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**Q10. y ^{3}â€“7y+6**

**Sol :**

Given, f(y) = y^{3}â€“7y+6

The constant term in f(y) is 6

The factors are Â±1, Â±2, Â±3, Â±6

Let , y â€“ 1 = 0

â‡’ y = 1

f(1) = (1)^{3}â€“7(1)+6

= 1 â€“ 7 + 6

= 0

So, (y â€“ 1) is the factor of f(y)

Similarly, (y â€“ 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

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

â‡’ y^{3}â€“7y+6 = k(y â€“ 1)( y â€“ 2)(y + 3) â€”â€”â€”â€“ 1

Substitute k = 0 in eq 1

â‡’ 0 â€“ 0 + 6 = k(-1)(-2)(3)

â‡’ 6 = 6k

â‡’ k = 1

y^{3}â€“7y+6 = (1)(y â€“ 1)( y â€“ 2)(y + 3)

y^{3}â€“7y+6 = (y â€“ 1)( y â€“ 2)(y + 3)

Hence, y^{3}â€“7y+6 = (y â€“ 1)( y â€“ 2)(y + 3)

**Q11. x ^{3}â€“10x^{2}â€“53xâ€“42**

**Sol :**

Given , f(x) = x^{3}â€“10x^{2}â€“53xâ€“42

The constant in f(x) is -42

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

Let, x + 1 = 0

â‡’ x = -1

f(-1) = (âˆ’1)^{3}â€“10(âˆ’1)^{2}â€“53(âˆ’1)â€“42

= -1 â€“ 10 + 53 â€“ 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,

x^{2} â€“ 11x â€“ 42

x + 1 x^{3} â€“ 10x^{2} â€“ 53x â€“ 42

x^{3 } + x^{2}

(-) (-)

-11x^{2} â€“ 53x

-11x^{2} â€“ 11x

(+) (+)

-42x â€“ 42

-42x â€“ 42

(+) (+)

0

â‡’ x^{3} â€“ 10x^{2} â€“ 53x â€“ 42 = (x + 1) (x^{2} â€“ 11x â€“ 42)

Now,

x^{2} â€“ 11x â€“ 42 = x^{2} â€“ 14x + 3x â€“ 42

= x(x â€“ 14) + 3(x â€“ 14)

= (x + 3)(x â€“ 14)

Hence, x^{3} â€“ 10x^{2} â€“ 53x â€“ 42 = (x + 1) (x + 3)(x â€“ 14)

**Q12. y ^{3}â€“2y^{2}â€“29yâ€“42**

Sol :

Given, f(x) = y^{3}â€“2y^{2}â€“29yâ€“42

The constant in f(x) is -42

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

Let, y + 2 = 0

â‡’ y = -2

f(-2) = (âˆ’2)^{3}â€“2(âˆ’2)^{2}â€“29(âˆ’2)â€“42

= -8 -8 + 58 â€“ 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

y^{2} â€“ 4y â€“ 21

y + 2 y^{3} â€“ 2y^{2} â€“ 29y â€“ 42

y^{3} + 2y^{2}

(-) (-)

-4y^{2} â€“ 29y

-4y^{2} â€“ 8y

(+) (+)

-21y â€“ 42

-21y â€“ 42

(+) (+)

0

â‡’ y^{3} â€“ 2y^{2} â€“ 29y â€“ 42 = (y + 2) (y^{2} â€“ 4y â€“ 21)

Now,

y^{2} â€“ 4y â€“ 21 = y^{2} â€“ 7y + 3y â€“ 21

= y(y â€“ 7) +3(y â€“ 7)

= (y â€“ 7)(y + 3)

Hence, y^{3} â€“ 2y^{2} â€“ 29y â€“ 42 = (y + 2) (y â€“ 7)(y + 3)

**Q13. 2y ^{3}â€“5y^{2}â€“19y+42**

Sol :

Given, f(x) = 2y^{3}â€“5y^{2}â€“19y+42

The constant in f(x) is +42

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

Let, y â€“ 2 = 0

â‡’ y = 2

f(2) = 2(2)^{3}â€“5(2)^{2}â€“19(2)+42

= 16 â€“ 20 â€“ 38 + 42

= 0

So, (y â€“ 2) is the factor of f(y)

Now, divide f(y) with (y â€“ 2) to get other factors

By, long division method

2y^{2} â€“ y â€“ 21

y â€“ 2 2y^{3} â€“ 5y^{2} -19y + 42

2y^{3} â€“ 4y^{2}

(-) (+)

-y^{2} â€“ 19y

-y^{2} + 2y

(+) (-)

-21y + 42

-21y + 42

(+) (-)

0

â‡’ 2y^{3} â€“ 5y^{2} -19y + 42 = (y â€“ 2) (2y^{2} â€“ y â€“ 21)

Now,

2y^{2} â€“ y â€“ 21

The factors are (y + 3) (2y â€“ 7)

Hence, 2y^{3} â€“ 5y^{2} -19y + 42 = (y â€“ 2) (y + 3) (2y â€“ 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 Â±1, Â±2, Â±4, Â±5, Â±10, Â±20

Let, x + 1 = 0

â‡’ x = -1

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

= -1 + 13 â€“ 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

x^{2} + 12x + 20

x + 1 x^{3} + 13x^{2} +32x + 20

x^{3} + x^{2}

(-) (-)

12x^{2} + 32x

12x^{2} + 12x

(-) (-)

20x â€“ 20

20x â€“ 20

(-) (-)

0

â‡’ x^{3} + 13x^{2} +32x + 20 = (x + 1)( x^{2} + 12x + 20)

Now,

x^{2} + 12x + 20 = x^{2} + 10x + 2x + 20

= x(x + 10) + 2(x + 10)

The factors are (x + 10) and (x + 2)

Hence, x^{3} + 13x^{2} +32x + 20 = (x + 1)(x + 10)(x + 2)

**Q15. x ^{3}â€“3x^{2}â€“9xâ€“5**

Sol :

Given, f(x) = x^{3}â€“3x^{2}â€“9xâ€“5

The constant in f(x) is -5

The factors of -5 are Â±1, Â±5

Let, x + 1 = 0

â‡’ x = -1

f(-1) = (âˆ’1)^{3}â€“3(âˆ’1)^{2}â€“9(âˆ’1)â€“5

= -1 â€“ 3 + 9 â€“ 5

= 0

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

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

By, long division

x^{2} â€“ 4x â€“ 5

x + 1 x^{3} â€“ 3x^{2} â€“ 9x â€“ 5

x^{3} + x^{2}

(-) (-)

-4x^{2} â€“ 9x

-4x^{2} â€“ 4x

(+) (+)

-5x â€“ 5

-5x â€“ 5

(+) (+)

0

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

Now,

x^{2} â€“ 4x â€“ 5 = x^{2} â€“ 5x + x â€“ 5

= x(x â€“ 5) + 1(x â€“ 5)

The factors are (x â€“ 5) and (x + 1)

Hence, x^{3} â€“ 3x^{2} â€“ 9x â€“ 5 = (x + 1)(x â€“ 5)(x + 1)

**Q16. 2y3+y2â€“2yâ€“1**

Sol :

Given , f(y) = 2y^{3}+y^{2}â€“2yâ€“1

The constant term is 2

The factors of 2 are Â±1, Â± 1/2

Let, y â€“ 1= 0

â‡’ y = 1

f(1) = 2(1)^{3}+(1)^{2}â€“2(1)â€“1

= 2 + 1 â€“ 2 â€“ 1

= 0

So, (y â€“ 1) is the factor of f(y)

Divide f(y) with (y â€“ 1) to get other factors

By, long division

2y^{2} + 3y + 1

y â€“ 1 2y^{3} + y^{2 } â€“ 2y â€“ 1

2y^{3} â€“ 2y^{2}

(-) (+)

3y^{2} â€“ 2y

3y^{2} â€“ 3y

(-) (+)

y â€“ 1

y â€“ 1

(-) (+)

0

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

Now,

2y^{2} + 3y + 1 = 2y^{2} + 2y + y + 1

= 2y(y + 1) + 1(y + 1)

= (2y + 1) ( y + 1) are the factors

Hence, 2y^{3} + y^{2 } â€“ 2y â€“ 1 = (y â€“ 1) (2y + 1) ( y + 1)

**Q17. x ^{3}â€“2x^{2}â€“x+2**

Sol :

Let, f(x) = x^{3}â€“2x^{2}â€“x+2

The constant term is 2

The factors of 2 are Â±1, Â±12

Let, x â€“ 1= 0

â‡’ x = 1

f(1) = (1)^{3}â€“2(1)^{2}â€“(1)+2

= 1 â€“ 2 â€“ 1 + 2

= 0

So, (x â€“ 1) is the factor of f(x)

Divide f(x) with (x â€“ 1) to get other factors

By, long division

x^{2} â€“ x â€“ 2

x â€“ 1 x^{3} â€“ 2x^{2} â€“ y + 2

x^{3} â€“ x^{2}

(-) (+)

-x^{2} â€“ x

-x^{2} + x

(+) (-)

â€“ 2x + 2

â€“ 2x + 2

(+) (-)

0

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

Now,

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

= x(x â€“ 2) + 1(x â€“ 2)

=(x â€“ 2)(x + 1) are the factors

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

**Q18. Factorize each of the following polynomials :**

**1. x ^{3}+13x^{2}+31xâ€“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}+31xâ€“45 given that x + 9 is a factor

let, f(x) = x^{3}+13x^{2}+31xâ€“45

given that (x + 9) is the factor

divide f(x) with (x + 9) to get other factors

by , long division

x^{2} + 4x â€“ 5

x + 9 x^{3} + 13x^{2} + 31x â€“ 45

x^{3 } + 9x^{2}

(-) (-)

4x^{2} + 31x

4x^{2} + 36x

(-) (-)

-5x â€“ 45

-5x â€“ 45

(+) (+)

0

â‡’ x^{3} + 13x^{2} + 31x â€“ 45 = (x + 9)( x^{2} + 4x â€“ 5)

Now,

x^{2} + 4x â€“ 5 = x^{2} + 5x â€“ x â€“ 5

= x(x + 5) -1(x + 5)

= (x + 5) (x â€“ 1) are the factors

Hence, x^{3} + 13x^{2} + 31x â€“ 45 = (x + 9)(x + 5)(x â€“ 1)

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

let, f(x) = 4x3+20x^{2}+33x+18

given that 2x + 3 is a factor

divide f(x) with (2x + 3) to get other factors

by, long division

2x^{2} + 7x + 6

2x + 3 4x^{3 }+ 20x^{2} + 33x + 18

4x^{3} + 6x^{2}

(-) (-)

14x^{2} â€“ 33x

14x^{2} â€“ 21x

(-) (+)

12x + 18

12x + 18

(-) (-)

0

â‡’ 4x^{3 }+ 20x^{2} + 33x + 18 = (2x + 3) (2x^{2} + 7x + 6)

Now,

2x^{2} + 7x + 6 = 2x^{2} + 4x + 3x + 6

= 2x(x + 2) + 3(x + 2)

= (2x + 3)(x + 2) are the factors

Hence, 4x^{3 }+ 20x^{2} + 33x + 18 = (2x + 3)(2x + 3)(x + 2)

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