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**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,

x^{2} – 11x – 42

x + 1 x^{3} – 10x^{2} – 53x – 42

x^{3 } + x^{2}

(-) (-)

-11x^{2} – 53x

-11x^{2} – 11x

(+) (+)

-42x – 42

-42x – 42

(+) (+)

0

⇒ x^{3} – 10x^{2} – 53x – 42 = (x + 1) (x^{2} – 11x – 42)

Now,

x^{2} – 11x – 42 = x^{2} – 14x + 3x – 42

= x(x – 14) + 3(x – 14)

= (x + 3)(x – 14)

Hence, x^{3} – 10x^{2} – 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

y^{2} – 4y – 21

y + 2 y^{3} – 2y^{2} – 29y – 42

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

(-) (-)

-4y^{2} – 29y

-4y^{2} – 8y

(+) (+)

-21y – 42

-21y – 42

(+) (+)

0

⇒ y^{3} – 2y^{2} – 29y – 42 = (y + 2) (y^{2} – 4y – 21)

Now,

y^{2} – 4y – 21 = y^{2} – 7y + 3y – 21

= y(y – 7) +3(y – 7)

= (y – 7)(y + 3)

Hence, y^{3} – 2y^{2} – 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

2y^{2} – y – 21

y – 2 2y^{3} – 5y^{2} -19y + 42

2y^{3} – 4y^{2}

(-) (+)

-y^{2} – 19y

-y^{2} + 2y

(+) (-)

-21y + 42

-21y + 42

(+) (-)

0

⇒ 2y^{3} – 5y^{2} -19y + 42 = (y – 2) (2y^{2} – y – 21)

Now,

2y^{2} – y – 21

The factors are (y + 3) (2y – 7)

Hence, 2y^{3} – 5y^{2} -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

x^{2} + 12x + 20

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

x^{3} + x^{2}

(-) (-)

12x^{2} + 32x

12x^{2} + 12x

(-) (-)

20x – 20

20x – 20

(-) (-)

0

⇒ 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}–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

x^{2} – 4x – 5

x + 1 x^{3} – 3x^{2} – 9x – 5

x^{3} + x^{2}

(-) (-)

-4x^{2} – 9x

-4x^{2} – 4x

(+) (+)

-5x – 5

-5x – 5

(+) (+)

0

⇒ x^{3} – 3x^{2} – 9x – 5 = (x + 1)( x^{2} – 4x – 5)

Now,

x^{2} – 4x – 5 = x^{2} – 5x + x – 5

= x(x – 5) + 1(x – 5)

The factors are (x – 5) and (x + 1)

Hence, x^{3} – 3x^{2} – 9x – 5 = (x + 1)(x – 5)(x + 1)

**Q16. 2y3+y2–2y–1**

Sol :

Given , f(y) = 2y^{3}+y^{2}–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

2y^{2} + 3y + 1

y – 1 2y^{3} + y^{2 } – 2y – 1

2y^{3} – 2y^{2}

(-) (+)

3y^{2} – 2y

3y^{2} – 3y

(-) (+)

y – 1

y – 1

(-) (+)

0

⇒ 2y^{3} + y^{2 } – 2y – 1 = (y – 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 } – 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, ±12

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

x^{2} – x – 2

x – 1 x^{3} – 2x^{2} – y + 2

x^{3} – x^{2}

(-) (+)

-x^{2} – x

-x^{2} + x

(+) (-)

– 2x + 2

– 2x + 2

(+) (-)

0

⇒ x^{3} – 2x^{2} – y + 2 = (x – 1) (x^{2} – x – 2)

Now,

x^{2} – x – 2 = x^{2} – 2x + x – 2

= x(x – 2) + 1(x – 2)

=(x – 2)(x + 1) are the factors

Hence, x^{3} – 2x^{2} – 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

x^{2} + 4x – 5

x + 9 x^{3} + 13x^{2} + 31x – 45

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

(-) (-)

4x^{2} + 31x

4x^{2} + 36x

(-) (-)

-5x – 45

-5x – 45

(+) (+)

0

⇒ x^{3} + 13x^{2} + 31x – 45 = (x + 9)( x^{2} + 4x – 5)

Now,

x^{2} + 4x – 5 = x^{2} + 5x – x – 5

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

= (x + 5) (x – 1) are the factors

Hence, x^{3} + 13x^{2} + 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) = 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} – 33x

14x^{2} – 21x

(-) (+)

12x + 18

12x + 18

(-) (-)

0

⇒ 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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- RD Sharma Solutions Ex-6.5, (Part -1), Factorization Of Polynomials, Class 9, Maths
- RD Sharma Solutions Ex-6.4, (Part -2), Factorization Of Polynomials, Class 9, Maths
- RD Sharma Solutions Ex-6.4, (Part -1), Factorization Of Polynomials, Class 9, Maths
- RD Sharma Solutions Ex-6.3, Factorization Of Polynomials, Class 9, Maths