## RD Sharma Solutions Class 8 Chapter 7 Exercise 7.8

**Resolve each of the following quadratic equation trinomials into factors:**

**Q-1. Â 2x ^{2 }+ 5x + 3**

**Solution.** The given expression is 2x^{2 }+ 5x + 3.

(Co-efficient of x^{2 }= 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 x^{2} and the constant term, i.e., 2 Ã— 3 = 6.

Now,

2 + 3 = 5

And

2Â Ã— 3 = 6

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

2x^{2 }+ 5x + 3 = 2x^{2 }+ 2x + 3x + 3

= ( 2x^{2 }+ 2x ) + ( 3x + 3 )

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

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

**Q-2. 2x ^{2 }– 3x – 2**

**Solution.**

The given expression is 2x^{2 }– 3x – 2.

(Co-efficient of x^{2 }= 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 x^{2} and the constant term, i.e., 2 Ã— (-2) = -4

Now,

( -4 ) + 1 = -3

And

( -4 ) Ã— 1 = -4

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

2x^{2 }– 3x â€“ 2 = 2x^{2 }â€“ 4x + x â€“ 2

= ( 2x^{2 }â€“ 4x ) + ( x â€“ 2 )

= 2x( x â€“ 2 ) + 1( x- 2 )

= ( x â€“ 2 )( 2x + 1 )

**Q-3. 3x ^{2 }+ 10x + 3**

**Solution.**

The given expression is 3x^{2 }+ 10x + 3.

(Co-efficient of x^{2 }= 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 x^{2} and the constant term, i.e., 3 Ã— 3 = 9

Now,

9 + 1 = 10

And

9 Ã— 1 = 9

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

3x^{2 }+ 10x + 3 = 3x^{2 }+ 9x + x + 3

= ( 3x^{2 }+ 9x ) + ( x + 3 )

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

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

**Q-4. 7x â€“ 6 â€“ 2x ^{2}**

**Solution.**

The given expression is 7x â€“ 6 â€“ 2x^{2}.

(Co-efficient of x^{2 }= -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 x^{2} and the constant term, i.e., ( -2 ) Ã— ( -6 ) = 12

Now,

4 + 3 = 7

And

4 Ã— 3 = 12

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

7x â€“ 6 â€“ 2x^{2 }= -2x^{2 }+ 4x + 3x – 6

= ( -2x^{2 }+ 4x ) + ( 3x â€“ 6 )

= 2x( 2 – x ) – 3( 2 – x )

= ( 2x – 3 )( 2 – x )

**Q-5. 7x ^{2 }– 19x – 6**

**Solution.**

The given expression is 7x^{2 }– 19x – 6.

(Co-efficient of x^{2 }= 7, co-efficient of x = -19Â 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 x^{2} and the constant term, i.e., 7 Ã— ( -6 ) = 9

Now,

(-21) + 2 = -19

And

(-21) Ã— 2 = -42

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

7x^{2 }– 19x â€“ 6 = 7x^{2 }– 21x + 2x – 6

= ( 7x^{2 }– 21x ) + ( 2x – 6 )

= 7x( x – 3 ) + 2( x – 3 )

= ( x – 3 )( 7x + 2 )

**Â **

**Â **

**Q-6. Â 28 â€“ 31x â€“ 5x ^{2}**

**Solution.**

The given expression is 28 â€“ 31x â€“ 5x^{2}.

(Co-efficient of x^{2 }= -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 x^{2} and the constant term, i.e., ( -5 ) Ã— ( 28 ) = -140

Now,

( -35 ) + 4 = -31

And

( -35 ) Ã— 4 = -140

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

28 â€“ 31x â€“ 5x^{2 }= -5x^{2 }– 35x + 4x + 28

= ( -5x^{2 }– 35x ) + ( 4x + 28 )

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

= ( 4 â€“ 5x )( x + 7 )

**Q-7. Â 3 + 23y â€“ 8y ^{2}**

**Solution.**

The given expression is 3 + 23y â€“ 8y^{2}.

(Co-efficient of y^{2 }= -8, co-efficient of y = 23Â 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 x^{2} and the constant term, i.e., ( -8 ) Ã—Â 3 = -24

Now,

( -1 ) + 24 = 23

And

( -1 ) Ã— 24 = -24

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

3 + 23y â€“ 8y^{2 }= -8y^{2 }– y + 24y + 3

= ( -8y^{2 }– y ) + ( 24y + 3 )

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

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

**Q-8. 11x ^{2 }– 54x + 63**

**Solution.**

The given expression is 11x^{2 }– 54x + 63.

(Co-efficient of x^{2 }= 11, co-efficient of x = -54Â 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 x^{2} and the constant term, i.e., 11 Ã— 63 = 693

Now,

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

And

(-33) Ã— ( -21 ) = 693

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

11x^{2 }– 54x + 63 = 11x^{2 }– 33x – 21x + 63

= ( 11x^{2 }– 33x ) + ( -21x + 63 )

= 11x( x – 3 ) – 21( x – 3 )

= ( x – 3 )( 11x – 21 )

**Q-9. Â 7x – 6x ^{2 }+ 20**

**Solution.**

The given expression is 7x – 6x^{2 }+ 20.

(Co-efficient of x^{2 }= -6, co-efficient of x = 7Â 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 x^{2} and the constant term, i.e., ( -6 ) Ã— 20 = -120

Now,

( 15 ) + ( -8 )= 7

And

( 15 ) Ã— ( -8 ) = -120

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

7x – 6x^{2 }+ 20 = -6x^{2 }+ 15x – 8x + 20

= ( -6x^{2 }+ 15x ) + ( -8x + 20 )

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

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

**Q-10. Â 3x ^{2 }+ 22x + 35**

**Solution.**

The given expression is 3x^{2 }+ 22x + 35.

(Co-efficient of x^{2 }= 3, co-efficient of x = 22Â 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 x^{2} and the constant term, i.e., 3 Ã— 35 = 105

Now,

( 15 ) + ( 7 )= 22

And

( 15 ) Ã— ( 7 ) = 105

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

3x^{2 }+ 22x + 35 = 3x^{2 }+ 15x + 7x + 35

= ( 3x^{2 }+ 15x ) + ( 7x + 35 )

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

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

**Q-11. 12x ^{2 }– 17xy + 6y^{2}**

**Solution.**

The given expression is 12x^{2 }– 17xy + 6y^{2}.

(Co-efficient of x^{2 }= 12, co-efficient of x = -17yÂ and the constant term = 6y^{2 })

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 x^{2} and the constant term, i.e., 12 Ã— 6y^{2} = 72y^{2}

Now,

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

And

( -9y ) Ã— ( -8y ) = 72y^{2}

Replacing the middle term -17xy by -9xy â€“ 8xy, we have:

12x^{2 }– 17xy + 6y^{2 }= 12x^{2 }– 9xy â€“ 8xy + 6y^{2}

= (12x^{2 }– 9xy ) – (8xy + 6y^{2 })

= 3x( 4x â€“ 3y ) -2y( 4x â€“ 3y )

= (4x â€“ 3y )( 3x â€“ 2y )

**Q-12. 6x ^{2 }– 5xy – 6y^{2}**

**Solution.** The given expression is 6x^{2 }– 5xy – 6y^{2}.

(Co-efficient of x^{2 }= 6, co-efficient of x = -5yÂ and the constant term = -6y^{2 })

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 x^{2} and the constant term, i.e., 6 Ã— ( -6y^{2} ) = -36y^{2}

Now,

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

And

( -9y ) Ã— ( 4y ) = -36y^{2}

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

6x^{2 }– 5xy – 6y^{2 }= 6x^{2 }– 9xy + 4xy – 6y^{2}

= (6x^{2 }– 9xy ) + ( 4xy – 6y^{2 })

= 3x( 2x â€“ 3y ) + 2y( 2x â€“ 3y )

= (2x â€“ 3y )( 3x + 2y )

**Q-13. Â 6x ^{2 }– 13xy + 2y^{2}**

**Solution.**

The given expression is 6x^{2 }– 13xy + 2y^{2}.

(Co-efficient of x^{2 }= 6, co-efficient of x = -13yÂ and the constant term = 2y^{2 })

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 x^{2} and the constant term, i.e., 6 Ã— ( 2y^{2} ) = 12y^{2}

Now,

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

And

( -12y ) Ã— ( -y ) = 12y^{2}

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

6x^{2 }– 13xy + 2y^{2 }= 6x^{2 }– 12xy â€“ xy + 2y^{2}

= (6x^{2 }– 12xy ) – ( xy – 2y^{2 })

= 6x( x â€“ 2y ) – y( x â€“ 2y )

= (x â€“ 2y )( 6x – y )

**Q-14. 14x ^{2 }+ 11xy – 15y^{2}**

**Solution. **

The given expression is 14x^{2 }+ 11xy – 15y^{2}.

(Co-efficient of x^{2 }= 14, co-efficient of x = 11yÂ and the constant term = -15y^{2 })

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 x^{2} and the constant term, i.e., 14 Ã— ( -15y^{2} ) = -210y^{2}

Now,

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

And

( 21y ) Ã— ( -10y ) = -210y^{2}

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

14x^{2 }+ 11xy – 15y^{2 }= 14x^{2 }– 10xy + 21xy – 15y^{2}

= (14x^{2 }– 10xy ) + ( 21 xy – 15y^{2 })

= 2x( 7x â€“ 5y ) + 3y( 7x â€“ 5y )

= (7x â€“ 5y )( 2x + 3y )

**Q-15. 6a ^{2 }+ 17ab â€“ 3b^{2}**

**Solution.**

The given expression is 6a^{2 }+ 17ab â€“ 3b^{2}.

(Co-efficient of a^{2 }= 6, co-efficient of a = 17bÂ and the constant term = -3b^{2 })

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 a^{2} and the constant term, i.e., 6 Ã— ( -3b^{2} ) = -18b^{2}

Now,

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

And

( 18b ) Ã— ( -b ) = -18b^{2}

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

6a^{2 }+ 17ab â€“ 3b^{2 }= 6a^{2 }-ab + 18ab â€“ 3b^{2}

^{Â Â Â Â Â Â Â Â Â Â Â Â }Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = (6a^{2 }– ab ) + ( 18ab â€“ 3b^{2 })

= a( 6a â€“ b ) + 3b( 6a â€“ b )

= ( a + 3b )( 6a – b )

**Q-16. 36a ^{2 }+ 12abc â€“ 15b^{2}c^{2}**

**Solution. **

The given expression is 36a^{2 }+ 12abc â€“ 15b^{2}c^{2}.

(Co-efficient of a^{2 }= 36, co-efficient of a = 12bcÂ and the constant term = -15b^{2 }c^{2})

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 a^{2} and the constant term, i.e., 36 Ã— ( -15b^{2} c^{2}) = -540b^{2}c^{2}

Now,

( -18bc ) + 30bc= 12bc

And

( -18bc ) Ã— ( 30bc ) = -540b^{2} c^{2}

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

36a^{2 }+ 12abc â€“ 15b^{2}c^{2 }= 36a^{2 }-18abc + 30abc â€“ 15b^{2}c^{2}

= (36a^{2 }-18abc ) + (30abc â€“ 15b^{2}c^{2 })

= 18a( 2a â€“ bc ) + 15bc( 2a â€“ bc )

= 3( 6a + 5bc )( 2a – bc )

**Â **

**Â **

**Q-17. 15x ^{2 }â€“ 16xyz â€“ 15y^{2}z^{2}**

**Solution.**

The given expression is 15x^{2 }â€“ 16xyz â€“ 15y^{2}z^{2}.

(Co-efficient of x^{2 }= 15, co-efficient of x = -16yzÂ and the constant term = -15y^{2 }z^{2})

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 x^{2} and the constant term, i.e., 15 Ã— ( -15y^{2} z^{2}) = -225y^{2}z^{2}

Now,

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

And

( -25yz ) Ã— ( 9yz ) = -225y^{2} z^{2}

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

15x^{2 }â€“ 16xyz â€“ 15y^{2}z^{2 }= 15x^{2 }– 25xyz + 9xyz â€“ 15y^{2}z^{2}

= ( 15x^{2 }-25xyz ) + ( 9xyz â€“ 15y^{2}z^{2})

= 5x( 3x â€“ 5yz ) + 3yz( 3x â€“ 5yz )

= (3x â€“ 5yz )( 5x + 3yz )

**Q-18. ( x â€“ 2y ) ^{2} â€“ 5( x â€“ 2y ) + 6**

**Solution.**

The given expression is a^{2} â€“ 5a + 6.

Assuming a = x â€“ 2y, we have:

( x â€“ 2y )^{2} â€“ 5( x â€“ 2y ) + 6 = a^{2} â€“ 5a + 6

(Co-efficient of a^{2} = 1, co-efficient of a = -5 and the constant term = 6)

Now, we will split the co-efficient ofÂ aÂ into two parts such that their sum is -5 and their product equals to the product of the co-efficient of a^{2} and the constant term, i.e.,Â 1 Ã— 6 = 6.

Clearly,

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

And,

( -2 ) Ã— ( -3 ) = 6

Replacing the middle term -5a by -2a â€“ 3a, we have:

a^{2} â€“ 5a + 6 = a^{2} â€“ 2a â€“ 3a + 6

= (a^{2} â€“ 2a ) â€“ ( 3a â€“ 6 )

= a( a â€“ 2 ) â€“ 3 ( a â€“ 2 )

= ( a â€“ 2 )( a â€“ 3 )

Replacing a by ( x â€“ 2y ), we get:

( a â€“ 3 )( a â€“ 2 ) = ( x â€“ 2y â€“ 3 )( x â€“ 2y â€“ 2 )

**Q-19. ( 2a â€“ b ) ^{2} + 2( 2a â€“ b ) â€“ 8**

**Solution.**

Assuming x = 2a â€“ b, we have:

( 2a â€“ b )^{2} + 2( 2a â€“ b ) â€“ 8 = x^{2} + 2x â€“ 8

The given expression becomes x^{2} + 2x â€“ 8

(Co-efficient of x^{2} = 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 x^{2} and the constant term, i.e., 1 Ã— ( -8 ) = -8

Clearly,

( -2 ) + 4 = 2

And,

( -2 ) Ã— 4 = -8

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

x^{2} + 2x â€“ 8 = x^{2} – 2x + 4x â€“ 8

= ( x^{2} – 2x ) + ( 4x â€“ 8 )

= x( x â€“ 2 ) + 4 ( x â€“ 2 )

= ( x â€“ 2 )( x + 4 )

Replacing x by 2a â€“ b, we get:

( x + 4 )( x â€“ 2 ) = ( 2a â€“ b + 4 )( 2a â€“ b â€“ 2 )