Solutions for Chapter 6 MCQs

RD Sharma ki Class 9 Chapter 6 MCQs factorization of polynomials ke concept ko strong banane ke liye perfect hai. Yahan har question ka step-by-step solution diya gaya hai jo aapko concept clear karne mein help karega. Polynomials ka factorization ek fundamental topic hai jo aage ke chapters mein bhi kaam aayega.

Is chapter mein total 20 MCQs hain jo different difficulty levels par based hain. Beginners se lekar advanced level tak ke students in questions se benefit le sakte hain. Har question ka detailed explanation diya gaya hai taaki aap method samajh sako.

Mark the correct alternative in each of the following:
Question 1.
If x – 2 is a factor of x² + 3 ax – 2a, then a =
(a) 2
(b) -2
(c) 1
(d) -1
Solution:
∴ x – 2 is a factor of
f(x) = x² + 3 ax – 2a
∴ Remainder = 0
Let x – 2 = 0, then x = 2
Now f(2) = (2)² + 3a x 2 – 2a
= 4 + 6a – 2a = 4 + 4a
∴ Remainder = 0
∴ 4 + 4a = 0 ⇒ 4a = -4
⇒ a = −44 = -1
∴ a= -1 (d)

Question 2.
If x³ + 6x² + 4x + k is exactly divisible by x + 2, then k =
(a) -6
(b) -7
(c) -8
(d) -10
Solution:
f(x) – x³ + 6x² + 4x + k is divisible by x + 2
∴ Remainder = 0
Let x + 2 = 0, then x = -2
∴ f(-2) = (-2)³ + 6(-2)² + 4(-2) + k
= -8 + 24-8 + k = 8 + k
∴ x + 2 is a factor
∴ Remainder = 0.
⇒ 8 + k= 0 ⇒ k = -8
k = -8 (c)

Question 3.
If x – a is a factor of x³ – 3x² a + 2a²x + b, then the value of b is
(a) 0
(b) 2
(c) 1
(d) 3
Solution:
∴ x – a is a factor of x³ – 3x² a + 2a²x + b
Let f(x) = x³ – 3x² a + 2a²x+ b
and x – a = 0, then x = a
f(a) = a³ – 3a².a + 2a².a + b
= a³ – 3a³ + 2a³ + b = b
∵ x – a is a factor of f(x)
∴ b = 0 (a)

Question 4.
If x¹⁴⁰ + 2x¹⁵¹ + k is divisible by x + 1, then the value of k is
(a) 1
(b) -3
(c) 2
(d) -2
Solution:
∴ x + 1 is a factor of f(x) = x¹⁴⁰ + 2x¹⁵¹ + k
∴ The remainder will be zero
Let x + 1 = 0, then x = -1
∴ f(-1) = (-1)¹⁴⁰ + 2(-1)¹⁵¹ + k
= 1 + 2 x (-1) + k {∵ 140 is even and 151 is odd}
=1-2+k=k-1
∵ Remainder = 0
∴ k – 1=0 ⇒ k=1 (a)

Yahan ek important concept hai ki even powers mein negative numbers positive ho jaate hain, jabki odd powers mein negative hi rehte hain. Is concept ko yaad rakhna bohot zaroori hai MCQs solve karte waqt.

Question 5.
If x + 2 is a factor of x² + mx + 14, then m =
(a) 7
(b) 2
(c) 9
(d) 14
Solution:
x + 2 is a factor of(x) = x² + mx + 14
Let x + 2 = 0, then x = -2
f(-2) = (-2)² + m{-2) + 14
= 4 – 2m + 14 = 18 – 2m
∴ x + 2 is a factor of f(x)
∴ Remainder = 0
⇒ 18 – 2m = 0
2m = 18 ⇒ m = 182 = 9 (c)

Question 6.
If x – 3 is a factor of x² – ax – 15, then a =
(a) -2
(b) 5
(c) -5
(d) 3
Solution:
x – 3 is a factor of(x) = x² – ax – 15
Let x – 3 = 0, then x = 3
∴ f(3) = (3)² – a(3) – 15
= 9 -3a- 15
= -6 -3a
∴ x – 3 is a factor
∴ Remainder = 0
-6 – 3a = 0 ⇒ 3a = -6
∴ a = −63 = -2 (a)

Question 7.
If x⁵¹ + 51 is divided by x + 1, the remainder is
(a) 0
(b) 1
(c) 49
(d) 50
Solution:
Left(x) = x⁵¹ + 51 is divisible by x + 1
Let x+1=0, then x = -1
∴ f(-1) = (-1)⁵¹ + 51 =-1+51 (∵ power 51 is an odd integer)
= 50 (d)

Is question mein odd power ka concept use hua hai. Jab bhi negative number ka odd power hota hai, result negative aata hai. Yeh basic rule hai jo har student ko yaad rakhna chahiye.

Question 8.
If x+ 1 is a factor of the polynomial 2x² + kx, then k =
(a) -2
(b) -3
(c) 4
(d) 2
Solution:
∴ x + 1 is a factor of the polynomial 2x² + kx
Let x+1=0, then x = -1
Now f(x) = 2x² + kx
∴ Remainder =f(-1) = 0
= 2(-1)² + k(-1)
= 2 x 1 + k x (-1) = 2 – k
∴ x + 1 is a factor of f(x)
∴ Remainder = 0
∴ 2 – k = 0 ⇒ k = 2 (d)

Question 9.
If x + a is a factor of x⁴ – a²x² + 3x – 6a, then a =
(a) 0
(b) -1
(c) 1
(d) 2
Solution:
x + a is a factor o f(x) = x⁴– a²x² + 3x – 6a
Let x + a = 0, then x = -a
Now, f(-a) – (-a)⁴ -a²(-a)² + 3 (-a) – 6a
= a⁴-a⁴-3a-6a = -9a
∴ x + a is a factor of f(x)
∴Remainder = 0
∴ -9a = 0 ⇒ a = 0 (a)

Yeh question thoda complex hai kyunki fourth power aur square terms dono involved hain. Dhyan se calculation karni padti hai aise questions mein.

Question 10.
The value of k for which x – 1 is a factor of 4x³ + 3x² – 4x + k, is
(a) 3
(b) 1
(c) -2
(d) -3
Solution:
x- 1 is a factor of f(x) = 4x³ + 3x² – 4x + k
Let x – 1 = 0, then x = 1
f(1) = 4(1 )³ + 3(1)² – 4 x 1 + k
= 4+3-4+k=3+k
∴ x- 1 is a factor of f(x)
∴ Remainder = 0
∴ 3 + k = 0 ⇒ k = -3 (d)

Question 11.
If x+2 and x-1 are the factors of x³+ 10x² + mx + n, then the values of m and n are respectively
(a) 5 and-3
(b) 17 and-8
(c) 7 and-18
(d) 23 and -19
Solution:
x+ 2 and x – 1 are the factors of
f(x) = x³ + 10x² + mx + n
Let x + 2 = 0, then x = -2
∴ f(-2) = (-2)³ + 10(-2)² + m(-2) + n
= -8 + 40 – 2m + n = 32 – 2m + n
∴ x + 2 is a factor of f(x)
∴ Remainder = 0
∴ 32 – 2m + n = 0 ⇒ 2m – n = 32 …(i)
Again x – 1 is a factor of f(x)
Let x-1=0, then x= 1
∴ f(1) = (1)³ + 10(1)² + m x 1 +n
= 1 + 10+ m + n =m + n+11
∴ x- 1 is a factor of f(x)
∴ m + n+ 11=0 ⇒ m+n =-11 …(ii)
Adding (i) and (ii),
3m = 32 – 11 = 21
⇒ m = 213 = 7
and n = -11 – m = -11 – 7 = -18
∴ m= 7, n = -18 (c)

Is question mein do factors diye gaye hain, isliye do equations banaani padi. Simultaneous equations solve karna ek important skill hai mathematics mein.

Question 12.
Let f(x) be a polynomial such that f( −12 )= 0, then a factor of f(x) is
(a) 2x – 1
(b) 2x + 1
(c) x- 1
(d) x + 1
Solution:

Agar f(-1/2) = 0 hai, toh iska matlab hai ki (x + 1/2) ek factor hai, jo ki (2x + 1) ke equivalent hai. Yeh factor theorem ka direct application hai.

Question 13.
When x³ – 2x² + ax – b is divided by x² – 2x-3, the remainder is x – 6. The value of a and b are respectively.
(a) -2, -6
(b) 2 and -6
(c) -2 and 6
(d) 2 and 6
Solution:
Let f(x) = x³ – 2x² + ax – b
and Dividing f(x) by x² – 2x + 3
Remainder = x – 6
Let p(x) = x³ – 2x² + ax – b – (x – 6) or x³ – 2x² + x(a – 1) – b + 6 is divisible by x² – 2x+ 3 exactly
Now, x²-2x-3 = x²-3x + x- 3
= x(x – 3) + 1(x – 3)
= (x – 3) (x + 1)
∴ x – 3 and x + 1 are the factors of p(x)
Let x – 3 = 0, then x = 3
∴ p(3) = (3)³ – 2(3)² + (a-1)x3-b + 6
= 27-18 + 3a-3-b + 6
= 33-21+3 a-b
= 12 + 3 a-b
∴ x – 3 is a factor
∴ 12 + 3a-b = 0 ⇒ 3a-b = -12 …(i)
Again let x + 1 = 0, then x = -1
∴p(-1) = (-1)³ – 2(-l)² + (a – 1) X (-1) – b + 6
= -1-2 + 1 – a – 6 + 6
= 4 – a- b
∴ x + 1 is a factor
4-a-b = 0 ⇒ a + b = 4 …(ii)
Adding (i) and (ii),
4 a = -12 + 4 = -8 ⇒ a = −84 = -2
From (ii),
and -2 + b = 4⇒ b = 4 + 2 = 6
∴ a = -2, b = 6 (c)

Yeh question polynomial division ke concept par based hai. Remainder theorem ka use karke hum unknown coefficients find kar sakte hain.

Question 14.
One factor of x⁴ + x² – 20 is x² + 5, the other is
(a) x² – 4
(b) x – 4
(c) x²-5
(d) x + 4
Solution:

Agar x⁴ + x² – 20 = (x² + 5)(x² – 4) hai, toh dusra factor x² – 4 hoga. Yeh simple division se pata chal jaata hai.

Question 15.
If (x – 1) is a factor of polynomial fix) but not of g(x), then it must be a factor of
(a) f(x) g(x)
(b) -f(x) + g(x)
(c) f(x) – g(x)
(d) {f(x) + g(x)}g(x)
Solution:
∴ (x – 1) is a factor of a polynomial f(x)
But not of a polynomial g(x)
∴ (x – 1) will be the factor of the product of f(x) and g(x) (a)

Factor properties ke according, agar ek factor kisi polynomial ka hai toh woh uske product mein bhi factor hoga.

Question 16.
(x + 1) is a factor of xⁿ + 1 only if
(a) n is an odd integer
(b) n is an even integer
(c) n is a negative integer
(d) n is a positive integer
Solution:
∴ (x + 1) is a factor of xⁿ+ 1
Let x + 1 = 0, then x = -1
∴ f(x) = xⁿ + 1
and f(-1) = (-1)ⁿ + 1
But (-1)ⁿ is positive if n is an even integer and negative if n is an odd integer and (-1)ⁿ +1=0 {∵ x + 1 is a factor of(x)}
(-1)ⁿ must be negative
∴ n is an odd integer (a)

Yahan even aur odd powers ka concept use hua hai. Odd powers mein negative sign preserve hota hai.

Question 17.
If x² + x + 1 is a factor of the polynomial 3x³ + 8x² + 8x + 3 + 5k, then the value of k is
(a) 0
(b) 25
(c) 52
(d) -1
Solution:
x² + x + 1 is a factor of
f(x) = 3x³ + 8x² + 8x + 3 + 5k
Now dividing by x² + x + 1, we get

Is question mein polynomial long division use karna pada hai kyunki quadratic factor diya gaya hai.

Question 18.
If (3x – 1)⁷ = a₇x_⁷ + a₆x⁶ + a₅x⁵ + … + a₁x + a₀, then a₇ + a₆ + a₅ + … + a₁ + a₀ =
(a) 0
(b) 1
(c) 128
(d) 64
Solution:
f(x) = [3(1) – 1]⁷ = a₇x⁷ + a₆x⁶ + a₅x⁵ + … + a₁x + a₀
Let x = 1, then
f(1) = (3x- 1)⁷ = a₇(1)⁷ + a₆(1)⁶ + a₅(1)⁵ + … + a₁ x 1 + a₀
⇒ (3 – 1)⁷ = a₇ x 1 + a₆ x 1+ a₅ x 1 + … + a₁x 1 +a₀
⇒ (2)⁷ = a₇ + a₆ + a₅ + … + a₁ +a₀
∴ a₇ + a₆ + a₅ + … + a₁ + a₀= 128 (c)

Yeh binomial expansion ka interesting application hai. x = 1 substitute karke saare coefficients ka sum mil jaata hai.

Question 19.
If both x – 2 and x – 12 are factors of px² + 5x + r, then
(a) p = r
(b) p + r = 0
(c) 2p + r = 0
(d) p + 2r = 0
Solution:

Do factors diye gaye hain, isliye do equations banegi aur unhe solve karke p aur r ke beech relation find karna hai.

Question 20.
If x² – 1 is a factor of ax⁴ + bx³ + cx² + dx + e, then
(a) a + c + e- b + d
(b)a + b + e = c + d
(c)a + b + c = d+ e
(d)b + c + d= a + e
Solution:
X² – 1 is a factor of ax⁴ + bx³ + cx² + dx + e
⇒ (x + 1), (x – 1) are the factors of ax⁴ + bx³ + cx² + dx + e
Let f(x) = ax⁴ + bx³ + cx² + dx + e
and x + 1 = 0 then x = -1
∴ f(-1) = a(-1)⁴ + b(-1)³ + c(-1)² + d(-1) + e
= a- b + c- d+ e
∴ x + 1 is a factor of f(x)
∴ a-b + c- d+e = 0
⇒ a + c + e = b + d (a)

x² – 1 = (x+1)(x-1) hai, toh dono factors separately use karne padte hain. Yeh difference of squares ka concept hai.

RD Sharma Class 9 Solutions Chapter 6 MCQs complete kar chuke hain. Ye saare questions CBSE Class 9 Maths exam ke liye bohot important hain. Factorization ke concepts ko strong banane ke liye regular practice zaroori hai.

In MCQs mein jo main concepts cover hue hain woh hain:

• Factor Theorem: Agar (x-a) polynomial f(x) ka factor hai toh f(a) = 0 hoga
• Remainder Theorem: Polynomial ko (x-a) se divide karne par remainder f(a) hota hai
• Even aur Odd Powers: Negative numbers ke even powers positive, odd powers negative hote hain
• Polynomial Division: Long division method se quotient aur remainder find karna
• Multiple Factors: Jab do ya zyada factors diye hon toh simultaneous equations solve karna

Ye concepts aage ke chapters mein bhi use hote hain, especially quadratic equations aur coordinate geometry mein. Practice ke liye aur RD Sharma Class 9 Solutions Chapter 6 ke exercises bhi solve kar sakte hain.

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