Class 9 Exam  >  Class 9 Notes  >  RD Sharma Solutions for Class 9 Mathematics  >  RD Sharma Solutions Ex-6.4, (Part -1), Factorization Of Polynomials, Class 9, Maths

RD Sharma Solutions Ex-6.4, (Part -1), Factorization Of Polynomials, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics PDF Download

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)

Q1. f(x) = x3–6x2+11x–6,g(x) = x–3

Sol :

Here , f(x) = x3–6x2+11x–6

g(x) = x – 3

To prove that g(x) is the factor of f(x) ,

we should show ⇒ f(3) = 0

here , x – 3 = 0

⇒ x = 3

Substitute the value of x in f(x)

f(3) = 33–6∗(3)2+11(3)–6

= 27 – (6*9) + 33 – 6

= 27 – 54 + 33 – 6

= 60 – 60

= 0

Since, the result is 0 g(x) is the factor of f(x)


Q2. f(x) = 3x4+17x3+9x2–7x–10,g(x) = x+5

Sol :

Here , f(x) = 3x4+17x3+9x2–7x–10

g(x) = x + 5

To prove that g(x) is the factor of f(x) ,

we should show ⇒ f(-5) = 0

here , x + 5 = 0

⇒ x = -5

Substitute the value of x in f(x)

f(−5)=3(−5)4+17(−5)3+9(−5)2–7(−5)–10

= (3 * 625) + (12 * (-125)) +(9*25) + 35 – 10

= 1875 – 2125 + 225 + 35 – 10

= 2135 – 2135

= 0

Since, the result is 0 g(x) is the factor of f(x)


Q3. f(x) = x5+3x4–x3–3x2+5x+15,g(x) = x+3

Sol :

Here , f(x) = x5+3x4–x3–3x2+5x+15

g(x) = x + 3

To prove that g(x) is the factor of f(x) ,

we should show ⇒ f(-3) = 0

here , x + 3 = 0

⇒ x = -3

Substitute the value of x in f(x)

f(−3)=(−3)5+3(−3)4–(−3)3–3(−3)2+5(−3)+15

= -243 + 243 + 27 – 27 – 15 + 15

= 0

Since, the result is 0 g(x) is the factor of f(x)


Q4. f(x) = x3–6x2–19x+84,g(x)=x–7

Sol :

Here , f(x) = x3–6x2–19x+84

g(x) = x – 7

To prove that g(x) is the factor of f(x) ,

we should show ⇒ f(7) = 0

here , x – 7 = 0

⇒ x = 7

Substitute the value of x in f(x)

f(7) = 73–6(7)2–19(7)+84

= 343 – (6*49) – (19*7) + 84

= 342 – 294 – 133 + 84

= 427 – 427

= 0

Since, the result is 0 g(x) is the factor of f(x)


Q5. f(x) = 3x3+x2–20x+12,g(x) = 3x–2

Sol :

Here , f(x)=3x3+x2–20x+12

g(x) = 3x – 2

To prove that g(x) is the factor of f(x) ,

we should show  ⇒ RD Sharma Solutions Ex-6.4, (Part -1), Factorization Of Polynomials, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

here , 3x – 2 = 0

⇒ 3x = 2

⇒ x = 2/3

Substitute the value of x in f(x)

RD Sharma Solutions Ex-6.4, (Part -1), Factorization Of Polynomials, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics
RD Sharma Solutions Ex-6.4, (Part -1), Factorization Of Polynomials, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

= 0

Since, the result is 0 g(x) is the factor of f(x)


Q6. f(x) = 2x3–9x2+x+13,g(x) = 3–2x

Sol :

Here , f(x) = 2x3–9x2+x+13

g(x) = 3 – 2x

To prove that g(x) is the factor of f(x) ,

To prove that g(x) is the factor of f(x) ,

we should show ⇒ f(3/2) = 0

here , 3 – 2x = 0

⇒ -2x = -3

⇒ 2x = 3

⇒ x = 3/2

Substitute the value of x in f(x)

RD Sharma Solutions Ex-6.4, (Part -1), Factorization Of Polynomials, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Taking L.C.M

RD Sharma Solutions Ex-6.4, (Part -1), Factorization Of Polynomials, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Since, the result is 0 g(x) is the factor of f(x)


Q7. f(x) = x3–6x2+11x–6,g(x) = x2–3x+2

Sol :

Here , f(x) = x3–6x2+11x–6

g(x) = x2–3x+2

First we need to find the factors of x2–3x+2

⇒ x2–2x–x+2

⇒ x(x – 2) -1(x – 2)

⇒ (x – 1) and (x – 2) are the factors

To prove that g(x) is the factor of f(x) ,

The results of f(1) and f(2) should be zero

Let , x – 1 = 0

x = 1

substitute the value of x in f(x)

f(1) = 13–6(1)2+11(1)–6

= 1 – 6 + 11 – 6

= 12 – 12

= 0

Let , x – 2 = 0

x = 2

substitute the value of x in f(x)

f(2) = 23–6(2)2+11(2)–6

= 8 – (6 * 4) + 22 – 6

= 8 – 24 + 22 – 6

= 30 – 30

= 0

Since, the results are 0 g(x) is the factor of f(x)


Q8. Show that (x – 2) , (x + 3) and (x – 4) are the factors of x3–3x2–10x+24

Sol :

Here , f(x) = x3–3x2–10x+24

The factors given are (x – 2) , (x + 3) and (x – 4)

To prove that g(x) is the factor of f(x) ,

The results of f(2) , f(-3) and f(4) should be zero

Let , x – 2 = 0

⇒ x = 2

Substitute the value of x in f(x)

f(2) = 23–3(2)2–10(2)+24

= 8 – (3 * 4) – 20 + 24

= 8 – 12 – 20 + 24

= 32 – 32

= 0

Let , x + 3 = 0

⇒ x = -3

Substitute the value of x in f(x)

f(-3) = (−3)3–3(−3)2–10(−3)+24

= -27 – 3(9) + 30 + 24

= -27 – 27 + 30 + 24

= 54 – 54

= 0

Let , x – 4 = 0

⇒ x = 4

Substitute the value of x in f(x)

f(4) = (4)3–3(4)2–10(4)+24

= 64 – (3*16) – 40 + 24

= 64 – 48 – 40 + 24

= 84 – 84

= 0

Since, the results are 0 g(x) is the factor of f(x)


Q9. Show that (x + 4) , (x – 3) and (x – 7) are the factors of x3–6x2–19x+84

Sol :

Here , f(x) = x3–6x2–19x+84

The factors given are (x + 4) , (x – 3) and (x – 7)

To prove that g(x) is the factor of f(x) ,

The results of f(-4) , f(3) and f(7) should be zero

Let, x + 4 = 0

⇒ x = -4

Substitute the value of x in f(x)

f(-4) = (−4)3–6(−4)2–19(−4)+84

= -64 – (6 * 16) – ( 19 * (-4)) + 84

= -64 – 96 + 76 + 84

= 160 – 160

= 0

Let, x – 3 = 0

⇒ x = 3

Substitute the value of x in f(x)

f(3) = (3)3–6(3)2–19(3)+84

= 27 – (6 * 9) – ( 19 * 3) + 84

= 27 – 54 – 57 + 84

= 111 – 111

= 0

Let, x – 7 = 0

⇒ x = 7

Substitute the value of x in f(x)

f(7) = (7)3–6(7)2–19(7)+84

= 343 – (6 * 49) – ( 19 * 7) + 84

= 343 – 294 – 133 + 84

= 427 – 427

= 0

Since, the results are 0 g(x) is the factor of f(x)


Q10. For what value of a is (x – 5) a factor of x3–3x2+ax–10

Sol :

Here, f(x) = x3–3x2+ax–10

By factor theorem

If (x – 5) is the factor of f(x) then , f(5) = 0

⇒ x – 5 = 0

⇒ x = 5

Substitute the value of x in f(x)

f(5) = 53–3(5)2+a(5)–10

= 125 – (3 * 25) + 5a – 10

= 125 – 75 + 5a – 10

= 5a + 40

Equate f(5) to zero

f(5) = 0

⇒ 5a + 40 = 0

⇒ 5a = -40

⇒  RD Sharma Solutions Ex-6.4, (Part -1), Factorization Of Polynomials, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

= -8

When a= -8 , (x – 5) will be factor of f(x)


Q11. Find the value of a such that (x – 4) is a factor of 5x3–7x2–ax–28

Sol :

Here, f(x) = 5x3–7x2–ax–28

By factor theorem

If (x – 4) is the factor of f(x) then , f(4) = 0

⇒ x – 4 = 0

⇒ x = 4

Substitute the value of x in f(x)

f(4) = 5(4)3–7(4)2–a(4)–28

= 5(64) – 7(16) – 4a – 28

= 320 – 112 – 4a – 28

= 180 – 4

Equate f(4) to zero, to find a

f(4) = 0

⇒ 180 – 4a = 0

⇒ -4a = -180

⇒ 4a = 180

⇒ a = 180/4

⇒ a = 45

When a = 45 , (x – 4) will be factor of f(x)


Q12. Find the value of a, if (x + 2) is a factor of 4x4+2x3–3x2+8x+5a

Sol :

Here, f(x) = 4x4+2x3–3x2+8x+5a

By factor theorem

If (x + 2) is the factor of f(x) then , f(-2) = 0

⇒ x + 2 = 0

⇒ x = -2

Substitute the value of x in f(x)

f(-2) = 4(−2)4+2(−2)3–3(−2)2+8(−2)+5a

= 4(16) + 2(-8) – 3( 4) – 16 + 5a

= 64 – 16 – 12 – 16 + 5a

= 5a + 20

equate f(-2) to zero

f(-2) = 0

⇒ 5a + 20 = 0

⇒ 5a = -20

⇒  RD Sharma Solutions Ex-6.4, (Part -1), Factorization Of Polynomials, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

⇒ a = -4

When a = -4 , (x + 2) will be factor of f(x)


Q13. Find the value of k if x – 3 is a factor of k2x3–kx2+3kx–k

Sol :

Let f(x) = k2x3–kx2+3kx–k

From factor theorem if x – 3 is the factor of f(x) then f(3) = 0

⇒ x – 3 = 0

⇒ x = 3

Substitute the value of x in f(x)

f(3) = k2(3)3–k(3)2+3k(3)–k

= 27k2–9k+9k – k

= 27k2–k

= k(27k – 1)

Equate f(3) to zero, to find k

⇒ f(3) = 0

⇒ k(27k – 1) = 0

⇒ k = 0 and 27k – 1 = 0

⇒ k = 0 and 27k = 1

 

⇒ k = 0 and k = 12/7

When k = 0 and 12/7 , (x – 3) will be the factor of f(x)

The document RD Sharma Solutions Ex-6.4, (Part -1), Factorization Of Polynomials, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.
All you need of Class 9 at this link: Class 9
91 docs

Top Courses for Class 9

FAQs on RD Sharma Solutions Ex-6.4, (Part -1), Factorization Of Polynomials, Class 9, Maths - RD Sharma Solutions for Class 9 Mathematics

1. What is the importance of factorization of polynomials in mathematics?
Ans. Factorization of polynomials is important in mathematics as it helps in simplifying complex expressions, solving equations, finding roots, and understanding the behavior of functions. It allows us to break down a polynomial into its factors, making it easier to analyze and work with.
2. How can I factorize a polynomial using the common factor method?
Ans. To factorize a polynomial using the common factor method, follow these steps: 1. Identify the common factor among all the terms of the polynomial. 2. Divide each term of the polynomial by the common factor. 3. Write the quotient obtained for each term. 4. The common factor becomes the factor outside the brackets, and the quotient becomes the factor inside the brackets. 5. Simplify the expression inside the brackets, if possible.
3. What are the different methods of factorizing polynomials?
Ans. There are several methods of factorizing polynomials, including: 1. Common Factor method: Identify and factor out the common factor from all the terms. 2. Grouping method: Group the terms in pairs and factor out the common factors from each pair. 3. Difference of Squares method: Factorize a polynomial of the form "a^2 - b^2" as "(a + b)(a - b)". 4. Perfect Square Trinomial method: Factorize a polynomial of the form "a^2 + 2ab + b^2" as "(a + b)^2". 5. Sum and Difference of Cubes method: Factorize a polynomial of the form "a^3 + b^3" or "a^3 - b^3" using the respective formulas.
4. Can all polynomials be factorized?
Ans. Not all polynomials can be factorized into linear factors. Some polynomials, such as prime polynomials, cannot be further factorized. However, most polynomials can be factorized using various methods, as mentioned earlier.
5. How can factorization of polynomials be applied in real-life situations?
Ans. Factorization of polynomials finds applications in various real-life situations, such as: 1. Finance: Factorization helps in calculating compound interest and loan payments. 2. Engineering: Factorization is used in solving equations and modeling real-world systems. 3. Computer Science: Factorization plays a crucial role in cryptography, where it is used to secure data and communications. 4. Statistics: Factorization is used in data analysis and regression analysis to simplify equations and identify relationships between variables. 5. Physics: Factorization helps in solving equations and understanding the behavior of physical systems.
91 docs
Download as PDF
Explore Courses for Class 9 exam

Top Courses for Class 9

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

RD Sharma Solutions Ex-6.4

,

ppt

,

Factorization Of Polynomials

,

Maths | RD Sharma Solutions for Class 9 Mathematics

,

Class 9

,

Free

,

Class 9

,

Viva Questions

,

RD Sharma Solutions Ex-6.4

,

(Part -1)

,

mock tests for examination

,

Maths | RD Sharma Solutions for Class 9 Mathematics

,

Semester Notes

,

Summary

,

Exam

,

MCQs

,

Previous Year Questions with Solutions

,

Sample Paper

,

Extra Questions

,

past year papers

,

study material

,

Objective type Questions

,

shortcuts and tricks

,

Class 9

,

Factorization Of Polynomials

,

practice quizzes

,

Important questions

,

Factorization Of Polynomials

,

(Part -1)

,

(Part -1)

,

RD Sharma Solutions Ex-6.4

,

pdf

,

Maths | RD Sharma Solutions for Class 9 Mathematics

,

video lectures

;