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**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 Â±1, Â±2, Â±3, Â±6

Let, x + 1 = 0

â‡’ x = -1

Substitute the value of x in f(x)

f(-1) = (âˆ’1)^{3}+6(âˆ’1)^{2}+11(âˆ’1)+6

= -1 + 6 -11 + 6

= 12 â€“ 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.

âˆ´ f(x) = k(x + 1)(x + 2)(x + 3)

â‡’ x^{3}+6x^{2}+11x+6 = k(x + 1)(x + 2)(x + 3)

Substitute x = 0 on both the sides

â‡’ 0 + 0 + 0 + 6 = k(0 +1)(0 + 2)(0 + 3)

â‡’ 6 = k(1*2*3)

â‡’ 6 = 6k

â‡’ k = 1

Substitute k value in f(x) = k(x + 1)(x + 2)(x + 3)

â‡’ f(x) = (1)(x + 1)(x + 2)(x + 3)

â‡’ f(x) = (x + 1)(x + 2)(x + 3)

âˆ´ x^{3}+6x^{2}+11x+6 = (x + 1)(x + 2)(x + 3)

**Q2. x3+2x2â€“xâ€“2**

Sol:

Given, f(x) = x^{3}+2x^{2}â€“xâ€“2

The constant term in f(x) is -2

The factors of (-2) are Â±1, Â±2

Let , x â€“ 1 = 0

â‡’ x = 1

Substitute the value of x in f(x)

f(1) = (1)3+2(1)2â€“1â€“2

= 1 + 2 â€“ 1 â€“ 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.

âˆ´ f(x) = k(x â€“ 1)(x + 2)(x + 1)

x^{3}+2x^{2}â€“xâ€“2 = k(x â€“ 1)(x + 2)(x + 1)

Substitute x = 0 on both the sides

0 + 0 â€“ 0 â€“ 2 = k(-1)(1)(2)

â‡’ â€“ 2 = -2k

â‡’ k = 1

Substitute k value in f(x) = k(x â€“ 1)(x + 2)(x + 1)

f(x) = (1)(x â€“ 1)(x + 2)(x + 1)

â‡’ f(x) = (x â€“ 1)(x + 2)(x + 1)

So, x^{3}+2x^{2}â€“xâ€“2 = (x â€“ 1)(x + 2)(x + 1)

**Q3. x ^{3}â€“6x^{2}+3x+10**

**Sol:**

Let, f(x) = x^{3}â€“6x^{2}+3x+10

The constant term in f(x) is 10

The factors of 10 are Â±1, Â±2, Â±5, Â±10

Let , x + 1 = 0

â‡’ x = -1

Substitute the value of x in f(x)

f(-1) = (âˆ’1)^{3}â€“6(âˆ’1)^{2}+3(âˆ’1)+10

= -1 â€“ 6 â€“ 3 + 10

= 0

Similarly , the other factors (x â€“ 2) and (x â€“ 5) of f(x)

Since, f(x) is a polynomial having a degree 3, it cannot have more than three linear factors.

âˆ´ f(x) = k(x + 1)(x â€“ 2)(x â€“ 5 )

Substitute x = 0 on both sides

â‡’ x^{3}â€“6x^{2}+3x+10 = k(x + 1)(x â€“ 2)(x â€“ 5)

â‡’ 0 â€“ 0 + 0 + 10 = k(1)(-2)(-5)

â‡’ 10 = k(10)

â‡’ k = 1

Substitute k = 1 in f(x) = k(x + 1)(x â€“ 2)(x â€“ 5)

f(x) = (1)(x + 1)(x â€“ 2)(x â€“ 5)

so, x^{3}â€“6x^{2}+3x+10 = (x + 1)(x â€“ 2)(x â€“ 5)

**Q4. x ^{4}â€“7x^{3}+9x2+7xâ€“10**

**Sol:**

Given, f(x) = x^{4}â€“7x^{3}+9x^{2}+7xâ€“10

The constant term in f(x) is 10

The factors of 10 are Â±1, Â±2, Â±5, Â±10

Let , x â€“ 1 = 0

â‡’ x = 1

Substitute the value of x in f(x)

f(x) = 1^{4}â€“7(1)^{3}+9(1)^{2}+7(1)â€“10

= 1 â€“ 7 + 9 + 7 â€“ 10

= 10 â€“ 10

= 0

(x â€“ 1) is the factor of f(x)

Simarly, the other factors are (x + 1) ,(x â€“ 2) , (x â€“ 5)

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

So, f(x) = k(x â€“ 1)(x + 1)(x â€“ 2)(x â€“ 5)

â‡’ x^{4}â€“7x^{3}+9x^{2}+7xâ€“10 = k(x â€“ 1)(x + 1)(x â€“ 2)(x â€“ 5)

Put x = 0 on both sides

0 â€“ 0 + 0 â€“ 10 = k(-1)(1)(-2)(-5)

â€“ 10 = k(-10)

â‡’ k = 1

Substitute k = 1 in f(x) = k(x â€“ 1)(x + 1)(x â€“ 2)(x â€“ 5)

f(x) = (1)(x â€“ 1)(x + 1)(x â€“ 2)(x â€“ 5)

= (x â€“ 1)(x + 1)(x â€“ 2)(x â€“ 5)

So, x^{4}â€“7x^{3}+9x^{2}+7xâ€“10 = (x â€“ 1)(x + 1)(x â€“ 2)(x â€“ 5)

**Q5. x ^{4}â€“2x^{3}â€“7x^{2}+8x+12**

Sol:

Given , f(x) = x^{4}â€“2x^{3}â€“7x^{2}+8x+12

The constant term f(x) is equal is 12

Tha factors of 12 are Â±1, Â±2, Â±3, Â±4, Â±6, Â±12

Let, x + 1 = 0

â‡’ x = -1

Substitute the value of x in f(x)

f(-1) = (âˆ’1)^{4}â€“2(âˆ’1)^{3}â€“7(âˆ’1)^{2}+8(âˆ’1)+12

= 1 + 2 â€“ 7 â€“ 8 + 12

= 0

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

Similarly, (x + 2), (x â€“ 2), (x â€“ 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.

â‡’ f(x) = k(x + 1)(x + 2)(x â€“ 3)(x â€“ 2)

â‡’ x^{4}â€“2x^{3}â€“7x^{2}+8x+12 = k(x + 1)(x + 2)(x â€“ 3)(x â€“ 2)

Substitute x = 0 on both sides,

â‡’ 0 â€“ 0 â€“ 0 + 12 = k(1)(2)(-2)(-3)

â‡’ 12 = k12

â‡’ k = 1

Substitute k = 1 in f(x) = k(x â€“ 2)(x + 1)(x + 2)(x â€“ 3)

f(x) = (x â€“ 2)(x + 1)(x + 2)(x â€“ 3)

so, x^{4}â€“2x^{3}â€“7x^{2}+8x+12 = (x â€“ 2)(x + 1)(x + 2)(x â€“ 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 Â±1, Â±2, Â±3, Â±4, Â±6, Â±8, Â±12, Â±24

Let, x + 1 = 0

â‡’ x = -1

Substitute the value of x in f(x)

f(-1) = (âˆ’1)4+10(âˆ’1)3+35(âˆ’1)2+50(âˆ’1)+24

= 1 â€“ 10 + 35 â€“ 50 + 24

= 0

â‡’ (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.

â‡’ f(x) = k(x + 1)(x + 2)(x + 3)(x + 4)

â‡’ x^{4}+10x^{3}+35x^{2}+50x+24 = k(x + 1)(x + 2)(x + 3)(x + 4)

Substitute x = 0 on both sides

â‡’ 0 + 0 + 0 + 0 + 24 = k(1)(2)(3)(4)

â‡’ 24 = k(24)

â‡’ 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}â€“7x^{3}â€“13x^{2}+63xâ€“45**

Sol :

Given, f(x) = 2x^{4}â€“7x^{3}â€“13x^{2}+63xâ€“45

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

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

Let, x â€“ 1 = 0

â‡’ x= 1

f(1) = 2(1)^{4}â€“7(1)^{3}â€“13(1)^{2}+63(1)â€“45

= 2 â€“ 7 â€“ 13 + 63 â€“ 45

= 0

Let, x â€“ 3 = 0

â‡’ x = 3

f(3) = 2(3)^{4}â€“7(3)^{3}â€“13(3)^{2}+63(3)â€“45

= 162 â€“ 189 â€“ 117 + 189 â€“ 45

= 0

So, (x â€“ 1) and (x â€“ 3) are the roots of f(x)

â‡’ x^{2} â€“ 4x + 3 is the factor of f(x)

Divide f(x) with x^{2} â€“ 4x + 3 to get other three factors

By long division,

2x^{2} + x â€“ 15

x^{2} â€“ 4x + 3 2x^{4} â€“ 7x^{3} â€“ 13x^{2 } + 63x â€“ 45

2x^{4 } â€“ 8x^{3 } + 6x^{2}

(-) (+) (-)

x^{3} â€“ 19x^{2 } + 63x

x^{3 }â€“ 4x^{2} + 3x

(-) (+) (-)

â€“ 15x^{2} + 60x â€“ 45

(+) (-) (+)

0

â‡’ 2x^{4}â€“7x^{3}â€“13x^{2}+63xâ€“45 = (x^{2} â€“ 4x + 3)(2x^{2} + x â€“ 15)

â‡’ 2x^{4}â€“7x^{3}â€“13x^{2}+63xâ€“45 = (x â€“ 1) (x â€“ 3)(2x^{2} + x â€“ 15)

Now,

2x^{2} + x â€“ 15 = 2x^{2}+ 6x â€“ 5x â€“ 15

= 2x(x + 3) â€“ 5 (x + 3)

= (2x â€“ 5) (x + 3)

So, 2x^{4}â€“7x^{3}â€“13x^{2}+63xâ€“45 = (x â€“ 1)(x â€“ 3)(x + 3)(2x â€“ 5)

**Q8. 3x ^{3}âˆ’x2â€“3x+1**

Sol :

Given , f(x) = 3x3âˆ’x2â€“3x+1

The factors of constant term 1 is Â±1

The factors of the coefficient of x2 = 3

The possible rational roots are Â±1 ,1/3

Let, x â€“ 1 = 0

â‡’ x = 1

f(1) = 3(1)^{3}âˆ’(1)^{2}â€“3(1)+1

= 3 â€“ 1 â€“ 3 + 1

= 0

So, x â€“ 1 is tha factor of f(x)

Now, divide f(x) with (x â€“ 1) to get other factors

By long division method,

3x^{2} + 2x â€“ 1

x â€“ 1 3x^{3} â€“ x^{2 }â€“ 3x + 1

3x^{3} â€“ x^{2}

(-) (+)

2x^{2} â€“ 3x

2x^{2} â€“ 2x

(-) (+)

-x + 1

-x + 1

(+) (-)

0

â‡’ 3x^{3}âˆ’x^{2}â€“3x+1 = (x â€“ 1)( 3x^{2} + 2x â€“ 1)

Now,

3x^{2} + 2x â€“ 1 = 3x^{2} + 3x â€“ x â€“ 1

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

= (3x â€“ 1)(x + 1)

Hence , 3x^{3}âˆ’x^{2}â€“3x+1 = (x â€“ 1) (3x â€“ 1)(x + 1)

**Q9. x3âˆ’23x ^{2}+142xâ€“120**

**Sol :**

Let, f(x) = x^{3}âˆ’23x^{2}+142xâ€“120

The constant term in f(x) is -120

The factors of -120 are Â±1, Â±2, Â±3, Â±4, Â±5, Â±6, Â±8, Â±10, Â±12, Â±15, Â±20, Â±24, Â±30, Â±40, Â±60, Â±120

Let, x â€“ 1 = 0

â‡’ x = 1

f(1) = (1)^{3}âˆ’23(1)^{2}+142(1)â€“120

= 1 â€“ 23 + 142 â€“ 120

= 0

So, (x â€“ 1) is the factor of f(x)

Now, divide f(x) with (x â€“ 1) to get other factors

By long division,

x^{2} â€“ 22x + 120

x â€“ 1 x^{3} â€“ 23x^{2} + 142x â€“ 120

x^{3} â€“ x^{2}

(-) (+)

-22x^{2} + 142x

-22x^{2} + 22x

(+) (-)

120x â€“ 120

120x â€“ 120

(-) (+)

0

â‡’ x^{3} â€“ 23x^{2} + 142x â€“ 120 = (x â€“ 1) (x^{2} â€“ 22x + 120)

Now,

x^{2} â€“ 22x + 120 = x^{2} â€“ 10x â€“ 12x + 120

= x( x â€“ 10) â€“ 12( x â€“ 10)

= (x â€“ 10) (x â€“ 12)

Hence, x^{3} â€“ 23x^{2} + 142x â€“ 120 = (x â€“ 1) (x â€“ 10) (x â€“ 12