Class 9 RD Sharma Solutions – Chapter 6 Factorisation of Polynomials- Exercise 6.4 | Set 1

In Class 9 mathematics, the chapter on Factorisation of Polynomials is crucial for understanding algebraic expressions and their manipulation. Exercise 6.4 focuses on practical problems related to factorization enhancing students’ skills in breaking down polynomials into simpler components. This exercise helps students grasp the concept of factorization and apply it to solve polynomial equations efficiently.

Factorization of Polynomials

The Factorisation of polynomials involves expressing a polynomial as a product of its factors. This process is essential for solving polynomial equations and simplifying the algebraic expressions. By breaking down a polynomial into its constituent factors one can solve equations find roots and simplify expressions more easily. Common techniques include factoring by the grouping using the special identities and applying the quadratic formula.

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

Question 1. f(x) = x³ – 6x² + 11x – 6, g(x) = x – 3

Solution: 

Given: f(x) = x³ – 6x² + 11x – 6, g(x) = x – 3

Here,

x – 3 = 0

x = 3 

To prove: g(x) is a factor of f(x), that is f(3) = 0 

On substituting the value of x in f(x), we get

 f(3) = 3³ – 6 × 3² + 11 × 3 – 6

= 27 – (6 × 9) + 33 – 6

= 27 – 54 + 33 – 6

= 60 – 60

= 0 

Since, the result is 0, 

Hence, proved that g(x) is a factor of f(x).

Question 2. f(x) = 3x⁴ + 17x³ + 9x² – 7x – 10, g(x) = x + 5

Solution: 

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

Here,

x + 5 = 0

x = -5

To prove: g(x) is a factor of f(x), that is f(-5) = 0

On substituting the value of x in f(x), we get

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

= 3 × 625 + 17 × (-125) + 9 × 25 + 35 – 10

= 1875 – 2125 + 225 + 35 – 10

= 2135 – 2135

= 0 

Since, the result is 0, 

Hence, proved that g(x) is a factor of f(x).

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

Solution: 

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

Here,

x + 3 = 0 

x = -3

To prove: g(x) is a factor of f(x), that is f(-3) = 0

On substituting the value of x in f(x), we get

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, 

Hence, proved that g(x) is a factor of f(x).

Question 4. f(x) = x³ – 6x² – 19x + 84, g(x) = x – 7

Solution: 

Given: f(x) = x³ – 6x² – 19x + 84, g(x) = x – 7

Here,

x – 7 = 0  

x = 7

To prove: g(x) is a factor of f(x), that is f(7) = 0

On substituting the value of x in f(x), we get

f(7) = 7³ – 6 × 7² – 19 × 7 + 84     

= 343 – (6 × 49) – (19 × 7) + 84

= 343 – 294 – 133 + 84

= 427 – 427

= 0

Since, the result is 0, 

Hence, proved that g(x) is a factor of f(x).

Question 5. f(x) = 3x³ + x² – 20x + 12, g(x) = 3x – 2

Solution: 

Given: f(x) = 3x³ + x² – 20x + 12, g(x) = 3x – 2

Here,

3x – 2 = 0  

x = 2/3

To prove: g(x) is a factor of f(x), that is f(2/3) = 0

On substituting the value of x in f(x), we get

[Tex]f(\frac23)=3(\frac23)^3+(\frac23)^2-20(\frac23)+12[/Tex]

[Tex]= 3(\frac8{27})+\frac49-\frac{40}3+12[/Tex]

[Tex]=\frac89+\frac49-\frac{40}3+12[/Tex]

[Tex]= \frac{12}9-\frac{40}3+12[/Tex]

Taking L.C.M 

 = [Tex]\frac{12-120+108}{9}[/Tex]

 [Tex]=\frac{120-120}{9}[/Tex]

 = 0 

Since, the result is 0, 

Hence, proved that g(x) is a factor of f(x).

Question 6. f(x) = 2x³ – 9x² + x + 13, g(x) = 3 – 2x

Solution: 

Given: f(x) = 2x³ – 9x² + x + 13, g(x) = 3 – 2x

Here,

3 – 2x = 0  

x = 3/2

To prove: g(x) is a factor of f(x), that is f(3/2) = 0

On substituting the value of x in f(x), we get

[Tex]f(\frac32)=2(\frac32)^3-9(\frac32)^2+(\frac32)+13[/Tex]

[Tex]=2(\frac{27}8)-9(\frac94)+\frac34+13[/Tex]

[Tex]=\frac{27}4-\frac{81}4 +\frac{3}{2}+12[/Tex]

[Tex]=\frac{21-81+6+48}{4}[/Tex]

 [Tex]= \frac{81-81}4[/Tex]

= 0 

Since, the result is 0, 

Hence, proved that g(x) is a factor of f(x).

Question 7. f(x) = x³ – 6x² + 11x – 6, g(x) = x² – 3x + 2

Solution: 

Given: f(x) = x³ – 6x² + 11x – 6, g(x) = x² – 3x + 2

Here,

x² – 3x + 2 = 0  

On factorizing the above, we get

⇒ x² – 3x + 2 = 0

⇒ x² – 2x – x + 2 = 0

⇒ x(x – 2) – 1(x – 2)

⇒ (x – 1)(x – 2) are the factors 

To prove: g(x) is a factor of f(x), that is f(1) and f(2) should be 0

Let x = 1 

On substituting the value of x in f(x), we get

f(1) = 1³ – 6 × 1² + 11 × 1 – 6

= 1 – 6 + 11 – 6

= 12 – 12

= 0 

Let x = 2

On substituting the value of x in f(x), we get

f(x) = 2³ – 6 × 2² + 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)

Question 8. Show that (x – 2), (x + 3) and (x – 4) are the factors of x³ – 3x² – 10x + 24

Solution: 

Given:

 f(x) = x³ – 3x² – 10x + 24 

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

To prove: g(x) is a factor of f(x), that is f(2), f(-3), f(4) should be 0

Here, x – 2 = 0

Let, x = 2

On substituting the value of x in f(x), we get

 f(2) = 2³ – 3 × 2² – 10 × 2 + 24 

= 8 – (3 × 4) – 20 + 24

= 8 – 12 – 20 + 24

= 32 – 32

= 0 

Here, x + 3 = 0

Let, x = -3

On substituting the value of x in f(x), we get

f(-3) = (-3)³ – 3 × (-3)² – 10 × (-3) + 24

= -27 -3(9) + 30 + 24

= -27- 27 + 30 + 24

= 54 – 54

= 0  

Here, x – 4 = 0

Let, x = 4

On substituting the value of x in f(x), we get

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)

Question 9. Show that (x + 4), (x – 3) and (x – 7) are the factors of x³ – 6x² – 19x + 84

Solution: 

Given:

f(x) = x³ – 6x²– 19x + 84

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

To prove: g(x) is a factor of f(x), that is f(4), f(3), f(-7) should be 0

Here, x + 4 = 0

Let, x = -4

On substituting the value of x in f(x), we get

f(-4) = (-4)³ – 6(-4)² – 19(-4) + 84

= -64 – (6 × 16) – 19 × (-4) + 84

= -64 – 96 + 76 + 84

= 160 – 160

= 0

Here, x – 3 = 0

Let, x = 3

On substituting the value of x in f(x), we get

f(3) = (3)³ – 6(3)² – 19(3) + 84

= 27 – 6(9) – 19 × 3 + 84

= 27 – 54 – 57 + 84

= 111 – 111

= 0  

Here, x – 7 = 0

Let, x = 7

On substituting the value of x in f(x), we get

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)

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

Solution: 

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

On substituting the value of x in f(x), we get

f(5) = 5³ – 3 × 5² + 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

⇒ a = -40/5

⇒ a = -8

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

Question 11. Find the value of such that (x – 4) is a factor of  5x³ – 7x² – ax – 28    

Solution: 

Given: f(x) = 5x³– 7x²– ax – 28

Using factor theorem

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

⇒ x – 4 = 0

⇒ x = 4

On substituting the value of x in f(x), we get

f(4) = 5(4)³ – 7(4)² – a × 4 – 28

= 5(64) – 7(16) – 4a – 28

= 320 – 112 – 4a – 28

= 180 – 4a

Equate f(4) to zero, to find a

f(4) = 0

⇒ 180 – 4a = 0 

⇒ a = 180/4

⇒ a = 45     

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

Question 12. Find the value of a, if (x + 2) is a factor of 4x⁴ + 2x³ – 3x² + 8x + 5a 

Solution: 

Given: f(x) = 4x⁴+ 2x³– 3x²+ 8x + 5a 

Using factor theorem 

If (x + 2) is the factor of f(x), then f(-2) should be zero

⇒ x + 2 =0 

⇒ x = -2

On substituting the value of x in f(x), we get

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 0 

f(-2) = 4(-2)²+ 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 0

f(-2) = 0

⇒ 5a + 20 = 0

⇒ 5a = – 20

⇒ a = -4

When a = -4, then (x + 2) is the factor of f(x)

Read More:

• Factorization of Polynomial
• What are Polynomials?

Conclusion

Exercise 6.4 in RD Sharma’s Class 9 textbook provides the comprehensive practice on factorising polynomials using the various techniques. By solving these problems, students gain a deeper understanding of the polynomial factorisation and improve their algebraic manipulation skills. Mastery of these techniques is essential for the progressing in the higher algebra and mathematics.

FAQs on Factorisation of Polynomials

What is the importance of factorising polynomials in algebra?

The Factorising polynomials simplifies complex algebraic expressions making it easier to solve equations and find roots.

What are the common methods used for factorising polynomials?

The Common methods include the factoring by the grouping using the special identities and applying the quadratic formula.

How can I check if my factorisation is correct?

We can verify factorisation by the expanding the factors and checking if we obtain the original polynomial.

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Article Tags :

• Class 9
• Mathematics
• RD Sharma Solutions
• School Learning
• Maths-Class-9
• RD Sharma Class-9

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