RD Sharma Solutions Class 9 Factorization Of Polynomials Exercise 6.5

RD Sharma Solutions Class 9 Chapter 6 Exercise 6.5

RD Sharma Class 9 Solutions Chapter 6 Ex 6.5 Download

Using factor theorem, factorize each of the following polynomials :

Q1. \(x^{3} + 6x^{2} + 11x + 6\)

Sol:

Given polynomial, f(x) = \(x^{3} + 6x^{2} + 11x + 6\)

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)

=> \(x^{3} + 6x^{2} + 11x + 6\) = 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)

\(∴\) \(x^{3} + 6x^{2} + 11x + 6\) = (x + 1)(x + 2)(x + 3)

Q2. \(x^{3} + 2x^{2} – x – 2\)

Sol:

Given, f(x) = \(x^{3} + 2x^{2} – x – 2\)

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 )

\(x^{3} + 2x^{2} – x – 2\) = 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, \(x^{3} + 2x^{2} – x – 2\) = (x – 1)(x + 2)(x + 1 )

Q3. \(x^{3} – 6x^{2} +  3x + 10\)

Sol:

Let, f(x) = \(x^{3} – 6x^{2} + 3x + 10\)

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

=> \(x^{3} – 6x^{2} + 3x + 10\) = 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, \(x^{3} – 6x^{2} +  3x + 10\) = (x + 1)(x – 2)(x – 5 )

Q4. \(x^{4} – 7x^{3} + 9x^{2} + 7x – 10\)

Sol:

Given, f(x) = \(x^{4} – 7x^{3} + 9x^{2} + 7x – 10\)

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) = \(1^{4} – 7(1)^{3} + 9(1)^{2} + 7(1) – 10\)

= 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)

=> \(x^{4} – 7x^{3} + 9x^{2} + 7x – 10\) = 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, \(x^{4} – 7x^{3} + 9x^{2} + 7x – 10\) = (x – 1)(x + 1)(x – 2)(x – 5)

Q5. \(x^{4} – 2x^{3} – 7x^{2} + 8x + 12\)

Sol:

Given , f(x) = \(x^{4} – 2x^{3} – 7x^{2} + 8x + 12\)

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)

=> \(x^{4} – 2x^{3} – 7x^{2} + 8x + 12\) = 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, \(x^{4} – 2x^{3} – 7x^{2} + 8x + 12\) = (x – 2)(x + 1)(x + 2)(x – 3)

Q6. \(x^{4} + 10x^{3}  + 35x^{2} + 50x + 24\)

Sol:

Given, f(x) = \(x^{4} + 10x^{3}  + 35x^{2} + 50x + 24\)

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)

=>   \(x^{4} + 10x^{3}  + 35x^{2} + 50x + 24\) = 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, \(x^{4} + 10x^{3}  + 35x^{2} + 50x + 24\) = (x + 1)(x + 2)(x + 3)(x + 4)

Q7.  \(2x^{4} – 7x^{3}  – 13x^{2} + 63x – 45\)

Sol :

Given, f(x) = \(2x^{4} – 7x^{3}  –  13x^{2} + 63x – 45\)

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

The factors of the coefficient of  x^{4} is 2. Hence possible rational roots of f(x) are

±1, ±3, ±5, ±9, ±15, ±45, \(\pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{5}{2}, \pm \frac{9}{2}, \pm \frac{15}{2}, \pm \frac{45}{2}\)

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)

=> \(x^{2}\) – 4x + 3 is the factor of f(x)

Divide f(x) with   \(x^{2}\) – 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

– 15x2 + 60x – 45

(+)           (-)         (+)

0

=>  \(2x^{4} – 7x^{3}  – 13x^{2} + 63x – 45\) = (x^{2} – 4x + 3)( 2x^{2} + x – 15)

=> \(2x^{4} – 7x^{3}  – 13x^{2} + 63x – 45\) = (x – 1) (x – 3)( 2x^{2} + x – 15)

Now,

\(2x^{2}\) + x – 15 = 2x^{2} + 6x – 5x  – 15

= 2x(x + 3) – 5 (x + 3)

= (2x – 5) (x + 3)

So, \(2x^{4} – 7x^{3}  – 13x^{2} + 63x – 45\) = (x – 1)(x – 3)(x + 3)(2x – 5)

Q8. \(3x^{3}- x^{2} – 3x + 1\)

Sol :

Given , f(x) = \(3x^{3}- x^{2} – 3x + 1\)

The factors of constant term 1 is ±1

The factors of the coefficient of \( x^{2}\) = 3

The possible rational roots are ±1 , \(\frac{1}{3}\)

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

=> \(3x^{3}- x^{2} – 3x + 1\) = (x – 1)( 3x2 + 2x – 1)

Now,

3x2 + 2x – 1 = 3x2 + 3x – x  – 1

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

= (3x – 1)(x + 1)

Hence , \(3x^{3}- x^{2} – 3x + 1\) = (x – 1) (3x – 1)(x + 1)

Q9. \(x^{3}- 23x^{2} + 142x – 120\)

Sol :

Let, f(x) = \(x^{3}- 23x^{2} + 142x – 120\)

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)

Q10. \(y^{3} – 7y + 6\)

Sol :

Given, f(y) = \(y^{3} – 7y + 6\)

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)

=> \(y^{3} – 7y + 6\) = 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

\(y^{3} – 7y + 6\) = (1)(y – 1)( y – 2)(y + 3)

\(y^{3} – 7y + 6\) = (y – 1)( y – 2)(y + 3)

Hence,  \(y^{3} – 7y + 6\) = (y – 1)( y – 2)(y + 3)

Q11. \(x^{3} – 10x^{2} – 53x – 42\)

Sol :

Given , f(x) = \(x^{3} – 10x^{2} – 53x – 42\)

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. \(y^{3} – 2y^{2} – 29y – 42\)

Sol :

Given, f(x) =  \(y^{3} – 2y^{2} – 29y – 42\)

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. \(2y^{3} – 5y^{2} – 19y + 42\)

Sol :

Given, f(x) =  \(2y^{3} – 5y^{2} – 19y + 42\)

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. \(x^{3} + 13x^{2} + 32x + 20\)

Sol:

Given, f(x) =  \(x^{3} + 13x^{2} + 32x + 20\)

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. \(x^{3} – 3x^{2} – 9x – 5\)

Sol :

Given, f(x) = \(x^{3} – 3x^{2} – 9x – 5\)

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. \(2y^{3} + y^{2} – 2y – 1\)

Sol :

Given , f(y) = \(2y^{3} + y^{2} – 2y – 1\)

The constant term is 2

The factors of 2 are ±1, ± \(\frac{1}{2}\)

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.  \(x^{3} – 2x^{2} – x + 2\)

Sol :

Let, f(x) = \(x^{3} – 2x^{2} – x +2\)

The  constant term is 2

The factors of 2 are ±1, ±\(\frac{1}{2}\)

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. \(x^{3} + 13x^{2} + 31x – 45\) given that x + 9 is a factor
  2. \(4x^{3} + 20x^{2} + 33x + 18\) given that 2x + 3 is a factor

Sol :

1. \(x^{3} + 13x^{2} + 31x – 45\) given that x + 9 is a factor

let, f(x) = \(x^{3} + 13x^{2} + 31x – 45\)

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. \(4x^{3} + 20x^{2} + 33x + 18\) given that 2x + 3 is a factor

let, f(x) =  \(4x^{3} + 20x^{2} + 33x + 18\)

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)