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.

**What is a polynomial?**

Let x be a variable, n be a positive integer and a_0, a_1, a_2,…., a_n be constants. Then a_nx_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.

**Types of Polynomials:**

Monomial: A polynomial containing only one term.

Binomial: A polynomial containing two terms.

Trinomial: A polynomial containing three terms.

Students can download RD Sharma Solution for class 9 all chapters enlisted in the textbook.

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

### Access Answers to Maths RD Sharma 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 – 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**

**(v) 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 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
- Remainder Theorem
- Factorisation of polynomials by using the factor theorem