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USE CODE: SELF10

Chapter 7: Factorization Exercise – 7.7

Question: 1

Factories

x2 + 12x – 45

Solution:

To factories x2 + 12x – 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:

x2 + 12x – 45

= x2 – 3x + 15x – 45

= (x2 – 3x) + (15x – 45)

= x(x – 3) + 15(x – 3)

= (x – 3) (x + 15)

Question: 2

Factories

40 + 3x – x2

Solution:

We have:

40 + 3x – x2

= -(x2 – 3x – 40)

To factories (x2 – 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 – 8x, we get:

40 + 3x – x2 = -(x2 – 3x – 40)

= -(x2 + 5x – 8x – 40)

= - [(x2 + 5x) – (8x + 40)]

= - [x(x + 5) – 8 (x + 5)]

= -(x – 8)(x + 5)

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

Question: 3

Factories

a2 + 3a – 88

Solution:

To factories a2 + 3a – 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 – 8a, we get:

a2 + 3a – 88 = a2 + 11a – 8a – 88

= (a2 + 11a) – (8a + 88)

= a(a + 11 ) – 8(a + 11)

= (a – 8)(a + 11)

Question: 4

Factories

a2 – 14a – 51

Solution:

To factories a2 – 14a – 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 – 17a, we get:

a2 – 14a – 51 = a2 + 3a -17a – 51

= (a2 + 3a) – (17a + 51)

= a(a + 3) – 17(a + 3)

= (a – 17) (a + 3)

Question: 5

Factories

x2 + 14x + 45

Solution:

To factories x2 + 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:

x2 + 14x + 45 = x2 + 9x + 5x + 45

= (x2 + 9x) + (5x + 45)

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

= (x + 5)(x + 9)

Question: 6

Factories

x2 – 22x + 120

Solution:

To factories x2 – 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:

x2 - 22x + 12 = x2 – 12x – 10x + 120

= (x2 – 12x) + (-10x + 120)

= x(x – 12) – 10(x – 12)

= (x – 10)(x – 12)

Question: 7

Factories

x2 – 11x – 42

Solution:

To factories x2 – 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:

x2 - 11x – 42 = x2 - 14x + 3x – 42

= (x2 – 14x) + (3x – 42)

= x(x – 14) + 3(x – 14)

= (x – 14)(x + 3)

Question: 8

Factories

a2 – 2a – 3

Solution:

To factories a2 – 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 – a + 3a, we get:

a2 + 2a – 3 = a2 – a + 3a – 3

= (a2 – a) + (3a – 3)

= a(a – 1) + 3(a – 1)

Question: 9

Factories

a2 + 14a + 48

Solution:

To factories a2 + 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:

a2 + 14a + 48 = a2 + 8a + 6a +48

= (a2 + 8a) + (6a + 48)

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

= (a + 6)(a + 8)

Question: 10

Factories

x2 – 4x – 21

Solution:

To factories x2 – 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:

x2 – 4x – 21 = x2 – 7x + 3x – 21

= (x2 – 7x) + (3x – 21)

= x(x – 7) + 3(x – 7)

= (x – 7) (x + 3)

Question: 11

Factories

y2 + 5y – 36

Solution:

To factories y2 + 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:

y2 + 5y – 36 = y2 – 4y + 9y – 36

= (y2 – 4y) + (9y – 36)

= y(y – 4) + 9(y – 4)

= (y – 4)(y – 4)

Question: 12

Factories

(a2 - 54)2 - 36

Solution:

(a2 - 54)2 - 36

= (a2 - 5a)2 - 62

= [(a2 - 5a) - 6] [(a2 - 5a) + 6]

= (a2 - 5a - 6) (a2 - 5a + 6)

In order to factories a2 - 5a - 6, we will find two numbers p and q such that p + q = - 5 and pq = -6

Now,

(-6) + 1 = - 5 And (- 6) x 1 = - 6

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

a2 - 5a - 6 = a2 - 6a + a - 6

= (a2 - 6a) + (a - 6)

= a(a - 6) + (a -6)

= (a + 1) (a - 6)

Now, In order to factories a2 - 5a + 6, we will find two numbers p and q such that p + q = - 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:

a2 – 5a + 6 = a2 – 2a – 3a + 6

= (a2 – 2a) – (3a – 6)

= a (a – 2) – 3(a – 2)

= (a – 3) (a – 2)

∴ (a2 – 5a – 6) (a2 – 5a + 6)

= (a – 6) (a + 1) (a – 3) (a – 2)

= (a + 1) (a – 2) (a – 3) (a – 6)

Question: 13

Factories

(a + 7)(a – 10) + 16

Solution:

(a + 7)(a – 10) + 16

= a2 – 10a + 7a – 70 + 16

= a2 ­- 3a – 54

To factories a2 – 3a – 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 – 9a + 6a, we get:

a2 – 3a – 54 = a2 – 9a + 6a – 54

= (a2 - 9a) + (6a – 54)

= a(a – 9) + 6(a – 9)

= (a + 6)(a – 9)

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