Class 9 Exam  >  Class 9 Notes  >  RD Sharma Solutions for Class 9 Mathematics  >  RD Sharma Solutions Ex-6.5, (Part -2), Factorization Of Polynomials, Class 9, Maths

RD Sharma Solutions Ex-6.5, (Part -2), Factorization Of Polynomials, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics PDF Download

Q10. y3–7y+6

Sol :

Given, f(y) = y3–7y+6

The constant term in f(y) is 6

The factors are ±1, ±2, ±3, ±6

Let , y – 1 = 0

⇒ y = 1

f(1) = (1)3–7(1)+6

= 1 – 7 + 6

= 0

So, (y – 1) is the factor of f(y)

Similarly, (y – 2) and (y + 3) are also the factors

Since, f(y) is a polynomial which has degree 3, it cannot have more than 3 linear factors

⇒ f(y) = k(y – 1)( y – 2)(y + 3)

⇒ y3–7y+6 = k(y – 1)( y – 2)(y + 3) ———– 1

Substitute k = 0 in eq 1

⇒ 0 – 0 + 6 = k(-1)(-2)(3)

⇒ 6 = 6k

⇒ k = 1

y3–7y+6 = (1)(y – 1)( y – 2)(y + 3)

y3–7y+6 = (y – 1)( y – 2)(y + 3)

Hence,  y3–7y+6 = (y – 1)( y – 2)(y + 3)


Q11. x3–10x2–53x–42

Sol :

Given , f(x) = x3–10x2–53x–42

The constant in f(x) is -42

The factors of -42 are ±1, ±2, ±3, ±6, ±7, ±14, ±21,±42

Let, x + 1 = 0

⇒ x = -1

f(-1) = (−1)3–10(−1)2–53(−1)–42

= -1 – 10 + 53 – 42

= 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 – 11x – 42

x + 1   x3 – 10x2 – 53x – 42

x + x2

(-)     (-)

-11x2   – 53x

-11x2   – 11x

(+)           (+)

-42x – 42

-42x – 42

(+)       (+)

0

⇒ x3 – 10x2 – 53x – 42 = (x + 1) (x2 – 11x – 42)

Now,

x2 – 11x – 42 = x2 – 14x + 3x – 42

= x(x – 14) + 3(x – 14)

= (x + 3)(x – 14)

Hence, x3 – 10x2 – 53x – 42 = (x + 1) (x + 3)(x – 14)


Q12. y3–2y2–29y–42

Sol :

Given, f(x) =  y3–2y2–29y–42

The constant in f(x) is -42

The factors of -42 are ±1, ±2, ±3, ±6, ±7, ±14, ±21,±42

Let, y + 2 = 0

⇒ y = -2

f(-2) =  (−2)3–2(−2)2–29(−2)–42

= -8 -8 + 58 – 42

= 0

So, ( y + 2) is the factor of f(y)

Now, divide f(y) with (y + 2) to get other factors

By, long division

y2 – 4y – 21

y + 2   y3 – 2y2 – 29y – 42

y3 + 2y2

                     (-)      (-)

-4y2 – 29y

-4y2 – 8y

(+)      (+)

-21y – 42

-21y – 42

(+)        (+)

0

⇒ y3 – 2y2 – 29y – 42 = (y + 2) (y2 – 4y – 21)

Now,

y2 – 4y – 21 = y2 – 7y + 3y – 21

= y(y – 7) +3(y – 7)

= (y – 7)(y + 3)

Hence, y3 – 2y2 – 29y – 42 = (y + 2) (y – 7)(y + 3)

Q13. 2y3–5y2–19y+42

Sol :

Given, f(x) =  2y3–5y2–19y+42

The constant in f(x) is +42

The factors of 42 are ±1, ±2, ±3, ±6, ±7, ±14, ±21,±42

Let, y – 2 = 0

⇒ y = 2

f(2) = 2(2)3–5(2)2–19(2)+42

= 16 – 20 – 38 + 42

= 0

So, (y – 2) is the factor of f(y)

Now, divide f(y) with (y – 2) to get other factors

By, long division method

2y2 – y – 21

y – 2    2y3 – 5y2  -19y + 42

2y3 – 4y2

(-)       (+)

-y2 – 19y

-y2 + 2y

(+)      (-)

-21y + 42

-21y + 42

(+)       (-)

0

⇒ 2y3 – 5y2  -19y + 42 = (y – 2) (2y2 – y – 21)

Now,

2y2 – y – 21

The factors are (y + 3) (2y – 7)

Hence,  2y3 – 5y2  -19y + 42 = (y – 2) (y + 3) (2y – 7)


Q14. x3+13x2+32x+20

Sol:

Given, f(x) =  x3+13x2+32x+20

The constant in f(x) is 20

The factors of 20 are ±1, ±2, ±4, ±5, ±10, ±20

Let, x + 1 = 0

⇒ x = -1

f(-1) =  (−1)3+13(−1)2+32(−1)+20

= -1 + 13 – 32 + 20

= 0

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

Divide f(x) with (x + 1) to get other factors

By, long division

x2 + 12x + 20

x + 1  x3 + 13x2 +32x + 20

x3  + x2

(-)     (-)

12x2 + 32x

12x2 + 12x

(-)         (-)

20x – 20

20x – 20

(-)          (-)

0

⇒ x3 + 13x2 +32x + 20 = (x + 1)( x2 + 12x + 20)

Now,

x2 + 12x + 20 = x2 + 10x + 2x + 20

= x(x + 10) + 2(x + 10)

The factors are (x + 10) and (x + 2)

Hence, x3 + 13x2 +32x + 20 = (x + 1)(x + 10)(x + 2)


Q15. x3–3x2–9x–5

Sol :

Given, f(x) = x3–3x2–9x–5

The constant in f(x) is -5

The factors of -5 are ±1, ±5

Let, x + 1 = 0

⇒ x = -1

f(-1) = (−1)3–3(−1)2–9(−1)–5

= -1 – 3 + 9 – 5

= 0

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

Divide f(x) with (x + 1) to get other factors

By, long division

x2 – 4x – 5

x + 1   x3 – 3x2 – 9x – 5

x3 + x2

(-)   (-)

-4x2 – 9x

-4x2 – 4x

(+)      (+)

-5x – 5

-5x – 5

(+)     (+)

0

⇒ x3 – 3x2 – 9x – 5 = (x + 1)( x2 – 4x – 5)

Now,

x2 – 4x – 5 = x2 – 5x + x – 5

= x(x – 5) + 1(x – 5)

The factors are (x – 5) and (x + 1)

Hence, x3 – 3x2 – 9x – 5 = (x + 1)(x – 5)(x + 1)


Q16. 2y3+y2–2y–1

Sol :

Given , f(y) = 2y3+y2–2y–1

The constant term is 2

The factors of 2 are ±1, ± 1/2

Let, y – 1= 0

⇒ y = 1

f(1) = 2(1)3+(1)2–2(1)–1

= 2 + 1 – 2 – 1

= 0

So, (y – 1) is the factor of f(y)

Divide f(y) with (y – 1) to get other factors

By, long division

2y2 + 3y + 1

y – 1  2y3 + y – 2y – 1

2y3 – 2y2

(-)     (+)

3y2 – 2y

3y2 – 3y

(-)       (+)

y – 1

y – 1

(-)   (+)

0

⇒ 2y3 + y – 2y – 1 = (y – 1) (2y2 + 3y + 1)

Now,

2y2 + 3y + 1  = 2y2 + 2y + y + 1

= 2y(y + 1) + 1(y + 1)

= (2y + 1) ( y + 1)  are the factors

Hence,  2y3 + y – 2y – 1 = (y – 1) (2y + 1) ( y + 1)


Q17.  x3–2x2–x+2

Sol :

Let, f(x) = x3–2x2–x+2

The  constant term is 2

The factors of 2 are ±1, ±12

Let, x – 1= 0

⇒ x = 1

f(1) = (1)3–2(1)2–(1)+2

= 1 – 2 – 1 + 2

= 0

So, (x – 1) is the factor of f(x)

Divide f(x) with (x – 1) to get other factors

By, long division

x2 – x – 2

x – 1    x3 – 2x2 – y + 2

x3  – x2

(-)      (+)

-x2  – x

-x2 + x

(+)      (-)

– 2x + 2

– 2x + 2

(+)      (-)

0

⇒  x3 – 2x2 – y + 2 = (x – 1) (x2 – x – 2)

Now,

x2 – x – 2 = x2 – 2x + x  – 2

= x(x – 2) + 1(x – 2)

=(x – 2)(x + 1)  are the factors

Hence,  x3 – 2x2 – y + 2 = (x – 1)(x + 1)(x – 2)

Q18.  Factorize each of the following polynomials :

1. x3+13x2+31x–45 given that x + 9 is a factor
 2. 4x3+20x2+33x+18 given that 2x + 3 is a factor

Sol :

1. x3+13x2+31x–45 given that x + 9 is a factor

let, f(x) = x3+13x2+31x–45

given that (x + 9) is the factor

divide f(x) with (x + 9) to get other factors

by , long division

x2 + 4x – 5

x + 9    x3 + 13x2 + 31x – 45

x + 9x2

(-)       (-)

4x2  + 31x

4x2  + 36x

(-)        (-)

-5x  – 45

-5x – 45

(+)       (+)

0

⇒ x3 + 13x2 + 31x – 45 = (x + 9)( x2 + 4x – 5)

Now,

x2 + 4x – 5 = x2 + 5x – x  – 5

= x(x + 5) -1(x + 5)

= (x + 5) (x – 1) are the factors

Hence, x3 + 13x2 + 31x – 45 = (x + 9)(x + 5)(x – 1)

2. 4x3+20x2+33x+18 given that 2x + 3 is a factor

let, f(x) =  4x3+20x2+33x+18

given that 2x + 3 is a factor

divide f(x) with (2x + 3) to get other factors

by, long division

2x2 + 7x + 6

2x + 3    4x+ 20x2 + 33x + 18

4x3 + 6x2

(-)      (-)

14x2 – 33x

14x2 – 21x

(-)         (+)

12x  + 18

12x  + 18

(-)           (-)

0

⇒ 4x+ 20x2 + 33x + 18 = (2x + 3) (2x2 + 7x + 6)

Now,

2x2 + 7x + 6 = 2x2 + 4x + 3x + 6

= 2x(x + 2) + 3(x + 2)

= (2x + 3)(x + 2) are the factors

Hence, 4x+ 20x2 + 33x + 18 = (2x + 3)(2x + 3)(x + 2)

The document RD Sharma Solutions Ex-6.5, (Part -2), 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.
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FAQs on RD Sharma Solutions Ex-6.5, (Part -2), Factorization Of Polynomials, Class 9, Maths - RD Sharma Solutions for Class 9 Mathematics

1. What is the concept of factorization of polynomials?
Ans. Factorization of polynomials refers to the process of expressing a polynomial as a product of its factors. It involves finding the factors of a given polynomial by dividing it with other polynomials until no further division is possible.
2. How can we factorize a quadratic polynomial?
Ans. To factorize a quadratic polynomial, we can use the middle-term splitting method or the quadratic formula. In the middle-term splitting method, we split the middle term of the quadratic polynomial into two terms such that their sum is equal to the coefficient of the linear term. We then group the terms and find the common factors to factorize the polynomial.
3. Can all polynomials be factorized?
Ans. No, not all polynomials can be factorized. Polynomials with complex roots or irreducible factors cannot be factorized. However, quadratic polynomials can always be factorized, either into linear factors or into a product of two quadratic factors.
4. What is the importance of factorization of polynomials?
Ans. Factorization of polynomials is important in various areas of mathematics and science. It helps in simplifying complex algebraic expressions, solving polynomial equations, finding roots of polynomials, and understanding the behavior of polynomial functions. It also aids in solving problems related to areas, volumes, and geometry by expressing them in terms of factors.
5. Can factorization of polynomials be applied in real-life situations?
Ans. Yes, factorization of polynomials is applicable in real-life situations. It is used in fields such as engineering, physics, economics, and computer science. For example, in engineering, factorization is used to simplify electrical circuits and analyze their behavior. In economics, it is used to model and solve problems related to production, cost, and revenue functions. In computer science, factorization is used in cryptography and coding theory.
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