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1. Factor trinomials of the form ax² + bx + c

(i) 2a^{2} + 11a + 12

(ii) 3a^{2} + 8a + 4

(iii) 3a^{2} – 13a + 14

(iv) 4a^{2} – 7a + 3

(v) 5a^{2} – 11a – 12

(vi) 7a^{2} – 15a – 18

## Solution:

(i) The given expression 2a^{2} + 11a + 12

Rewrite the equation

2a^{2} + 8a + 3a + 12

Group the first two terms and last two terms.

The first two terms are 2a^{2} + 8a and the second two terms are 3a + 12.

Take 2a^{ }common from the first two terms.

2a (a + 4)

Take 3 common from the second two terms.

3 (a + 4)

2a (a + 4) + 3 (a + 4)

Then, take (a + 4) common from the above expression.

(a + 4) (2a + 3)

The final answer is (a + 4) (2a + 3).

(ii) The given expression 3a^{2} + 8a + 4

Rewrite the equation

3a^{2} + 6a + 2a + 4

Group the first two terms and last two terms.

The first two terms are 3a^{2} + 6a and the second two terms are 2a + 4.

Take 3a^{ }common from the first two terms.

3a (a + 2)

Take 2 common from the second two terms.

2 (a + 2)

3a (a + 2) + 2 (a + 2)

Then, take (a + 2) common from the above expression.

(a + 2) (3a + 2)

The final answer is (a + 2) (3a + 2).

(iii) The given expression 3a^{2} – 13a + 14

Rewrite the equation

3a^{2} – 6a – 7a + 14

Group the first two terms and last two terms.

The first two terms are 3a^{2} – 6a and the second two terms are – 7a + 14.

Take 3a^{ }common from the first two terms.

3a (a – 2)

Take -7 common from the second two terms.

-7 (a – 2)

3a (a – 2) – 7 (a – 2)

Then, take (a – 2) common from the above expression.

(a – 2) (3a – 7)

The final answer is (a – 2) (3a – 7).

(iv) The given expression 4a^{2} – 7a + 3

Rewrite the equation

4a^{2} – 4a – 3a + 3

Group the first two terms and last two terms.

The first two terms are 4a^{2} – 4a and the second two terms are – 3a + 3.

Take 4a^{ }common from the first two terms.

4a (a – 1)

Take -3 common from the second two terms.

-3 (a – 1)

4a (a – 1) – 3 (a – 1)

Then, take (a – 1) common from the above expression.

(a – 1) (4a – 3)

The final answer is (a – 1) (4a – 3).

(v) The given expression 5a^{2} – 11a – 12

Rewrite the equation

5a^{2} – 15a + 4a – 12

Group the first two terms and last two terms.

The first two terms are 5a^{2} – 15a and the second two terms are 4a – 12.

Take 4a^{ }common from the first two terms.

5a (a – 3)

Take 4 common from the second two terms.

4 (a – 3)

5a (a – 3) + 4 (a – 3)

Then, take (a – 3) common from the above expression.

(a – 3) (5a + 4)

The final answer is (a – 3) (5a + 4).

(vi) The given expression 7a^{2} – 15a – 18

Rewrite the equation

7a^{2} – 21a + 6a – 18

Group the first two terms and last two terms.

The first two terms are 7a^{2} – 21a and the second two terms are 6a – 18.

Take 7a^{ }common from the first two terms.

7a (a – 3)

Take 6 common from the second two terms.

6 (a – 3)

7a (a – 3) + 6 (a – 3)

Then, take (a – 3) common from the above expression.

(a – 3) (7a + 6)

The final answer is (a – 3) (7a + 6).

2. Factor trinomials

(i) 7a^{2} + 6a – 1

(ii) 9a^{2} + 35a – 4

(iii) 2a^{2} – 5a + 3

(iv) 7a – 6 – 2a^{2
}(v) 11a^{2} – 54a + 63

(vi) a^{2} + 2a – 3

## Solution:

(i) The given expression 7a^{2} + 6a – 1

Rewrite the equation

7a^{2} + 7a – a – 1

Group the first two terms and last two terms.

The first two terms are 7a^{2} + 7a and the second two terms are – a – 1.

Take 7a^{ }common from the first two terms.

7a (a + 1)

Take -1 common from the second two terms.

-1 (a + 1)

7a (a + 1) – 1 (a + 1)

Then, take (a + 1) common from the above expression.

(a + 1) (7a – 1)

The final answer is (a + 1) (7a – 1).

(ii) The given expression 9a^{2} + 35a – 4

Rewrite the equation

9a^{2} + 36a – a – 4

Group the first two terms and last two terms.

The first two terms are 9a^{2} + 36a and the second two terms are – a – 4.

Take 9a^{ }common from the first two terms.

9a (a + 4)

Take -1 common from the second two terms.

-1 (a + 4)

9a (a + 4) – 1 (a + 4)

Then, take (a + 4) common from the above expression.

(a + 4) (9a – 1)

The final answer is (a + 4) (9a – 1).

(iii) The given expression 2a^{2} – 5a + 3

Rewrite the equation

2a^{2} – 2a – 3a + 3

Group the first two terms and last two terms.

The first two terms are 2a^{2} – 2a and the second two terms are – 3a + 3.

Take 2a^{ }common from the first two terms.

2a (a – 1)

Take -3 common from the second two terms.

-3 (a – 1)

2a (a – 1) – 3 (a – 1)

Then, take (a – 1) common from the above expression.

(a – 1) (2a – 3)

The final answer is (a – 1) (2a – 3).

(iv) The given expression 7a – 6 – 2a^{2}

Rewrite the equation

– 2a^{2 }+ 4a + 3a – 6

Group the first two terms and last two terms.

The first two terms are – 2a^{2 }+ 4a and the second two terms are 3a – 6.

Take -2a^{ }common from the first two terms.

-2a (a – 2)

Take 3 common from the second two terms.

3 (a – 2)

-2a (a – 2) + 3 (a – 2)

Then, take (a – 2) common from the above expression.

(a – 2) (-2a + 3)

The final answer is (a – 2) (-2a + 3).

(v) The given expression 11a^{2} – 54a + 63

Rewrite the equation

11a^{2} – 33a – 21a + 63

Group the first two terms and last two terms.

The first two terms are 11a^{2} – 33a and the second two terms are – 21a + 63.

Take 11a^{ }common from the first two terms.

11a (a – 3)

Take -21 common from the second two terms.

-21 (a – 3)

11a (a – 3) – 21 (a – 3)

Then, take (a – 3) common from the above expression.

(a – 3) (11a – 21)

The final answer is (a – 3) (11a – 21).

(vi) The given expression a^{2} + 2a – 3

Rewrite the equation

a^{2} + 3a – a – 3

Group the first two terms and last two terms.

The first two terms are a^{2} + 3a and the second two terms are – a – 3.

Take a^{ }common from the first two terms.

a (a + 3)

Take -1 common from the second two terms.

-1 (a + 3)

a (a + 3) – 1 (a + 3)

Then, take (a + 3) common from the above expression.

(a + 3) (a – 1)

The final answer is (a + 3) (a – 1).

3. Factorize the quadratic expression

(i) 2x^{2} + 5x + 2

(ii) 3a^{2} + 14a + 8

(iii) 2x^{2} + 7x + 6

(iv) 6a^{2} – a – 15

(v) 9s^{2} – s – 8

(vi) 12 + a – 6a^{2
}(vii) 6 + 5x – 6x^{2
}(viii) a^{2} + 8a – 105

## Solution:

(i) The given expression 2x^{2} + 5x + 2

Rewrite the equation

2x^{2} + 4x + x + 2

Group the first two terms and last two terms.

The first two terms are 2x^{2} + 4x and the second two terms are x + 2.

Take 2x^{ }common from the first two terms.

2x (x + 2)

Take 1 common from the second two terms.

1 (x + 2)

2x (x + 2) + 1 (x + 2)

Then, take (x + 2) common from the above expression.

(x + 2) (2x + 1)

The final answer is (x + 2) (2x + 1).

(ii) The given expression 3a^{2} + 14a + 8

Rewrite the equation

3a^{2} + 12a + 2a + 8

Group the first two terms and last two terms.

The first two terms are 3a^{2} + 12a and the second two terms are 2a + 8.

Take 3a^{ }common from the first two terms.

3a (a + 4)

Take 2 common from the second two terms.

2 (a + 4)

3a (a + 4) + 2 (a + 4)

Then, take (a + 4) common from the above expression.

(a + 4) (3a + 2)

The final answer is (a + 4) (3a + 2).

(iii) The given expression 2x^{2} + 7x + 6

Rewrite the equation

2x^{2} + 4x + 3x + 6

Group the first two terms and last two terms.

The first two terms are 2x^{2} + 4x and the second two terms are 3x + 6.

Take 2x^{ }common from the first two terms.

2x (x + 2)

Take 3 common from the second two terms.

3 (x + 2)

2x (x + 2) + 3 (x + 2)

Then, take (x + 2) common from the above expression.

(x + 2) (2x + 3)

The final answer is (x + 2) (2x + 3).

(iv) The given expression 6a^{2} – a – 15

Rewrite the equation

6a^{2} + 9a -10a – 15

Group the first two terms and last two terms.

The first two terms are 6a^{2} + 9a and the second two terms are -10a – 15.

Take 3a^{ }common from the first two terms.

3a (2a + 3)

Take -5 common from the second two terms.

-5 (2a + 3)

3a (2a + 3) -5 (2a + 3)

Then, take (2a + 3) common from the above expression.

(2a + 3) (3a – 5)

The final answer is (2a + 3) (3a – 5).

(v) The given expression 9s^{2} – s – 8

Rewrite the equation

9s^{2} – 9s + 8s – 8

Group the first two terms and last two terms.

The first two terms are 9s^{2} – 9s and the second two terms are 8s – 8.

Take 9s^{ }common from the first two terms.

9s (s – 1)

Take 8 common from the second two terms.

8 (s – 1)

9s (s – 1) + 8 (s – 1)

Then, take (s – 1) common from the above expression.

(s – 1) (9s + 8)

The final answer is (s – 1) (9s + 8).

(vi) The given expression 12 + a – 6a^{2}

Rewrite the equation

– 6a^{2 }+ 9a -8a + 12

Group the first two terms and last two terms.

The first two terms are – 6a^{2 }+ 9a and the second two terms are -8a + 12 .

Take -3a^{ }common from the first two terms.

-3a (2a – 3)

Take -4 common from the second two terms.

-4 (2a – 3)

-3a (2a – 3) – 4 (2a – 3)

Then, take (2a – 3) common from the above expression.

(2a – 3) (-3a – 4)

The final answer is (2a – 3) (-3a – 4).

(vii) The given expression 6 + 5x – 6x^{2}

Rewrite the equation

– 6x^{2 }+ 9x – 4x +6

Group the first two terms and last two terms.

The first two terms are – 6x^{2 }+ 9x and the second two terms are – 4x +6.

Take -3x^{ }common from the first two terms.

-3x (2x – 3)

Take -2 common from the second two terms.

-2 (2x – 3)

-3x (2x – 3) -2 (2x – 3)

Then, take (2x – 3) common from the above expression.

(2x – 3) (-3x – 2)

The final answer is (2x – 3) (-3x – 2).

(viii) The given expression a^{2} + 8a – 105

Rewrite the equation

a^{2} + 15a – 7a – 105

Group the first two terms and last two terms.

The first two terms are a^{2} + 15a and the second two terms are – 7a – 105.

Take a^{ }common from the first two terms.

a (a + 15)

Take -7 common from the second two terms.

-7 (a + 15)

a (a + 15) – 7 (a + 15)

Then, take (a + 15) common from the above expression.

(a + 15) (a – 7)

The final answer is (a + 15) (a – 7).