RD Sharma Solutions Ex-6.5, (Part -2), Factorization Of Polynomials, Class 9, Maths Class 9 Notes | EduRev

RD Sharma Solutions for Class 9 Mathematics

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Class 9 : RD Sharma Solutions Ex-6.5, (Part -2), Factorization Of Polynomials, Class 9, Maths Class 9 Notes | EduRev

The document RD Sharma Solutions Ex-6.5, (Part -2), Factorization Of Polynomials, Class 9, Maths Class 9 Notes | EduRev is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.
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Q10. y3–7y+6

Sol :

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)


Q11. x3–10x2–53x–42

Sol :

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

x + 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)


Q12. y3–2y2–29y–42

Sol :

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)

Q13. 2y3–5y2–19y+42

Sol :

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)


Q14. x3+13x2+32x+20

Sol:

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)


Q15. x3–3x2–9x–5

Sol :

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)


Q16. 2y3+y2–2y–1

Sol :

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 + y – 2y – 1

2y3 – 2y2

(-)     (+)

3y2 – 2y

3y2 – 3y

(-)       (+)

y – 1

y – 1

(-)   (+)

0

⇒ 2y3 + y – 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 + y – 2y – 1 = (y – 1) (2y + 1) ( y + 1)


Q17.  x3–2x2–x+2

Sol :

Let, f(x) = x3–2x2–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

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)

Q18.  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

Sol :

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

x + 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    4x+ 20x2 + 33x + 18

4x3 + 6x2

(-)      (-)

14x2 – 33x

14x2 – 21x

(-)         (+)

12x  + 18

12x  + 18

(-)           (-)

0

⇒ 4x+ 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, 4x+ 20x2 + 33x + 18 = (2x + 3)(2x + 3)(x + 2)

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