RD Sharma Solutions Class 9 Factorization Of Polynomials

Get free RD Sharma Solutions for Class 9 Maths 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 all chapters solutions enlisted in RD Sharma class 9 textbook.

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RD Sharma Solution class 9 Maths Chapter 6 Factorization of Polynomials 01
RD Sharma Solution class 9 Maths Chapter 6 Factorization of Polynomials 02
RD Sharma Solution class 9 Maths Chapter 6 Factorization of Polynomials 03
RD Sharma Solution class 9 Maths Chapter 6 Factorization of Polynomials 04
RD Sharma Solution class 9 Maths Chapter 6 Factorization of Polynomials 05
RD Sharma Solution class 9 Maths Chapter 6 Factorization of Polynomials 06
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RD Sharma Solution class 9 Maths Chapter 6 Factorization of Polynomials 08
RD Sharma Solution class 9 Maths Chapter 6 Factorization of Polynomials 09
RD Sharma Solution class 9 Maths Chapter 6 Factorization of Polynomials 10
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RD Sharma Solution class 9 Maths Chapter 6 Factorization of Polynomials 12
RD Sharma Solution class 9 Maths Chapter 6 Factorization of Polynomials 13
RD Sharma Solution class 9 Maths Chapter 6 Factorization of Polynomials 14
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RD Sharma Solution class 9 Maths Chapter 6 Factorization of Polynomials 17
RD Sharma Solution class 9 Maths Chapter 6 Factorization of Polynomials 18
RD Sharma Solution class 9 Maths Chapter 6 Factorization of Polynomials 19
RD Sharma Solution class 9 Maths Chapter 6 Factorization of Polynomials 20
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RD Sharma Solution class 9 Maths Chapter 6 Factorization of Polynomials 22
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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) 3x2 – 4x + 15

(ii) y2 + 2√3

(iii) 3√x + √2x

(iv) x – 4/x

(v) x12 + y3 + t50

Solution:

(i) 3x2 – 4x + 15

It is a polynomial of x.

(ii) y2 + 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) x12 + y3 + t50

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

Question 2: Write the coefficient of x2 in each of the following:

(i) 17 – 2x + 7x2

(ii) 9 – 12x + x3

(iii) ∏/6 x2 – 3x + 4

(iv) √3x – 7

Solution:

(i) 17 – 2x + 7x2

Coefficient of x2 = 7

(ii) 9 – 12x + x3

Coefficient of x2 =0

(iii) ∏/6 x2 – 3x + 4

Coefficient of x2 = ∏/6

(iv) √3x – 7

Coefficient of x2 = 0

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

(i) 7x3 + 4x2 – 3x + 12

(ii) 12 – x + 2x3

(iii) 5y – √2

(iv) 7

(v) 0

Solution:

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

(i) Degree of the polynomial 7x3 + 4x2 – 3x + 12 is 3

(ii) Degree of the polynomial 12 – x + 2x3 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 + x2 + 4

(ii) 3x – 2

(iii) 2x + x2

(iv) 3y

(v) t2 + 1

(v) 7t4 + 4t3 + 3t – 2

Solution:

(i) x + x2 + 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 + x2: It is a quadratic polynomial as its degree is 2.

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

(v) t2+ 1: It is a quadratic polynomial as its degree is 2.

(vi) 7t4 + 4t3 + 3t – 2: It is a biquadratic polynomial as its degree is 4.


Exercise 6.2 Page No: 6.8

Question 1: If f(x) = 2x3 – 13x2 + 17x + 12, find

(i) f (2)

(ii) f (-3)

(iii) f(0)

Solution:

f(x) = 2x3 – 13x2 + 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) = x2 – 1, x = 1,−1

(iii) g(x) = 3x2 – 2 , x = 2/√3 , −2/√3

(iv) p(x) = x3 – 6x2 + 11x – 6 , x = 1, 2, 3

(v) f(x) = 5x – π, x = 4/5

(vi) f(x) = x2 , 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) = x2 – 1, x = 1,−1

f(x) = x2 – 1

Given that x = (1 , -1)

Substitute x = 1 in f(x)

f(1) = 12 – 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) = x2 – 1

(iii) g(x) = 3x2 – 2 , x = 2/√3 , −2/√3

g(x) = 3x2 – 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 3x2–2.

(iv) p(x) = x3 – 6x2 + 11x – 6 , x = 1, 2, 3

p(1) = 13 – 6(1)2 + 11x 1 – 6 = 1 – 6 + 11 – 6 = 0

p(2) = 23 – 6(2)2 + 11×2 – 6 = 8 – 24 – 22 – 6 = 0

p(3) = 33 – 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) = x2 , x = 0

f(0) = 02 = 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) = x3 + 4x2 – 3x + 10, g(x) = x + 4

Solution:

f(x) = x3 + 4x2 – 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:

RD sharma class 9 maths chapter 6 ex 4.3 question 1 solution

Question 2: f(x) = 4x4 – 3x3 – 2x2 + x – 7, g(x) = x – 1

Solution:

f(x) = 4x4 – 3x3 – 2x2 + 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:

RD sharma class 9 maths chapter 6 ex 4.3 question 2 solution

Question 3: f(x) = 2x4 – 6X3 + 2x2 – x + 2, g(x) = x + 2

Solution:

f(x) = 2x4 – 6X3 + 2x2 – 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:

RD sharma class 9 maths chapter 6 ex 4.3 question 3 solution

Question 4: f(x) = 4x3 – 12x2 + 14x – 3, g(x) = 2x – 1

Solution:

f(x) = 4x3 – 12x2 + 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:

RD sharma class 9 maths chapter 6 ex 4.3 question 4 solution

Question 5: f(x) = x3 – 6x2 + 2x – 4, g(x) = 1 – 2x

Solution:

f(x) = x3 – 6x2 + 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:

RD sharma class 9 maths chapter 6 ex 4.3 question 5 solution

Question 6: f(x) = x4 – 3x2 + 4, g(x) = x – 2

Solution:

f(x) = x4 – 3x2 + 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:

RD sharma class 9 maths chapter 6 ex 4.3 question 6 solution

Question 7: f(x) = 9x3 – 3x2 + x – 5, g(x) = x – 2/3

Solution:

f(x) = 9x3 – 3x2 + 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:

RD sharma class 9 maths chapter 6 ex 4.3 question 7 solution


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) = x3 – 6x2 + 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) = 3X4 + 17x3 + 9x2 – 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) = x5 + 3x4 – x3 – 3x2 + 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) = x3 – 6x2 – 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) = 3x3 + x2 – 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) = 2x3 – 9x2 + 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) = x3 – 6x2 + 11x – 6, g(x) = x2 – 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 x2 – 3x + 2 = 0

x2 – 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 x3 – 3x2 – 10x + 24.

Solution:

Let f(x) = x3 – 3x2 – 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 x3 – 6x2 – 19x + 84.

Solution:

Let f(x) = x3 – 6x2 – 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: x3 + 6x2 + 11x + 6

Solution:

Let f(x) = x3 + 6x2 + 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: x3 + 2x2 – x – 2

Solution:

Let f(x) = x3 + 2x2 – 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: x3 – 6x2 + 3x + 10

Solution:

Let f(x) = x3 – 6x2 + 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: x4 – 7x3 + 9x2 + 7x- 10

Solution:

Let f(x) = x4 – 7x3 + 9x2 + 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: x4 – 2x3 – 7x2 + 8x + 12

Solution:

f(x) = x4 – 2x3 – 7x2 + 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: x4 + 10x3 + 35x2 + 50x + 24

Solution:

Let f(x) = x4 + 10x3 + 35x2 + 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: 2x4 – 7x3 – 13x2 + 63x – 45

Solution:

Let f(x) = 2x4 – 7x3 – 13x2 + 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) (2x3 – 5x2 -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) = 2x2 (x-3) + x(x-3) -15(x-3)

Taking (x-3) as common

= (x-3)(2x2 + x – 15)

= (x-3)(2x2+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) = 8x3 + ax2 – 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) = x3 + x2 – 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 x3 + 4x2 + 4x-3 if divided by x

Solution:

Using factor theorem,

Put x = 0

f(0) is the remainder.

Now,

f(0) = 03 + 4(0)2 + 4×0 -3 = -3

Therefore -3 is the remainder.

Question 5: If x+1 is a factor of x3 + a, then write the value of a.

Solution:

Let f(x) = x3 + a

If x+1 is a factor of x3 + a then f(-1) = 0

(-1)3 + a = 0

-1 + a = 0

or a = 1

Question 6: If f(x) = x^4 – 2x3 + 3x2 – 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 Exercises:

Get detailed solutions to all the questions listed under below exercises:

Exercise 6.1 Solutions

Exercise 6.2 Solutions

Exercise 6.3 Solutions

Exercise 6.4 Solutions

Exercise 6.5 Solutions

Exercise VSAQs Solutions

RD Sharma Solutions for Chapter 6 Factorization of Polynomials

In this chapter 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

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