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Class 9 Exam  >  Class 9 Notes  >  RD Sharma Solutions for Class 9 Mathematics  >  RD Sharma Solutions Ex-6.5, (Part -1), Factorization Of Polynomials, Class 9, Maths

RD Sharma Solutions Ex-6.5, (Part -1), Factorization Of Polynomials, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics PDF Download

Q1. x3+6x2+11x+6

Sol:

Given polynomial, f(x) = x3+6x2+11x+6

The constant term in f(x) is 6

The factors of 6 are ±1, ±2, ±3, ±6

Let, x + 1 = 0

⇒ x = -1

Substitute 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)

Similarly, (x + 2) and (x + 3) are also the factors of f(x)

Since, f(x) is a polynomial having a degree 3, it cannot have more than three linear factors.

∴ f(x) = k(x + 1)(x + 2)(x + 3)

⇒ x3+6x2+11x+6 = k(x + 1)(x + 2)(x + 3)

Substitute x = 0 on both the sides

⇒ 0 + 0 + 0 + 6 = k(0 +1)(0 + 2)(0 + 3)

⇒ 6 = k(1*2*3)

⇒ 6 = 6k

⇒ k = 1

Substitute k value in f(x) = k(x + 1)(x + 2)(x + 3)

⇒ f(x) = (1)(x + 1)(x + 2)(x + 3)

⇒ f(x) = (x + 1)(x + 2)(x + 3)

∴ x3+6x2+11x+6 = (x + 1)(x + 2)(x + 3)


Q2. x3+2x2–x–2

Sol:

Given, f(x) = x3+2x2–x–2

The constant term in f(x) is -2

The factors of (-2) are ±1, ±2

Let , x – 1 = 0

⇒ x = 1

Substitute the value of x in f(x)

f(1) = (1)3+2(1)2–1–2

= 1 + 2 – 1 – 2

= 0

Similarly , the other factors (x + 1) and (x + 2) of f(x)

Since, f(x) is a polynomial having a degree 3, it cannot have more than three linear factors.

∴ f(x) = k(x – 1)(x + 2)(x + 1)

x3+2x2–x–2 = k(x – 1)(x + 2)(x + 1)

Substitute x = 0 on both the sides

0 + 0 – 0 – 2 = k(-1)(1)(2)

⇒  – 2 = -2k

⇒  k = 1

Substitute k value in f(x)  = k(x – 1)(x + 2)(x + 1)

f(x) = (1)(x – 1)(x + 2)(x + 1)

⇒ f(x) =  (x – 1)(x + 2)(x + 1)

So, x3+2x2–x–2 = (x – 1)(x + 2)(x + 1)


Q3. x3–6x2+3x+10

Sol:

Let, f(x) = x3–6x2+3x+10

The constant term in f(x) is 10

The factors of 10 are ±1, ±2, ±5, ±10

Let , x + 1 = 0

⇒ x = -1

Substitute the value of x in f(x)

f(-1) = (−1)3–6(−1)2+3(−1)+10

= -1 – 6 – 3 + 10

= 0

Similarly , the other factors (x – 2) and (x – 5) of f(x)

Since, f(x) is a polynomial having a degree 3, it cannot have more than three linear factors.

∴ f(x) = k(x + 1)(x – 2)(x – 5 )

Substitute x = 0 on both sides

⇒ x3–6x2+3x+10 = k(x + 1)(x – 2)(x – 5)

⇒ 0 – 0 + 0 + 10 = k(1)(-2)(-5)

⇒ 10 = k(10)

⇒ k = 1

Substitute k = 1 in f(x) = k(x + 1)(x – 2)(x – 5)

f(x) = (1)(x + 1)(x – 2)(x – 5)

so, x3–6x2+3x+10 = (x + 1)(x – 2)(x – 5)


Q4. x4–7x3+9x2+7x–10

Sol:

Given, f(x) = x4–7x3+9x2+7x–10

The constant term in f(x) is 10

The factors of 10 are ±1, ±2, ±5, ±10

Let , x – 1 = 0

⇒ x = 1

Substitute the value of x in f(x)

f(x) = 14–7(1)3+9(1)2+7(1)–10

= 1 – 7 + 9 + 7 – 10

= 10 – 10

= 0

(x – 1)  is the factor of f(x)

Simarly, the other factors are (x + 1) ,(x – 2) , (x – 5)

Since, f(x) is a polynomial of degree 4. So, it cannot have more than four linear factor.

So, f(x) = k(x – 1)(x + 1)(x – 2)(x – 5)

⇒ x4–7x3+9x2+7x–10 = k(x – 1)(x + 1)(x – 2)(x – 5)

Put x = 0 on both sides

0 – 0 + 0 – 10  =  k(-1)(1)(-2)(-5)

– 10 = k(-10)

⇒ k = 1

Substitute k = 1 in  f(x) = k(x – 1)(x + 1)(x – 2)(x – 5)

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

= (x – 1)(x + 1)(x – 2)(x – 5)

So, x4–7x3+9x2+7x–10 = (x – 1)(x + 1)(x – 2)(x – 5)


Q5. x4–2x3–7x2+8x+12

Sol:

Given , f(x) = x4–2x3–7x2+8x+12

The constant term f(x) is equal is 12

Tha factors of 12 are ±1, ±2, ±3, ±4, ±6, ±12

Let, x + 1 = 0

⇒ x = -1

Substitute the value of x in f(x)

f(-1) = (−1)4–2(−1)3–7(−1)2+8(−1)+12

= 1 + 2 – 7 – 8 + 12

= 0

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

Similarly, (x + 2), (x – 2), (x – 3) are also the factors of f(x)

Since, f(x) is a polynomial of degree 4 , it cannot have more than four linear factors.

⇒ f(x) = k(x + 1)(x + 2)(x – 3)(x – 2)

⇒ x4–2x3–7x2+8x+12 = k(x + 1)(x + 2)(x – 3)(x – 2)

Substitute x = 0 on both sides,

⇒ 0 – 0 – 0 + 12 = k(1)(2)(-2)(-3)

⇒ 12 = k12

⇒ k = 1

Substitute k = 1 in f(x) = k(x – 2)(x + 1)(x + 2)(x – 3)

f(x) = (x – 2)(x + 1)(x + 2)(x – 3)

so, x4–2x3–7x2+8x+12 = (x – 2)(x + 1)(x + 2)(x – 3)

Q6. x4+10x3+35x2+50x+24

Sol:

Given, f(x) = x4+10x3+35x2+50x+24

The constant term in f(x) is equal to 24

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

Let, x + 1 = 0

⇒ x = -1

Substitute the value of x in f(x)

f(-1) = (−1)4+10(−1)3+35(−1)2+50(−1)+24

= 1 – 10 + 35 – 50 + 24

= 0

⇒ (x + 1) is the factor of f(x)

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

Since, f(x) is a polynomial of degree 4, it cannot have more than four linear factors.

⇒ f(x) = k(x + 1)(x + 2)(x + 3)(x + 4)

⇒ x4+10x3+35x2+50x+24 = k(x + 1)(x + 2)(x + 3)(x + 4)

Substitute x = 0 on both sides

⇒ 0 + 0 + 0 + 0 + 24 = k(1)(2)(3)(4)

⇒ 24 = k(24)

⇒ k = 1

Substitute k = 1 in f(x) = k(x + 1)(x + 2)(x + 3)(x + 4)

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

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

hence, x4+10x3+35x2+50x+24 = (x + 1)(x + 2)(x + 3)(x + 4)


Q7.  2x4–7x3–13x2+63x–45

Sol :

Given, f(x) = 2x4–7x3–13x2+63x–45

The factors of constant term -45 are ±1, ±3, ±5, ±9, ±15, ±45

The factors of the coefficient of  x4 is 2. Hence possible rational roots of f(x) are

RD Sharma Solutions Ex-6.5, (Part -1), Factorization Of Polynomials, Class 9, Maths | RD Sharma Solutions for Class 9 MathematicsRD Sharma Solutions Ex-6.5, (Part -1), Factorization Of Polynomials, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Let, x – 1 = 0

⇒ x= 1

f(1) = 2(1)4–7(1)3–13(1)2+63(1)–45

= 2 – 7 – 13 + 63 – 45

= 0

Let, x – 3 = 0

⇒ x = 3

f(3) = 2(3)4–7(3)3–13(3)2+63(3)–45

= 162 – 189 – 117 + 189 – 45

= 0

So, (x – 1) and (x – 3) are the roots of f(x)

⇒ x2 – 4x + 3 is the factor of f(x)

Divide f(x) with x2 – 4x + 3 to get other three factors

By long division,

2x2 + x – 15

x2  – 4x + 3    2x4 – 7x3 – 13x2  + 63x – 45

2x4  – 8x3  + 6x2

(-)      (+)        (-)

x3   – 19x2  + 63x

x3     – 4x2     + 3x

(-)        (+)          (-)

– 15x2 + 60x – 45

(+)           (-)         (+)

0

⇒  2x4–7x3–13x2+63x–45 = (x2 – 4x + 3)(2x2 + x – 15)

⇒ 2x4–7x3–13x2+63x–45 = (x – 1) (x – 3)(2x2 + x – 15)

Now,

2x2 + x – 15 = 2x2+ 6x – 5x  – 15

= 2x(x + 3) – 5 (x + 3)

= (2x – 5) (x + 3)

So, 2x4–7x3–13x2+63x–45 = (x – 1)(x – 3)(x + 3)(2x – 5)


Q8. 3x3−x2–3x+1

Sol :

Given , f(x) = 3x3−x2–3x+1

The factors of constant term 1 is ±1

The factors of the coefficient of x2 = 3

The possible rational roots are ±1 ,1/3

Let, x – 1 = 0

⇒ x = 1

f(1) = 3(1)3−(1)2–3(1)+1

= 3 – 1 – 3 + 1

= 0

So, x – 1 is tha factor of f(x)

Now, divide f(x) with (x – 1) to get other factors

By long division method,

3x2 + 2x – 1

x – 1   3x3 – x2 – 3x + 1

3x3 – x2

(-)       (+)

2x2 – 3x

2x2 – 2x

(-)       (+)

-x + 1

-x + 1

(+)  (-)

0

⇒ 3x3−x2–3x+1 = (x – 1)( 3x2 + 2x – 1)

Now,

3x2 + 2x – 1 = 3x2 + 3x – x  – 1

= 3x(x + 1) -1(x + 1)

= (3x – 1)(x + 1)

Hence , 3x3−x2–3x+1 = (x – 1) (3x – 1)(x + 1)


Q9. x3−23x2+142x–120

Sol :

Let, f(x) = x3−23x2+142x–120

The constant term in f(x) is -120

The factors of -120 are ±1, ±2, ±3, ±4, ±5, ±6, ±8, ±10, ±12, ±15, ±20, ±24, ±30, ±40, ±60, ±120

Let, x – 1 = 0

⇒ x = 1

f(1) = (1)3−23(1)2+142(1)–120

= 1 – 23  + 142 – 120

= 0

So, (x – 1) is the factor of f(x)

Now, divide f(x) with (x – 1) to get other factors

By long division,

x2 – 22x + 120

x – 1   x3 – 23x2 + 142x – 120

x3 –   x2

(-)    (+)

-22x2 + 142x

-22x2  + 22x

(+)            (-)

120x – 120

120x – 120

(-)           (+)

0

⇒ x3 – 23x2 + 142x – 120 = (x  – 1) (x2 – 22x + 120)

Now,

x2 – 22x + 120 = x2 – 10x – 12x  + 120

= x( x – 10) – 12( x – 10)

= (x – 10) (x – 12)

Hence, x3 – 23x2 + 142x – 120 = (x – 1) (x – 10) (x – 12

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FAQs on RD Sharma Solutions Ex-6.5, (Part -1), Factorization Of Polynomials, Class 9, Maths - RD Sharma Solutions for Class 9 Mathematics

1. What is the importance of factorizing polynomials?
Ans. Factorizing polynomials is important because it helps us simplify complex expressions and solve equations more easily. It allows us to break down a polynomial into its factors, which are easier to work with and understand. Factorization also helps in finding the roots or zeros of a polynomial equation.
2. How can we determine if a polynomial is factorizable?
Ans. A polynomial can be factorized if it can be expressed as a product of two or more polynomials. To determine if a polynomial is factorizable, we can use various techniques such as checking for common factors, using synthetic division, or applying factor theorem. If we can find any factors of the given polynomial, then it is factorizable.
3. What is the process of factorizing a polynomial?
Ans. The process of factorizing a polynomial involves breaking it down into its factors. This can be done by using various techniques such as grouping, finding common factors, applying difference of squares, using the quadratic formula, or using long division. The goal is to express the polynomial as a product of simpler polynomials or binomials.
4. Can all polynomials be factorized?
Ans. No, not all polynomials can be factorized. Some polynomials may not have any factors other than 1 and itself, making them prime polynomials. For example, the polynomial x^2 + 1 cannot be factorized further. However, most polynomials can be factorized using appropriate techniques.
5. How does factorization help in solving polynomial equations?
Ans. Factorization helps in solving polynomial equations by simplifying the expressions and allowing us to find the roots or zeros of the equation. Once a polynomial is factorized, we can set each factor equal to zero and solve for the variables. The solutions to these simpler equations give us the values of the variables that satisfy the original polynomial equation.
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