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

x³  + 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 + y²  – 2y – 1

2y³ – 2y²

(-)     (+)

3y² – 2y

3y² – 3y

(-)       (+)

y – 1

y – 1

(-)   (+)

0

⇒ 2y³ + y²  – 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 + y²  – 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

x³  + 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    4x³ + 20x² + 33x + 18

4x³ + 6x²

(-)      (-)

14x² – 33x

14x² – 21x

(-)         (+)

12x  + 18

12x  + 18

(-)           (-)

0

⇒ 4x³ + 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, 4x³ + 20x² + 33x + 18 = (2x + 3)(2x + 3)(x + 2)