## RD Sharma Solutions Class 8 Chapter 7 Exercise 7.7

Q1. x^{2} + 12x â€“ 45

Soln:

To factories x^{2} + 12x â€“ 45, we will find two numbers p and q such that p + q = 12 and pq = -45.

Now,

15 + (-3) = 12

And

15 x (-3) = -45

Splitting the middle term 12x in the given quadratic as -3x + 15x, we get:

x^{2} + 12x – 45

= x^{2} â€“ 3x + 15x â€“ 45

= (x^{2 }– 3x) + (15x – 45)

= x(x – 3) + 15(x – 3)

= (x – 3) (x + 15)

Q2. 40 + 3x â€“ x^{2}

Soln:

We have:

40 + 3x â€“ x^{2}

= -(x^{2} â€“ 3x – 40)

To factories (x^{2} â€“ 3x – 40), we fill find two number p and q such p + q = -3 and pq = -40

Now,

5 + (-8) = -3

And

5 x (-8) = -40

Splitting the middle term -3x in the given quadratic as 5x â€“ 8x, we get:

40 + 3x â€“ x^{2} = -(x^{2 }â€“ 3x – 40)

= -(x^{2 }+ 5x â€“ 8x â€“ 40)

= -[(x^{2} + 5x) â€“ (8x + 40)]

= – [x(x + 5) â€“ 8 (x + 5)]

= -(x â€“ 8)(x + 5)

= (x + 5)(-x + 8)

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Q3. a^{2 }+ 3a â€“ 88

Soln:

To factories a^{2 }+ 3a â€“ 88, we will find two numbers p and q such that p + q = 3 and pq = -88.

Now, 11 + (-8) = 3

And 11 x (-8) = -88

Splitting the middle term 3a in the given quadratic as 11a â€“ 8a, we get:

a^{2} + 3a â€“ 88 = a^{2 }+ 11a â€“ 8a â€“ 88

= (a^{2 }+ 11a) – (8a + 88)

= a(a + 11 ) â€“ 8(a + 11)

= (a – 8)(a + 11)

Q4. a^{2 }â€“ 14a â€“ 51

Soln:

To factories a^{2 }â€“ 14a â€“ 51, we will find two numbers p and q such that p + q = -14 and pq = -51

Now,

3 + (-17) = -14

and

3 x (-17) = -51

Splitting the middle term -14a in the given quadratic as 3a â€“ 17a, we get:

a^{2} â€“ 14a â€“ 51 = a^{2 }+ 3a -17a â€“ 51

= (a^{2} + 3a) â€“ (17a + 51)

= a(a + 3) â€“ 17(a + 3)3

= (a â€“ 17 ) ( aÂ + 3)

Q5. x^{2} + 14x + 45

Soln:

To factories x^{2} + 14x + 45, we will find two numbers p and q such that p + q = 14 and pq = 45

Now,

9 + 5 = 14

And

9 x 5 = 45

Splitting the middle term 14x in the given quadratic as 9x + 5x, we get:

x^{2} + 14x + 45 = x^{2} + 9x + 5x + 45

= (x^{2 }+ 9x) + (5x + 45)

= x(x + 9) + 5(x + 9)

= (x + 5)(x + 9)

Q6. x^{2} â€“ 22x + 120

Soln:

To factories x^{2} â€“ 22x + 120, we will find two numbers p and q such that p + q = -22 and pq = 120

Now, (-12) + (-10) = -22

And

(-12) x (-10) = 120

Splitting the middle term -22x in the given quadratic as -12x – 10x, we get:

x^{2} â€“ 22x + 12 = x^{2 }â€“ 12x â€“ 10x + 120

= (x^{2 }– 12x ) + (-10x + 120)

= x(x – 12) – 10(x – 12)

= (x â€“ 10 )(x â€“ 12)

Q7. x^{2} – 11x – 42

Soln:

To factories x^{2} – 11x – 42, we will find two numbers p and q such that p + q = -11 and pq = -42

Now,

3 + (-14) = -22

And

3 x (-14) = 42

Splitting the middle term -11x in the given quadratic as -14x + 3x, we get:

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

= (x^{2 }â€“ 14x) + (3x – 42)

= x(x – 14) + 3(x â€“ 14)

= (x – 14)(x + 3)

Q8. a^{2} â€“ 2a â€“ 3

Soln:

To factories a^{2} â€“ 2a – 3, we will find two numbers p and q such that p + q = 2 and pq = -3

Now,

3 + (-1) = 2

And

3 x (-1) = -3

Splitting the middle terms 2a in the given quadratic as â€“a + 3a, we get:

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

= (a^{2 }– a) + (3a – 3)

= a(a – 1) + 3(a – 1)

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Q9. a^{2} + 14a + 48

Soln:

To factories a^{2} + 14a + 48, we will find two numbers p and q such that p + q = 14 and pq = 48

Now,

8 + 6 = 14

And

8 x 6 = 48

Splitting the middle terms 14a in the given quadratic as 8a + 6a, we get:

a^{2} + 14a + 48 = a^{2} + 8a + 6a +48

= (a^{2} + 8a) + (6a + 48)

= a(a + 8) + 6(a + 8)

= (a + 6)(a + 8)

Q10. x^{2} â€“ 4x â€“ 21

Soln:

To factories x^{2} â€“ 4x – 21, we will find two numbers p and q such that p + q = -4 and pq = -21

Now,

3 + (-7) = -4

And

3 x (-7) = -21

Splitting the middle terms -4x in the given quadratic as -7x + 3x, we get:

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

= (x^{2} â€“ 7x) + (3x – 21)

= x(x – 7) + 3(x – 7)

= (x – 7) (x + 3)

Q11. y^{2} + 5y â€“ 36

Soln:

To factories y^{2} + 5y – 36, we will find two numbers p and q such that p + q = 5 and pq = -36

Now,

9 + (-4) = 5

And

9 x (-4) = -36

Splitting the middle terms 5y in the given quadratic as -7y + 9y, we get:

y^{2 }+ 5y â€“ 36 = y^{2} â€“ 4y + 9y â€“ 36

= (y^{2} â€“ 4y) + (9y – 36)

= y(y – 4) + 9(y – 4)

= (y â€“ 4)(y – 4)

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Q12. (a^{2} â€” 54)^{2} â€” 36

Soln:

(a^{2} â€” 54)^{2} â€” 36

= (a^{2} â€” 5a)^{2} â€” 6^{2}

= [(a^{2} â€” 5a) â€” 6] [(a^{2 }– 5a) + 6]

= (a^{2} â€” 5a â€” 6) (a^{2} â€” 5a + 6)

In order to factories a^{2} â€” 5a â€” 6, we will find two numbers p and q such that p+ q = â€”5 and pq = â€”6

Now,

(-6) + 1 = â€”5

and

(-6) x 1 = â€”6

Splitting the middle term – 5 in the given quadratic as – 6a + a, we get :

a^{2} â€” 5a â€” 6 = a^{2} â€” 6a + a â€” 6

= (a^{2} â€” 6a) + (a â€” 6)

= a(a â€” 6) + (a â€” 6)

= (a + 1) (a â€” 6)

Now, In order to factories a^{2} â€” 5a + 6, we will find two numbers p and q such that p + q = â€”5 and pq = 6

Clearly,

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

and

(-2) x (-3) = 6

Splitting the middle term – 5 in the given quadratic as – 2a – 3a, we get :

a^{2} – 5a + 6 = a^{2} – 2a – 3a + 6

= (a^{2} – 2a) – (3a – 6)

= a (a – 2) – 3(a – 2)

= (a – 3) (a – 2)

âˆ´Â (a^{2} – 5a – 6) (a^{2} – 5a + 6)

= (a – 6) (a + 1) (a – 3) (a – 2)

= (a + 1) (a – 2) (a – 3) (a – 6)

Q13. (a + 7)(a – 10) + 16

Soln:

(a + 7)(a – 10) + 16

= a^{2 }â€“ 10a + 7a â€“ 70 + 16

= a^{2 }Â- 3a â€“ 54

To factories a^{2 }â€“ 3a â€“ 54, we will find two numbers p and q such that p + q = -3 and pq = -54

Now,

6 + (-9) = -3

And 6 x (-9) = -54

Splitting the middle term -3a in the given quadratic as â€“ 9a + 6a, we get:

a^{2 }â€“ 3a â€“ 54 = a^{2} â€“ 9a + 6a â€“ 54

= (a^{2 }-9a) + (6a – 54)

= a(a – 9) + 6(a – 9)

= (a + 6)(a – 9)