## RD Sharma Solutions for Class 9 Maths Chapter 6 – Free PDF Download

Get free **RD Sharma Solutions for Class 9 Chapter 6** Factorization of Polynomials here. In this chapter, students will learn various terms such as polynomial, degree of polynomial, factors, multiples and zeros of a polynomial. Students might be wondering about What is a polynomial? So here is the answer for your question. Let x be a variable, n be a positive integer and a_{0}, a_{1}, a_{2},…., a_{n} be constants. Then a_{n}x_{n} + a_{(n-1)}x^{(n-1)} + a_{(n-2)}x^{(n-2)} + ….+ a_{1} x + a_{0} is known as polynomial in variable x. The coefficient a_n of the highest degree term is called the leading coefficient and a_{0} is called the constant term. Students can download RD Sharma Solution for Class 9 all chapters enlisted in the textbook.

The PDF of this exercise can be easily downloaded from the below links. BYJU’S subject matter experts have uniquely solved the exercise for your better understanding. By practising RD Sharma Solutions for Class 9, students can fulfil their wish by securing good marks in board exams.

## Download PDF of RD Sharma Solutions for Class 9 Maths Chapter 6 Factorization of Polynomials

### Access Answers to Maths RD Sharma Solutions for Class 9 Chapter 6 Factorization of Polynomials

### Exercise 6.1 Page No: 6.2

**Question 1: Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer:**

**(i) 3x ^{2} – 4x + 15**

**(ii) y ^{2} + 2√3**

**(iii) 3√x + √2x**

**(iv) x – 4/x**

**(v) x ^{12} + y^{3} + t^{50}**

**Solution:**

**(i)** 3x^{2} – 4x + 15

It is a polynomial of x.

**(ii)** y^{2} + 2√3

It is a polynomial of y.

**(iii)** 3√x + √2x

It is not a polynomial since the exponent of 3√x is a rational term.

**(iv)** x – 4/x

It is not a polynomial since the exponent of – 4/x is not a positive term.

**(v)** x^{12} + y^{3} + t^{50}

It is a three variable polynomial, x, y and t.

**Question 2: Write the coefficient of x ^{2} in each of the following:**

**(i) 17 – 2x + 7x ^{2}**

**(ii) 9 – 12x + x ^{3}**

**(iii) ∏/6 x ^{2} – 3x + 4**

**(iv) √3x – 7**

**Solution:**

**(i)** 17 – 2x + 7x^{2}

Coefficient of x^{2} = 7

**(ii)** 9 – 12x + x^{3}

Coefficient of x^{2 }=0

**(iii)** ∏/6 x^{2} – 3x + 4

Coefficient of x^{2 }= ∏/6

**(iv)** √3x – 7

Coefficient of x^{2 }= 0

**Question 3: Write the degrees of each of the following polynomials:**

**(i) 7x ^{3} + 4x^{2} – 3x + 12**

**(ii) 12 – x + 2x ^{3}**

**(iii) 5y – √2**

**(iv) 7**

**(v) 0**

**Solution**:

As we know, degree is the highest power in the polynomial

**(i)** Degree of the polynomial 7x^{3} + 4x^{2} – 3x + 12 is 3

**(ii)** Degree of the polynomial 12 – x + 2x^{3} is 3

**(iii)** Degree of the polynomial 5y – √2 is 1

**(iv)** Degree of the polynomial 7 is 0

**(v)** Degree of the polynomial 0 is undefined.

**Question 4: Classify the following polynomials as linear, quadratic, cubic and biquadratic polynomials:**

**(i) x + x ^{2} + 4**

**(ii) 3x – 2**

**(iii) 2x + x ^{2}**

**(iv) 3y**

**(v) t ^{2} + 1**

**(vi) 7t ^{4} + 4t^{3} + 3t – 2**

**Solution:**

**(i)** x + x^{2} + 4: It is a quadratic polynomial as its degree is 2.

**(ii)** 3x – 2 : It is a linear polynomial as its degree is 1.

**(iii)** 2x + x^{2}: It is a quadratic polynomial as its degree is 2.

**(iv)** 3y: It is a linear polynomial as its degree is 1.

**(v)** t^{2}+ 1: It is a quadratic polynomial as its degree is 2.

**(vi)** 7t^{4} + 4t^{3} + 3t – 2: It is a biquadratic polynomial as its degree is 4.

### Exercise 6.2 Page No: 6.8

**Question 1: If f(x) = 2x ^{3} – 13x^{2} + 17x + 12, find**

**(i) f (2)**

**(ii) f (-3)**

**(iii) f(0)**

**Solution:**

f(x) = 2x^{3} – 13x^{2} + 17x + 12

**(i)** f(2) = 2(2)^{3} – 13(2)^{ 2} + 17(2) + 12

= 2 x 8 – 13 x 4 + 17 x 2 + 12

= 16 – 52 + 34 + 12

= 62 – 52

= 10

**(ii)** f(-3) = 2(-3)^{3} – 13(-3)^{ 2} + 17 x (-3) + 12

= 2 x (-27) – 13 x 9 + 17 x (-3) + 12

= -54 – 117 -51 + 12

= -222 + 12

= -210

**(iii)** f(0) = 2 x (0)^{3} – 13(0)^{ 2} + 17 x 0 + 12

= 0-0 + 0+ 12

= 12

**Question 2: Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following cases:**

**(i) f(x) = 3x + 1, x = −1/3**

**(ii) f(x) = x ^{2} – 1, x = 1,−1**

**(iii) g(x) = 3x ^{2} – 2 , x = 2/√3 , −2/√3 **

**(iv) p(x) = x ^{3} – 6x^{2} + 11x – 6 , x = 1, 2, 3**

**(v) f(x) = 5x – π, x = 4/5**

**(vi) f(x) = x ^{2} , x = 0**

**(vii) f(x) = lx + m, x = −m/l**

**(viii) f(x) = 2x + 1, x = 1/2**

**Solution:**

**(i)** f(x) = 3x + 1, x = −1/3

f(x) = 3x + 1

Substitute x = −1/3 in f(x)

f( −1/3) = 3(−1/3) + 1

= -1 + 1

= 0

Since, the result is 0, so x = −1/3 is the root of 3x + 1

**(ii)** f(x) = x^{2} – 1, x = 1,−1

f(x) = x^{2} – 1

Given that x = (1 , -1)

Substitute x = 1 in f(x)

f(1) = 1^{2} – 1

= 1 – 1

= 0

Now, substitute x = (-1) in f(x)

f(-1) = (−1)^{2} – 1

= 1 – 1

= 0

Since , the results when x = 1 and x = -1 are 0, so (1 , -1) are the roots of the polynomial f(x) = x^{2 }– 1

**(iii)** g(x) = 3x^{2} – 2 , x = 2/√3 , −2/√3

g(x) = 3x^{2} – 2

Substitute x = 2/√3 in g(x)

g(2/√3) = 3(2/√3)^{2} – 2

= 3(4/3) – 2

= 4 – 2

= 2 ≠ 0

Now, Substitute x = −2/√3 in g(x)

g(2/√3) = 3(-2/√3)^{2} – 2

= 3(4/3) – 2

= 4 – 2

= 2 ≠ 0

Since, the results when x = 2/√3 and x = −2/√3) are not 0. Therefore (2/√3 , −2/√3 ) are not zeros of 3x^{2}–2.

**(iv)** p(x) = x^{3} – 6x^{2} + 11x – 6 , x = 1, 2, 3

p(1) = 1^{3} – 6(1)^{2} + 11x 1 – 6 = 1 – 6 + 11 – 6 = 0

p(2) = 2^{3} – 6(2)^{2} + 11×2 – 6 = 8 – 24 + 22 – 6 = 0

p(3) = 3^{3} – 6(3)^{2} + 11×3 – 6 = 27 – 54 + 33 – 6 = 0

Therefore, x = 1, 2, 3 are zeros of p(x).

**(v)** f(x) = 5x – π, x = 4/5

f(4/5) = 5 x 4/5 – π = 4 – π ≠ 0

Therefore, x = 4/5 is not a zeros of f(x).

**(vi)** f(x) = x^{2} , x = 0

f(0) = 0^{2} = 0

Therefore, x = 0 is a zero of f(x).

**(vii)** f(x) = lx + m, x = −m/l

f(−m/l) = l x −m/l + m = -m + m = 0

Therefore, x = −m/l is a zero of f(x).

**(viii)** f(x) = 2x + 1, x = ½

f(1/2) = 2x 1/2 + 1 = 1 + 1 = 2 ≠ 0

Therefore, x = ½ is not a zero of f(x).

### Exercise 6.3 Page No: 6.14

**In each of the following, using the remainder theorem, find the remainder when f(x) is divided by g(x) and verify the by actual division : (1 – 8)**

**Question 1: f(x) = x ^{3} + 4x^{2} – 3x + 10, g(x) = x + 4**

**Solution:**

f(x) = x^{3} + 4x^{2} – 3x + 10, g(x) = x + 4

Put g(x) =0

⇒ x + 4 = 0 or x = -4

Remainder = f(-4)

Now,

f(-4) = (-4)^{3} + 4(-4)^{2} – 3(-4) + 10 = -64 + 64 + 12 + 10 = 22

**Actual Division: **

**Question 2: f(x) = 4x ^{4} – 3x^{3} – 2x^{2} + x – 7, g(x) = x – 1**

**Solution:**

f(x) = 4x^{4} – 3x^{3} – 2x^{2} + x – 7

Put g(x) =0

⇒ x – 1 = 0 or x = 1

Remainder = f(1)

Now,

f(1) = 4(1)^{4} – 3(1)^{3} – 2(1)^{2} + (1) – 7 = 4 – 3 – 2 + 1 – 7 = -7

**Actual Division: **

**Question 3: f(x) = 2x ^{4} – 6X^{3} + 2x^{2} – x + 2, g(x) = x + 2**

**Solution:**

f(x) = 2x^{4} – 6X^{3} + 2x^{2} – x + 2, g(x) = x + 2

Put g(x) = 0

⇒ x + 2 = 0 or x = -2

Remainder = f(-2)

Now,

f(-2) = 2(-2)^{4} – 6(-2)^{3} + 2(-2)^{2} – (-2) + 2 = 32 + 48 + 8 + 2 + 2 = 92

**Actual Division: **

**Question 4: f(x) = 4x ^{3} – 12x^{2} + 14x – 3, g(x) = 2x – 1**

**Solution:**

f(x) = 4x^{3} – 12x^{2} + 14x – 3, g(x) = 2x – 1

Put g(x) =0

⇒ 2x -1 =0 or x = 1/2

Remainder = f(1/2)

Now,

f(1/2) = 4(1/2)^{3} – 12(1/2)^{2} + 14(1/2) – 3 = ½ – 3 + 7 – 3 = 3/2

**Actual Division: **

**Question 5: f(x) = x ^{3} – 6x^{2} + 2x – 4, g(x) = 1 – 2x**

**Solution:**

f(x) = x^{3} – 6x^{2} + 2x – 4, g(x) = 1 – 2x

Put g(x) = 0

⇒ 1 – 2x = 0 or x = 1/2

Remainder = f(1/2)

Now,

f(1/2) = (1/2)^{3} – 6(1/2)^{2} + 2(1/2) – 4 = 1 + 1/8 – 4 – 3/2 = -35/8

**Actual Division: **

**Question 6: f(x) = x ^{4} – 3x^{2} + 4, g(x) = x – 2**

**Solution:**

f(x) = x^{4} – 3x^{2} + 4, g(x) = x – 2

Put g(x) = 0

⇒ x – 2 = 0 or x = 2

Remainder = f(2)

Now,

f(2) = (2)^{4} – 3(2)^{2} + 4 = 16 – 12 + 4 = 8

**Actual Division: **

**Question 7: f(x) = 9x ^{3} – 3x^{2} + x – 5, g(x) = x – 2/3**

**Solution:**

f(x) = 9x^{3} – 3x^{2} + x – 5, g(x) = x – 2/3

Put g(x) = 0

⇒ x – 2/3 = 0 or x = 2/3

Remainder = f(2/3)

Now,

f(2/3) = 9(2/3)^{3} – 3(2/3)^{2} + (2/3) – 5 = 8/3 – 4/3 + 2/3 – 5/1 = -3

**Actual Division: **

### Exercise 6.4 Page No: 6.24

**In each of the following, use factor theorem to find whether polynomial g(x) is a factor of polynomial f(x) or, not: (1-7)**

**Question 1: f(x) = x ^{3} – 6x^{2} + 11x – 6; g(x) = x – 3**

**Solution:**

If g(x) is a factor of f(x), then the remainder will be zero that is g(x) = 0.

g(x) = x -3 = 0

or x = 3

Remainder = f(3)

Now,

f(3) = (3)^{3} – 6(3)^{2} +11 x 3 – 6

= 27 – 54 + 33 – 6

= 60 – 60

= 0

Therefore, g(x) is a factor of f(x)

**Question 2: f(x) = 3X ^{4} + 17x^{3} + 9x^{2} – 7x – 10; g(x) = x + 5**

**Solution:**

If g(x) is a factor of f(x), then the remainder will be zero that is g(x) = 0.

g(x) = x + 5 = 0, then x = -5

Remainder = f(-5)

Now,

f(3) = 3(-5)^{4} + 17(-5)^{3} + 9(-5)^{2} – 7(-5) – 10

= 3 x 625 + 17 x (-125) + 9 x (25) – 7 x (-5) – 10

= 1875 -2125 + 225 + 35 – 10

= 0

Therefore, g(x) is a factor of f(x).

**Question 3: f(x) = x ^{5} + 3x^{4} – x^{3} – 3x^{2} + 5x + 15, g(x) = x + 3**

**Solution**:

If g(x) is a factor of f(x), then the remainder will be zero that is g(x) = 0.

g(x) = x + 3 = 0, then x = -3

Remainder = f(-3)

Now,

f(-3) = (-3)^{5} + 3(-3)^{4} – (-3)^{3} – 3(-3)^{2} + 5(-3) + 15

= -243 + 3 x 81 -(-27)-3 x 9 + 5(-3) + 15

= -243 +243 + 27-27- 15 + 15

= 0

Therefore, g(x) is a factor of f(x).

**Question 4: f(x) = x ^{3} – 6x^{2} – 19x + 84, g(x) = x – 7**

**Solution**:

If g(x) is a factor of f(x), then the remainder will be zero that is g(x) = 0.

g(x) = x – 7 = 0, then x = 7

Remainder = f(7)

Now,

f(7) = (7)^{3} – 6(7)^{2} – 19 x 7 + 84

= 343 – 294 – 133 + 84

= 343 + 84 – 294 – 133

= 0

Therefore, g(x) is a factor of f(x).

**Question 5: f(x) = 3x ^{3} + x^{2} – 20x + 12, g(x) = 3x – 2**

**Solution:**

If g(x) is a factor of f(x), then the remainder will be zero that is g(x) = 0.

g(x) = 3x – 2 = 0, then x = 2/3

Remainder = f(2/3)

Now,

f(2/3) = 3(2/3) ^{3} + (2/3) ^{2} – 20(2/3) + 12

= 3 x 8/27 + 4/9 – 40/3 + 12

= 8/9 + 4/9 – 40/3 + 12

= 0/9

= 0

Therefore, g(x) is a factor of f(x).

**Question 6: f(x) = 2x ^{3} – 9x^{2} + x + 12, g(x) = 3 – 2x**

**Solution:**

If g(x) is a factor of f(x), then the remainder will be zero that is g(x) = 0.

g(x) = 3 – 2x = 0, then x = 3/2

Remainder = f(3/2)

Now,

f(3/2) = 2(3/2)^{3} – 9(3/2)^{2} + (3/2) + 12

= 2 x 27/8 – 9 x 9/4 + 3/2 + 12

= 27/4 – 81/4 + 3/2 + 12

= 0/4

= 0

Therefore, g(x) is a factor of f(x).

**Question 7: f(x) = x ^{3} – 6x^{2} + 11x – 6, g(x) = x^{2} – 3x + 2**

**Solution:**

If g(x) is a factor of f(x), then the remainder will be zero that is g(x) = 0.

g(x) = 0

or x^{2} – 3x + 2 = 0

x^{2} – x – 2x + 2 = 0

x(x – 1) – 2(x – 1) = 0

(x – 1) (x – 2) = 0

Therefore x = 1 or x = 2

Now,

f(1) = (1)^{3} – 6(1)^{2} + 11(1) – 6 = 1-6+11-6= 12- 12 = 0

f(2) = (2)^{3} – 6(2)^{2} + 11(2) – 6 = 8 – 24 + 22 – 6 = 30 – 30 = 0

⇒ f(1) = 0 and f(2) = 0

Which implies g(x) is factor of f(x).

**Question 8: Show that (x – 2), (x + 3) and (x – 4) are factors of x ^{3} – 3x^{2} – 10x + 24.**

**Solution:**

Let f(x) = x^{3} – 3x^{2} – 10x + 24

If x – 2 = 0, then x = 2,

If x + 3 = 0 then x = -3,

and If x – 4 = 0 then x = 4

Now,

f(2) = (2)^{3} – 3(2)^{2} – 10 x 2 + 24 = 8 – 12 – 20 + 24 = 32 – 32 = 0

f(-3) = (-3)^{3} – 3(-3)^{2} – 10 (-3) + 24 = -27 -27 + 30 + 24 = -54 + 54 = 0

f(4) = (4)^{3} – 3(4)^{2} – 10 x 4 + 24 = 64-48 -40 + 24 = 88 – 88 = 0

f(2) = 0

f(-3) = 0

f(4) = 0

Hence (x – 2), (x + 3) and (x – 4) are the factors of f(x)

**Question 9: Show that (x + 4), (x – 3) and (x – 7) are factors of x ^{3} – 6x^{2} – 19x + 84.**

**Solution:**

Let f(x) = x^{3} – 6x^{2} – 19x + 84

If x + 4 = 0, then x = -4

If x – 3 = 0, then x = 3

and if x – 7 = 0, then x = 7

Now,

f(-4) = (-4)^{3} – 6(-4)^{2} – 19(-4) + 84 = -64 – 96 + 76 + 84 = 160 – 160 = 0

f(-4) = 0

f(3) = (3)^{ 3} – 6(3)^{ 2} – 19 x 3 + 84 = 27 – 54 – 57 + 84 = 111 -111=0

f(3) = 0

f(7) = (7)^{ 3} – 6(7)^{ 2} – 19 x 7 + 84 = 343 – 294 – 133 + 84 = 427 – 427 = 0

f(7) = 0

Hence (x + 4), (x – 3), (x – 7) are the factors of f(x).

### Exercise 6.5 Page No: 6.32

**Using factor theorem, factorize each of the following polynomials:**

**Question 1: x ^{3} + 6x^{2} + 11x + 6**

**Solution**:

Let f(x) = x^{3} + 6x^{2} + 11x + 6

Step 1: Find the factors of constant term

Here constant term = 6

Factors of 6 are ±1, ±2, ±3, ±6

Step 2: Find the factors of f(x)

Let x + 1 = 0

⇒ x = -1

Put the value of x in f(x)

f(-1) = (−1)^{3} + 6(−1)^{2} + 11(−1) + 6

= -1 + 6 -11 + 6

= 12 – 12

= 0

So, (x + 1) is the factor of f(x)

Let x + 2 = 0

⇒ x = -2

Put the value of x in f(x)

f(-2) = (−2)^{3} + 6(−2)^{2} + 11(−2) + 6 = -8 + 24 – 22 + 6 = 0

So, (x + 2) is the factor of f(x)

Let x + 3 = 0

⇒ x = -3

Put the value of x in f(x)

f(-3) = (−3)^{3} + 6(−3)^{2} + 11(−3) + 6 = -27 + 54 – 33 + 6 = 0

So, (x + 3) is the factor of f(x)

Hence, f(x) = (x + 1)(x + 2)(x + 3)

**Question 2: x ^{3} + 2x^{2} – x – 2**

**Solution**:

Let f(x) = x^{3} + 2x^{2} – x – 2

Constant term = -2

Factors of -2 are ±1, ±2

Let x – 1 = 0

⇒ x = 1

Put the value of x in f(x)

f(1) = (1)^{3} + 2(1)^{2} – 1 – 2 = 1 + 2 – 1 – 2 = 0

So, (x – 1) is factor of f(x)

Let x + 1 = 0

⇒ x = -1

Put the value of x in f(x)

f(-1) = (-1)^{3} + 2(-1)^{2} – 1 – 2 = -1 + 2 + 1 – 2 = 0

(x + 1) is a factor of f(x)

Let x + 2 = 0

⇒ x = -2

Put the value of x in f(x)

f(-2) = (-2)^{3} + 2(-2)^{2} – (-2) – 2 = -8 + 8 + 2 – 2 = 0

(x + 2) is a factor of f(x)

Let x – 2 = 0

⇒ x = 2

Put the value of x in f(x)

f(2) = (2)^{3} + 2(2)^{2} – 2 – 2 = 8 + 8 – 2 – 2 = 12 ≠ 0

(x – 2) is not a factor of f(x)

Hence f(x) = (x + 1)(x- 1)(x+2)

**Question 3: x ^{3} – 6x^{2} + 3x + 10**

**Solution**:

Let f(x) = x^{3} – 6x^{2} + 3x + 10

Constant term = 10

Factors of 10 are ±1, ±2, ±5, ±10

Let x + 1 = 0 or x = -1

f(-1) = (-1)^{3} – 6(-1)^{2} + 3(-1) + 10 = 10 – 10 = 0

f(-1) = 0

Let x + 2 = 0 or x = -2

f(-2) = (-2)^{3} – 6(-2)^{2} + 3(-2) + 10 = -8 – 24 – 6 + 10 = -28

f(-2) ≠ 0

Let x – 2 = 0 or x = 2

f(2) = (2)^{3} – 6(2)^{2} + 3(2) + 10 = 8 – 24 + 6 + 10 = 0

f(2) = 0

Let x – 5 = 0 or x = 5

f(5) = (5)^{3} – 6(5)^{2} + 3(5) + 10 = 125 – 150 + 15 + 10 = 0

f(5) = 0

Therefore, (x + 1), (x – 2) and (x-5) are factors of f(x)

Hence f(x) = (x + 1) (x – 2) (x-5)

**Question 4: x ^{4} – 7x^{3} + 9x^{2} + 7x- 10**

**Solution:**

Let f(x) = x^{4} – 7x^{3} + 9x^{2} + 7x- 10

Constant term = -10

Factors of -10 are ±1, ±2, ±5, ±10

Let x – 1 = 0 or x = 1

f(1) = (1)^{4} – 7(1)^{3} + 9(1)^{2} + 7(1) – 10 = 1 – 7 + 9 + 7 -10 = 0

f(1) = 0

Let x + 1 = 0 or x = -1

f(-1) = (-1)^{4} – 7(-1)^{3} + 9(-1)^{2} + 7(-1) – 10 = 1 + 7 + 9 – 7 -10 = 0

f(-1) = 0

Let x – 2 = 0 or x = 2

f(2) = (2)^{4} – 7(2)^{3} + 9(2)^{2} + 7(2) – 10 = 16 – 56 + 36 + 14 – 10 = 0

f(2) = 0

Let x – 5 = 0 or x = 5

f(5) = (5)^{4} – 7(5)^{3} + 9(5)^{2} + 7(5) – 10 = 625 – 875 + 225 + 35 – 10 = 0

f(5) = 0

Therefore, (x – 1), (x + 1), (x – 2) and (x-5) are factors of f(x)

Hence f(x) = (x – 1) (x + 1) (x – 2) (x-5)

**Question 5: x ^{4} – 2x^{3} – 7x^{2} + 8x + 12**

**Solution**:

f(x) = x^{4} – 2x^{3} – 7x^{2} + 8x + 12

Constant term = 12

Factors of 12 are ±1, ±2, ±3, ±4, ±6, ±12

Let x – 1 = 0 or x = 1

f(1) = (1)^{4} – 2(1)^{3} – 7(1)^{2} + 8(1) + 12 = 1 – 2 – 7 + 8 + 12 = 12

f(1) ≠ 0

Let x + 1 = 0 or x = -1

f(-1) = (-1)^{4} – 2(-1)^{3} – 7(-1)^{2} + 8(-1) + 12 = 1 + 2 – 7 – 8 + 12 = 0

f(-1) = 0

Let x +2 = 0 or x = -2

f(-2) = (-2)^{4} – 2(-2)^{3} – 7(-2)^{2} + 8(-2) + 12 = 16 + 16 – 28 – 16 + 12 = 0

f(-2) = 0

Let x – 2 = 0 or x = 2

f(2) = (2)^{4} – 2(2)^{3} – 7(2)^{2} + 8(2) + 12 = 16 – 16 – 28 + 16 + 12 = 0

f(2) = 0

Let x – 3 = 0 or x = 3

f(3) = (3)^{4} – 2(3)^{3} – 7(3)^{2} + 8(3) + 12 = 0

f(3) = 0

Therefore, (x + 1), (x + 2), (x – 2) and (x-3) are factors of f(x)

Hence f(x) = (x + 1)(x + 2) (x – 2) (x-3)

**Question 6: x ^{4} + 10x^{3} + 35x^{2} + 50x + 24**

**Solution**:

Let f(x) = x^{4} + 10x^{3} + 35x^{2} + 50x + 24

Constant term = 24

Factors of 24 are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24

Let x + 1 = 0 or x = -1

f(-1) = (-1)^{4} + 10(-1)^{3} + 35(-1)^{2} + 50(-1) + 24 = 1 – 10 + 35 – 50 + 24 = 0

f(1) = 0

(x + 1) is a factor of f(x)

Likewise, (x + 2),(x + 3),(x + 4) are also the factors of f(x)

Hence f(x) = (x + 1) (x + 2)(x + 3)(x + 4)

**Question 7: 2x ^{4} – 7x^{3} – 13x^{2} + 63x – 45**

**Solution:**

Let f(x) = 2x^{4} – 7x^{3} – 13x^{2} + 63x – 45

Constant term = -45

Factors of -45 are ±1, ±3, ±5, ±9, ±15, ±45

Here coefficient of x^4 is 2. So possible rational roots of f(x) are

±1, ±3, ±5, ±9, ±15, ±45, ±1/2,±3/2,±5/2,±9/2,±15/2,±45/2

Let x – 1 = 0 or x = 1

f(1) = 2(1)^{4} – 7(1)^{3} – 13(1)^{2} + 63(1) – 45 = 2 – 7 – 13 + 63 – 45 = 0

f(1) = 0

f(x) can be written as,

f(x) = (x-1) (2x^{3} – 5x^{2} -18x +45)

or f(x) =(x-1)g(x) …(1)

Let x – 3 = 0 or x = 3

f(3) = 2(3)^{4} – 7(3)^{3} – 13(3)^{2} + 63(3) – 45 = = 162 – 189 – 117 + 189 – 45= 0

f(3) = 0

Now, we are available with 2 factors of f(x), (x – 1) and (x – 3)

Here g(x) = 2x^{2} (x-3) + x(x-3) -15(x-3)

Taking (x-3) as common

= (x-3)(2x^{2} + x – 15)

= (x-3)(2x^{2}+6x – 5x -15)

= (x-3)(2x-5)(x+3)

= (x-3)(x+3)(2x-5) ….(2)

From (1) and (2)

f(x) =(x-1) (x-3)(x+3)(2x-5)

### Exercise VSAQs Page No: 6.33

**Question 1: Define zero or root of a polynomial**

**Solution**:

zero or root, is a solution to the polynomial equation, f(y) = 0.

It is that value of y that makes the polynomial equal to zero.

**Question 2: If x = 1/2 is a zero of the polynomial f(x) = 8x ^{3} + ax^{2} – 4x + 2, find the value of a.**

**Solution: **

If x = 1/2 is a zero of the polynomial f(x), then f(1/2) = 0

8(1/2)^{3} + a(1/2)^{2} – 4(1/2) + 2 = 0

8 x 1/8 + a/4 – 2 + 2 = 0

1 + a/4 = 0

a = -4

**Question 3: Write the remainder when the polynomial f(x) = x ^{3} + x^{2} – 3x + 2 is divided by x + 1.**

**Solution: **

Using factor theorem,

Put x + 1 = 0 or x = -1

f(-1) is the remainder.

Now,

f(-1) = (-1)^{3} + (-1)^{2} – 3(-1) + 2

= -1 + 1 + 3 + 2

= 5

Therefore 5 is the remainder.

**Question 4: Find the remainder when x ^{3} + 4x^{2} + 4x-3 if divided by x**

**Solution: **

Using factor theorem,

Put x = 0

f(0) is the remainder.

Now,

f(0) = 0^{3} + 4(0)^{2} + 4×0 -3 = -3

Therefore -3 is the remainder.

**Question 5: **If x+1 is a factor of x^{3} + a, then write the value of a.

**Solution: **

Let f(x) = x^{3} + a

If x+1 is a factor of x^{3} + a then f(-1) = 0

(-1)^{3} + a = 0

-1 + a = 0

or a = 1

**Question 6: If f(x) = x^4 – 2x ^{3} + 3x^{2} – ax – b when divided by x – 1, the remainder is 6, then find the value of a+b.**

Solution:

From the statement, we have f(1) = 6

(1)^4 – 2(1)^{3} + 3(1)^{2} – a(1) – b = 6

1 – 2 + 3 – a – b = 6

2 – a – b = 6

a + b = -4

### RD Sharma Solutions for Class 9 Maths Chapter 6 Factorization of Polynomials

In the 6th chapter of Class 9 RD Sharma Solutions students will study important concepts on Factorization of Polynomials as listed below:

- Factorization of Polynomials introduction
- Terms and coefficients
- Degree of a polynomial
- Types of polynomials

- Monomial: A polynomial containing only one term.
- Binomial: A polynomial containing two terms.
- Trinomial: A polynomial containing three terms.

- Remainder Theorem
- Factorisation of polynomials by using the factor theorem