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USE CODE: SELF10

Chapter 6: Factorization of Polynomials Exercise – 6.5

Quesiton: 1

Using factor theorem, factorize of the polynomials:

x3 + 6x2 + 11x + 6

Solution:

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)

Quesiton: 2

Using factor theorem, factorize of the polynomials:

x3 + 2x2 – x – 2

Solution:

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)

Quesiton: 3

Using factor theorem, factorize of the polynomials:

x3 – 6x2 + 3x + 10

Solution:

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)

Quesiton: 4

Using factor theorem, factorize of the polynomials:

x4 –7x3 + 9x2 + 7x –10

Solution:

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)

Similarly, 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)

Quesiton: 5

Using factor theorem, factorize of the polynomials:

x4 – 2x3 – 7x2 + 8x + 12

Solution:

Given,

f(x) = x4 – 2x3 –7x2 + 8x + 12

The constant term f(x) is equal is 12

The 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 = 12K

=> 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)

Quesiton: 6

Using factor theorem, factorize of the polynomials:

x4 + 10x3 + 35x2 + 50x + 24

Solution:

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)

Quesiton: 7

Using factor theorem, factorize of the polynomials:

2x4–7x3–13x2 + 63x – 45

Solution:

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

± 1, ± 3, ± 5, ± 9, ± 15, ± 45, ± 1/2, ± 3/2, ± 5/2, ± 9/2, ± 15/2, ± 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)

=> 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

-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)

Quesiton: 8

Using factor theorem, factorize of the polynomials:

3x3 - x2 – 3x + 1

Solution:

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 the 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)

Quesiton: 9

Using factor theorem, factorize of the polynomials:

x3- 23x2 + 142x - 120

Solution:

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)

Quesiton: 10

Using factor theorem, factorize of the polynomials:

y3 – 7y + 6

Solution:

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)

Quesiton: 11

Using factor theorem, factorize of the polynomials:

x3 – 10x2 – 53x – 42

Solution:

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)

Quesiton: 12

Using factor theorem, factorize of the polynomials:

y3 – 2y2 – 29y – 42

Solution:

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)

Quesiton: 13

Using factor theorem, factorize of the polynomials:

2y3 – 5y2 – 19y + 42

Solution:

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)

Quesiton: 14

Using factor theorem, factorize of the polynomials:

x3 + 13x2 + 32x + 20

Solution:

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)

Quesiton: 15

Using factor theorem, factorize of the polynomials:

x3 – 3x2 – 9x – 5

Solution:

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)

Quesiton: 16

Using factor theorem, factorize of the polynomials:

2y3 + y2 – 2y – 1

Solution:

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)

Quesiton: 17

Using factor theorem, factorize of the polynomials:

x3 – 2x2 – x + 2

Solution:

Let, f(x) = x3 – 2x2 – x + 2

The constant term is 2

The factors of 2 are ±1, ± 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)

Quesiton: 18

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

Solution:

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)

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