RD Sharma Solutions Class 9 Factorization Of Polynomials Exercise 6.5

RD Sharma Class 9 Solutions Chapter 6 Ex 6.5 Free Download

RD Sharma Solutions Class 9 Chapter 6 Ex 6.5

Using factor theorem, factorize each of the following polynomials :

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

Sol:

As it is given that  polynomial, f(x) = \(x^{3} + 6x^{2} + 11x + 6\)

and the constant term in f(x) is 6

So, the factors of 6 are ±1, ±2, ±3, ±6

Now let, x + 1 = 0

=> x = -1

Put 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)

Likewise, (x + 2) and (x + 3) are the factors of f(x) too.

As, f(x) is a polynomialwhich have a degree 3, it can’t have more than 3 linear factors.

\(∴\) f(x) = k(x + 1)(x + 2)(x + 3)

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

Put 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

Put 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)

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

Q2. \(x^{3} + 2x^{2} – x – 2\)

Sol:

As it is given that, f(x) = \(x^{3} + 2x^{2} – x – 2\)

and the constant term in f(x) is -2

So, the factors of (-2) are ±1, ±2

Let , x – 1 = 0

=> x = 1

Put the value of x in f(x)

f(1) = \((1)^{3} + 2(1)^{2} – 1 – 2\)

= 1 + 2 – 1 – 2

= 0

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

As, f(x) is a polynomial having a degree 3, it can’t have more than 3 linear factors.

\(Therefore,\) f(x) = k(x – 1)(x + 2)(x + 1 )

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

Put x = 0 on both the sides

0 + 0 – 0 – 2 = k(-1)(1)(2)

=>     – 2 = -2k

=>       k = 1

Put 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 )

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

Q3. \(x^{3} – 6x^{2} +  3x + 10\)

Sol:

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

So, the constant term in f(x) is 10

The factors of 10 are ±1, ±2, ±5, ±10

Let us sassume , x + 1 = 0

=> x = -1

Put the value of x in f(x)

f(-1) = \((-1)^{3} – 6(-1)^{2} + 3(-1) + 10\)

= -1 – 6 – 3 + 10

= 0

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

As, f(x) is a polynomial having a degree 3, it can’t have more than 3 linear factors.

\(∴\) f(x) = k(x + 1)(x – 2)(x – 5 )

Put 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

Put 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:

As it is given that, f(x) = \(x^{4} – 7x^{3} + 9x^{2} + 7x – 10\)

And the constant term in f(x) is 10.

So, the factors of 10 are ±1, ±2, ±5, ±10

Now let , x – 1 = 0

=> x = 1

Put 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)

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

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

Then, 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

Put 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)

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

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

Sol:

As it is given that , f(x) = \(x^{4} – 2x^{3} – 7x^{2} + 8x + 12\)

As the constant term f(x) is equal is 12

and the factors of 12 are ±1, ±2, ±3, ±4, ±6, ±12

Let us assume, x + 1 = 0

=> x = -1

Put 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

and we get, x + 1 is factor of f(x)

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

As , 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)

Put x = 0 on both sides,

=> 0 – 0 – 0 + 12 = k(1)(2)(-2)(-3)

=> 12 = k12

=> k = 1

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

f(x) = (x – 2)(x + 1)(x + 2)(x – 3)

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

= (x – 2)(x + 1)(x + 2)(x – 3)

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

Sol:

As it is given that, f(x) = \(x^{4} + 10x^{3}  + 35x^{2} + 50x + 24\)

It is given that constant term in f(x) is equal to 24

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

Let us assume, x + 1 = 0

=> x = -1

Put 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)

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

As, f(x) is a polynomial of degree 4, it can’t have more than 4 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)

Put x = 0 on both sides

=> 0 + 0 + 0 + 0 + 24 = k(1)(2)(3)(4)

=> 24 = k(24)

=> k = 1

Put 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)

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

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

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

Sol :

As it is given that, f(x) = \(2x^{4} – 7x^{3}  –  13x^{2} + 63x – 45\)

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

The total 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 us assume, 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

Also, f(x) = (x-1) (2x3 – 5x2 -18x +45)

or, f(x) =(x-1)g(x)…… (a)

Let us assume, 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

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

Also, g(x) = 2x2 (x-3) + x(x-3) -15(x-3)

=(x-3)(2x2 + x – 15)

=(x-3)(2x2+6x – 5x -15)

= (x-3){(2x-5)(x+3)}

= (x-3)(x+3)(2x-5) ….(b)

From a and b, we get

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 :

As it is given that, 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 us suppose, x – 1 = 0

=> x = 1

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

= 3 – 1 – 3 + 1

= 0

hence, x – 1 is the factor of f(x)

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

Now as,

f(x) = 3x2(x-1) + 2x(x-1) – 1(x  – 1)

= (x-1) (3x2 + 2x – 1)

(x-1) (3x2 + 3x -x – 1)

= (3x – 1)(x + 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, f(x) = x2(x-1) – 22x(x-1) + 120(x-1)

(x-1)(x2 – 22x + 120) =(x-1)( x2 – 10x – 12x  + 120)

= (x-1){x( x – 10) – 12( x – 10)

= (x – 10) (x – 12)(x-1)

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

Q10. \(y^{3} – 7y + 6\)

Sol :

As it is given that, f(y) = \(y^{3} – 7y + 6\)

The constant term in f(y) is 6

The factors are ±1, ±2, ±3, ±6

Let us assume , 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)

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

Since, f(y) is a polynomial which has degree 3, it can’t 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

Put 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 :

As it is given that , f(x) = \(x^{3} – 10x^{2} – 53x – 42\)

And the constant in f(x) is -42

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

Let us assume, x + 1 = 0

=> x = -1

f(-1) = \((-1)^{3} – 10(-1)^{2} – 53(-1) – 42\)

= -1 – 10 + 53 – 42

= 0

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

Now,  f(x) = \(x^{3} – 10x^{2} – 53x – 42\) = (x2(x+1) – 11x(x+1) – 42(x+1))

(x+1)(x2 – 11x – 42) =(x+1)(x2 – 14x + 3x – 42)

=(x+1) {x(x – 14) + 3(x – 14)}

= (x+1)(x + 3)(x – 14)

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

Q12. \(y^{3} – 2y^{2} – 29y – 42\)

Sol :

As it is given that, 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 us assume, y + 2 = 0

=> y = -2

f(-2) =  \((-2)^{3} – 2(-2)^{2} – 29(-2) – 42\)

= -8 -8 + 58 – 42

= 0

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

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

(y+2)(y2 – 4y – 21) = (y+2)(y2 – 7y + 3y – 21)

=(y+2){ y(y – 7) +3(y – 7)}

= (y+2)(y – 7)(y + 3)

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

Q13. \(2y^{3} – 5y^{2} – 19y + 42\)

Sol :

As it is given that, 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 us sassume, y – 2 = 0

=> y = 2

f(2) = \(2(2)^{3} – 5(2)^{2} – 19(2) + 42\)

= 16 – 20 – 38 + 42

= 0

Hence, (y – 2) is the factor of f(y)

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

By, long division method

(y – 2)(2y2 – y – 21)

= (y – 2) (2y2 + 6y – 7y – 21)

=(y-2)(2y(y+3) – 7(y+3))

=(y-2)(2y-7)(y+3)

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:

As it is given that, 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 us assume, x + 1 = 0

=> x = -1

f(-1) =  \((-1)^{3} + 13(-1)^{2} + 32(-1) + 20\)

= -1 + 13 – 32 + 20

= 0

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

Now,

=> x3 + 13x2 +32x + 20 = (x + 1)( x2 + 12x + 20) =(x + 1)( x2 + 10x + 2x + 20) = (x + 1){ x(x + 10) + 2(x + 10)}

Hence the factors are x3 + 13x2 +32x + 20 = (x + 1)(x + 10)(x + 2)

Q15. \(x^{3} – 3x^{2} – 9x – 5\)

Sol :

As it is given that, f(x) = \(x^{3} – 3x^{2} – 9x – 5\)

The constant in f(x) is -5

The factors of -5 are ±1, ±5

Let us consider, x + 1 = 0

=> x = -1

f(-1) = \((-1)^{3} – 3(-1)^{2} – 9(-1) – 5\)

= -1 – 3 + 9 – 5

= 0

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

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

By, long division

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

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

Q16. \(2y^{3} + y^{2} – 2y – 1\)

Sol :

As it is given that, 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)

Now, f(x) m= (2y3 + y2  – 2y – 1)

=(y – 1 )(2y2 + 3y + 1)

=(y – 1 )(2y2 + 2y + y + 1) =(y – 1 ){2y(y + 1) + 1(y + 1)} = (y – 1) (2y + 1) ( y + 1)

Hence,  2y3 + y2  – 2y – 1 = (y – 1) (2y + 1) ( y + 1) are the factors

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)

Now,

f(x) = x3 – 2x2 – y + 2

= (x – 1)(x2 – x – 2) = (x – 1) (x2 – 2x + x  – 2) = (x – 1) (x2 – 2x + x  – 2) = (x – 1) {x(x – 2) + 1(x – 2)} = (x – 1)(x + 1)(x – 2)

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

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) = \(x^{3} + 13x^{2} + 31x – 45\)

As it is given that that (x + 9) is the factor

So, x3 + 13x2 + 31x – 45 = (x + 9)(x2 + 4x – 5) =(x + 9)(x2 + 5x – x  – 5) = (x + 9){(x(x + 5) -1(x + 5)} = (x + 9)(x + 5)(x – 1)

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

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

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

As it is given that that 2x + 3 is a factor

f(x) = 4x3 + 20x2 + 33x + 18 = (2x + 3) (2x2 + 7x + 6) = (2x + 3) ( 2x2 + 4x + 3x + 6) = (2x + 3) {2x(x + 2) + 3(x + 2)} = (2x + 3)(2x + 3)(x + 2)

Hence, 4x3 + 20x2 + 33x + 18 = (2x + 3)(2x + 3)(x + 2) are the factors

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