RD Sharma Solutions Class 8 Factorization Exercise 7.8

RD Sharma Solutions Class 8 Chapter 7 Exercise 7.8

RD Sharma Class 8 Solutions Chapter 7 Ex 7.8 PDF Download

Resolve each of the following quadratic equation trinomials into factors:

Q-1.  2x2 + 5x + 3

Solution. The given expression is 2x2 + 5x + 3.

(Co-efficient of x2 = 2, co-efficient of x = 5 and the constant term = 3)

We will split the co-efficient of x into two parts such that their sum is 5 and their product equals to the product of the co-efficient of x2 and the constant term, i.e., 2 × 3 = 6.

Now,

2 + 3 = 5

And

2  × 3 = 6

Replacing the middle term 5x by 2x + 3x, we have:

2x2 + 5x + 3 = 2x2 + 2x + 3x + 3

= ( 2x2 + 2x ) + ( 3x + 3 )

= 2x ( x + 1 ) + 3 ( x + 1 )

= ( 2x + 3 )( x + 1 )

Q-2. 2x2 – 3x – 2

Solution.

The given expression is 2x2 – 3x – 2.

(Co-efficient of x2 = 2, co-efficient of x = -3 and the constant term = -2)

We will split the co-efficient of x into two parts such that their sum is -3 and their product equals to the product of the co-efficient of x2 and the constant term, i.e., 2 × (-2) = -4

Now,

( -4 ) + 1 = -3

And

( -4 ) × 1 = -4

Replacing the middle term 3x by -4x + x, we have:

2x2 – 3x – 2 = 2x2 – 4x + x – 2

= ( 2x2 – 4x ) + ( x – 2 )

= 2x( x – 2 ) + 1( x- 2 )

= ( x – 2 )( 2x + 1 )

Q-3. 3x2 + 10x + 3

Solution.

The given expression is 3x2 + 10x + 3.

(Co-efficient of x2 = 3, co-efficient of x = 10 and the constant term = 3)

We will split the co-efficient of x into two parts such that their sum is 10 and their product equals to the product of the co-efficient of x2 and the constant term, i.e., 3 × 3 = 9

Now,

9 + 1 = 10

And

9 × 1 = 9

Replacing the middle term 10x by 9x + x, we have:

3x2 + 10x + 3 = 3x2 + 9x + x + 3

= ( 3x2 + 9x ) + ( x + 3 )

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

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

Q-4. 7x – 6 – 2x2

Solution.

The given expression is 7x – 6 – 2x2.

(Co-efficient of x2 = -2, co-efficient of x = 7 and the constant term = -6)

We will split the co-efficient of x into two parts such that their sum is 7 and their product equals to the product of the co-efficient of x2 and the constant term, i.e., ( -2 ) × ( -6 ) = 12

Now,

4 + 3 = 7

And

4 × 3 = 12

Replacing the middle term 7x by 4x + 3x, we have:

7x – 6 – 2x2 = -2x2 + 4x + 3x – 6

= ( -2x2 + 4x ) + ( 3x – 6 )

= 2x( 2 – x ) – 3( 2 – x )

= ( 2x – 3 )( 2 – x )

Q-5. 7x2 – 19x – 6

Solution.

The given expression is 7x2 – 19x – 6.

(Co-efficient of x2 = 7, co-efficient of x = -19  and the constant term = -6)

We will split the co-efficient of x into two parts such that their sum is -19 and their product equals to the product of the co-efficient of x2 and the constant term, i.e., 7 × ( -6 ) = 9

Now,

(-21) + 2 = -19

And

(-21) × 2 = -42

Replacing the middle term -19x by -21x + 2x, we have:

7x2 – 19x – 6 = 7x2 – 21x + 2x – 6

= ( 7x2 – 21x ) + ( 2x – 6 )

= 7x( x – 3 ) + 2( x – 3 )

= ( x – 3 )( 7x + 2 )

 

 

Q-6.  28 – 31x – 5x2

Solution.

The given expression is 28 – 31x – 5x2.

(Co-efficient of x2 = -5, co-efficient of x = -31 and the constant term = 28)

We will split the co-efficient of x into two parts such that their sum is -31 and their product equals to the product of the co-efficient of x2 and the constant term, i.e., ( -5 ) × ( 28 ) = -140

Now,

( -35 ) + 4 = -31

And

( -35 ) × 4 = -140

Replacing the middle term -31x by -35x + 4x, we have:

28 – 31x – 5x2 = -5x2 – 35x + 4x + 28

= ( -5x2 – 35x ) + ( 4x + 28 )

= -5x( x + 7 ) + 4( x + 7 )

= ( 4 – 5x )( x + 7 )

Q-7.  3 + 23y – 8y2

Solution.

The given expression is 3 + 23y – 8y2.

(Co-efficient of y2 = -8, co-efficient of y = 23  and the constant term = 3)

We will split the co-efficient of x into two parts such that their sum is 23 and their product equals to the product of the co-efficient of x2 and the constant term, i.e., ( -8 ) ×  3 = -24

Now,

( -1 ) + 24 = 23

And

( -1 ) × 24 = -24

Replacing the middle term 23y by -y + 24y, we have:

3 + 23y – 8y2 = -8y2 – y + 24y + 3

= ( -8y2 – y ) + ( 24y + 3 )

= -y( 8y + 1 ) + 3( 8y + 1 )

= (8y + 1 )( y + 3 )

Q-8. 11x2 – 54x + 63

Solution.

The given expression is 11x2 – 54x + 63.

(Co-efficient of x2 = 11, co-efficient of x = -54  and the constant term = 63)

We will split the co-efficient of x into two parts such that their sum is -19 and their product equals to the product of the co-efficient of x2 and the constant term, i.e., 11 × 63 = 693

Now,

( -33 ) + ( -21 )= -54

And

(-33) × ( -21 ) = 693

Replacing the middle term -54x by -33x – 21x, we have:

11x2 – 54x + 63 = 11x2 – 33x – 21x + 63

= ( 11x2 – 33x ) + ( -21x + 63 )

= 11x( x – 3 ) – 21( x – 3 )

= ( x – 3 )( 11x – 21 )

Q-9.  7x – 6x2 + 20

Solution.

The given expression is 7x – 6x2 + 20.

(Co-efficient of x2 = -6, co-efficient of x = 7  and the constant term = 20)

We will split the co-efficient of x into two parts such that their sum is -19 and their product equals to the product of the co-efficient of x2 and the constant term, i.e., ( -6 ) × 20 = -120

Now,

( 15 ) + ( -8 )= 7

And

( 15 ) × ( -8 ) = -120

Replacing the middle term 7x by 15x – 8x, we have:

7x – 6x2 + 20 = -6x2 + 15x – 8x + 20

= ( -6x2 + 15x ) + ( -8x + 20 )

= 3x( -2x + 5 ) + 4( -2x + 5 )

= (-2x + 5 )( 3x + 4 )

Q-10.  3x2 + 22x + 35

Solution.

The given expression is 3x2 + 22x + 35.

(Co-efficient of x2 = 3, co-efficient of x = 22  and the constant term = 35)

We will split the co-efficient of x into two parts such that their sum is -19 and their product equals to the product of the co-efficient of x2 and the constant term, i.e., 3 × 35 = 105

Now,

( 15 ) + ( 7 )= 22

And

( 15 ) × ( 7 ) = 105

Replacing the middle term 22x by 15x + 7x, we have:

3x2 + 22x + 35 = 3x2 + 15x + 7x + 35

= ( 3x2 + 15x ) + ( 7x + 35 )

= 3x( x + 5 ) + 7( x + 5 )

= ( x + 5 )( 3x + 7 )

Q-11. 12x2 – 17xy + 6y2

Solution.

The given expression is 12x2 – 17xy + 6y2.

(Co-efficient of x2 = 12, co-efficient of x = -17y  and the constant term = 6y2 )

We will split the co-efficient of x into two parts such that their sum is -17y and their product equals to the product of the co-efficient of x2 and the constant term, i.e., 12 × 6y2 = 72y2

Now,

( -9y ) + ( -8y )= -17y

And

( -9y ) × ( -8y ) = 72y2

Replacing the middle term -17xy by -9xy – 8xy, we have:

12x2 – 17xy + 6y2 = 12x2 – 9xy – 8xy + 6y2

= (12x2 – 9xy ) – (8xy + 6y2 )

= 3x( 4x – 3y ) -2y( 4x – 3y )

= (4x – 3y )( 3x – 2y )

Q-12. 6x2 – 5xy – 6y2

Solution. The given expression is 6x2 – 5xy – 6y2.

(Co-efficient of x2 = 6, co-efficient of x = -5y  and the constant term = -6y2 )

We will split the co-efficient of x into two parts such that their sum is -17y and their product equals to the product of the co-efficient of x2 and the constant term, i.e., 6 × ( -6y2 ) = -36y2

Now,

( -9y ) + ( 4y )= -5y

And

( -9y ) × ( 4y ) = -36y2

Replacing the middle term -5xy by -9xy + 4xy, we have:

6x2 – 5xy – 6y2 = 6x2 – 9xy + 4xy – 6y2

= (6x2 – 9xy ) + ( 4xy – 6y2 )

= 3x( 2x – 3y ) + 2y( 2x – 3y )

= (2x – 3y )( 3x + 2y )

Q-13.  6x2 – 13xy + 2y2

Solution.

The given expression is 6x2 – 13xy + 2y2.

(Co-efficient of x2 = 6, co-efficient of x = -13y  and the constant term = 2y2 )

We will split the co-efficient of x into two parts such that their sum is -13y and their product equals to the product of the co-efficient of x2 and the constant term, i.e., 6 × ( 2y2 ) = 12y2

Now,

( -12y ) + ( -y )= -13y

And

( -12y ) × ( -y ) = 12y2

Replacing the middle term -13xy by -12xy – xy, we have:

6x2 – 13xy + 2y2 = 6x2 – 12xy – xy + 2y2

= (6x2 – 12xy ) – ( xy – 2y2 )

= 6x( x – 2y ) – y( x – 2y )

= (x – 2y )( 6x – y )

Q-14. 14x2 + 11xy – 15y2

Solution.

The given expression is 14x2 + 11xy – 15y2.

(Co-efficient of x2 = 14, co-efficient of x = 11y  and the constant term = -15y2 )

We will split the co-efficient of x into two parts such that their sum is 11y and their product equals to the product of the co-efficient of x2 and the constant term, i.e., 14 × ( -15y2 ) = -210y2

Now,

( 21y ) + ( -10y )= 11y

And

( 21y ) × ( -10y ) = -210y2

Replacing the middle term -11xy by -10xy + 21 xy, we have:

14x2 + 11xy – 15y2 = 14x2 – 10xy + 21xy – 15y2

= (14x2 – 10xy ) + ( 21 xy – 15y2 )

= 2x( 7x – 5y ) + 3y( 7x – 5y )

= (7x – 5y )( 2x + 3y )

Q-15. 6a2 + 17ab – 3b2

Solution.

The given expression is 6a2 + 17ab – 3b2.

(Co-efficient of a2 = 6, co-efficient of a = 17b  and the constant term = -3b2 )

We will split the co-efficient of x into two parts such that their sum is 17b and their product equals to the product of the co-efficient of a2 and the constant term, i.e., 6 × ( -3b2 ) = -18b2

Now,

( 18b ) + ( -b )= 17b

And

( 18b ) × ( -b ) = -18b2

Replacing the middle term 17ab by -ab + 18ab, we have:

6a2 + 17ab – 3b2 = 6a2 -ab + 18ab – 3b2

                                  = (6a2 – ab ) + ( 18ab – 3b2 )

= a( 6a – b ) + 3b( 6a – b )

= ( a + 3b )( 6a – b )

Q-16. 36a2 + 12abc – 15b2c2

Solution.

The given expression is 36a2 + 12abc – 15b2c2.

(Co-efficient of a2 = 36, co-efficient of a = 12bc  and the constant term = -15b2 c2)

We will split the co-efficient of x into two parts such that their sum is 17b and their product equals to the product of the co-efficient of a2 and the constant term, i.e., 36 × ( -15b2 c2) = -540b2c2

Now,

( -18bc ) + 30bc= 12bc

And

( -18bc ) × ( 30bc ) = -540b2 c2

Replacing the middle term 12abc by -18abc + 30abc, we have:

36a2 + 12abc – 15b2c2 = 36a2 -18abc + 30abc – 15b2c2

= (36a2 -18abc ) + (30abc – 15b2c2 )

= 18a( 2a – bc ) + 15bc( 2a – bc )

= 3( 6a + 5bc )( 2a – bc )

 

 

Q-17. 15x2 – 16xyz – 15y2z2

Solution.

The given expression is 15x2 – 16xyz – 15y2z2.

(Co-efficient of x2 = 15, co-efficient of x = -16yz  and the constant term = -15y2 z2)

We will split the co-efficient of x into two parts such that their sum is -16yz and their product equals to the product of the co-efficient of x2 and the constant term, i.e., 15 × ( -15y2 z2) = -225y2z2

Now,

( -25yz ) + 9yz = -16yx

And

( -25yz ) × ( 9yz ) = -225y2 z2

Replacing the middle term -16xyz by -25xyz + 9xyz, we have:

15x2 – 16xyz – 15y2z2 = 15x2 – 25xyz + 9xyz – 15y2z2

= ( 15x2 -25xyz ) + ( 9xyz – 15y2z2)

= 5x( 3x – 5yz ) + 3yz( 3x – 5yz )

= (3x – 5yz )( 5x + 3yz )

Q-18. ( x – 2y )2 – 5( x – 2y ) + 6

Solution.

The given expression is a2 – 5a + 6.

Assuming a = x – 2y, we have:

( x – 2y )2 – 5( x – 2y ) + 6 = a2 – 5a + 6

(Co-efficient of a2 = 1, co-efficient of a = -5 and the constant term = 6)

Now, we will split the co-efficient of  a  into two parts such that their sum is -5 and their product equals to the product of the co-efficient of a2 and the constant term, i.e.,  1 × 6 = 6.

Clearly,

( -2 ) + ( -3 ) = -5

And,

( -2 ) × ( -3 ) = 6

Replacing the middle term -5a by -2a – 3a, we have:

a2 – 5a + 6 = a2 – 2a – 3a + 6

= (a2 – 2a ) – ( 3a – 6 )

= a( a – 2 ) – 3 ( a – 2 )

= ( a – 2 )( a – 3 )

Replacing a by ( x – 2y ), we get:

( a – 3 )( a – 2 ) = ( x – 2y – 3 )( x – 2y – 2 )

Q-19. ( 2a – b )2 + 2( 2a – b ) – 8

Solution.

Assuming x = 2a – b, we have:

( 2a – b )2 + 2( 2a – b ) – 8 = x2 + 2x – 8

The given expression becomes x2 + 2x – 8

(Co-efficient of x2 = 1 and that of x = 2; constant term = -8)

Now, we will split the co-efficient of x into two parts such that their sum is 2 and their product equals the product of the co-efficient of x2 and the constant term, i.e., 1 × ( -8 ) = -8

Clearly,

( -2 ) + 4 = 2

And,

( -2 ) × 4 = -8

Replacing the middle term 2x by -2x + 4x, we get:

x2 + 2x – 8 = x2 – 2x + 4x – 8

= ( x2 – 2x ) + ( 4x – 8 )

= x( x – 2 ) + 4 ( x – 2 )

= ( x – 2 )( x + 4 )

Replacing x by 2a – b, we get:

( x + 4 )( x – 2 ) = ( 2a – b + 4 )( 2a – b – 2 )