**Question 1:** Solve for factors of x^{2} + 12x – 45

**Solution:**

Given: x

^{2}+ 12x – 45To factorize the following expression we have to find two numbers a and b such that a + b = 12 and ab = – 45

We know that :

15 + (– 3) = 12 and 15 * (-3) = – 45

Splitting the middle term i.e. 12x in the given quadratic equation we get:

x

^{2}+ 12x – 45 = x^{2}+ 15x – 3x – 45= x (x + 15) – 3 (x + 15) [Taking the similar terms common]

= (x – 3) (x + 15) [as (x + 15) is common]

**Question 2:** Solve for factors of 40 + 3x – x^{2}

**Solution:**

Given: 40 + 3x – x

^{2}40 + 3x – x

^{2 }= -(x^{2}– 3x – 40)To factorize the following expression we have to find two numbers a and b such that a+b = -3 and ab = – 40

We know that :

5 + (-8) = -3 and 5 * (-8) = – 40

Splitting the middle term i.e. -3x in the given quadratic equation we get:

-(x

^{2}– 3x – 40) = -(x^{2}+ 5x – 8x – 40)= -(x (x + 5) – 8 (x + 5)) [Taking the similar terms common]

= -(x – 8) (x + 5) [as (x + 5) is common]

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

**Question 3:** Solve for factors of a^{2} + 3a – 88

**Solution:**

Given: a

^{2}+ 3a – 88To factorize the following expression we have to find two numbers a and b such that a+b = 3 and ab = -88

We know that :

11 + (-3) = 3 and 11 * (-8) = -88

Splitting the middle term i.e. 3a in the given quadratic equation we get:

a

^{2}+ 3a – 88 = a^{2}+ 11a – 8a – 88= a (a + 11) – 8 (a + 11) [Taking the similar terms common]

= (a – 8) (a + 11) [as (a + 11) is common]

**Question 4: **Solve for factors of a^{2} – 14a – 51

**Solution:**

Given: a

^{2}– 14a – 51To factorize the following expression we have to find two numbers a and b such that a+b = -14 and ab = -51

We know that :

3 + (-17) = -14 and 3 * (-17) = -51

Splitting the middle term i.e. -14a in the given quadratic equation we get:

a

^{2 }– 14a – 51 = a^{2}+ 3a – 17a – 51= a (a + 3) – 17 (a + 3) [Taking the similar terms common]

= (a – 17) (a + 3) [as (a + 3) is common]

**Question 5:** Solve for factors of x^{2} + 14x + 45

**Solution:**

Given: x

^{2}+ 14x + 45To factorize the following expression we have to find two numbers a and b such that a+b = 14 and ab = 45

We know that :

5 + 9 = 14 and 5 * 9 = 45

Splitting the middle term i.e. 14x in the given quadratic equation we get:

x

^{2}+ 14x + 45 = x^{2}+ 5x + 9x + 45= x (x + 5) – 9 (x + 5) [Taking the similar terms common]

= (x + 9) (x + 5) [as (x+5) is common]

**Question 6: **Solve for factors of x^{2 }– 22x + 120

**Solution:**

Given: x

^{2}– 22x + 120To factorize the following expression we have to find two numbers a and b such that a+b = -22 and ab = 120

We know that :

-12 + (-10) = -22 and -12 * (-10) = 120

Splitting the middle term i.e. -22x in the given quadratic equation we get:

x

^{2}– 22x + 120 = x^{2}– 12x – 10x + 120= x (x – 12) – 10 (x – 12) [Taking the similar terms common]

= (x – 10) (x – 12) [as (x – 12) is common]

**Question 7:** Solve for factors of x^{2} – 11x – 42

**Solution:**

Given: x

^{2}– 11x – 42To factorize the following expression we have to find two numbers a and b such that a+b = -11 and ab = -42

We know that :

3 + (-14) = -11 and 3 * (-14) = -42

Splitting the middle term i.e. -11x in the given quadratic equation we get:

x

^{2}– 11x – 42 = x^{2}+ 3x – 14x – 42= x (x + 3) – 14 (x + 3) [Taking the similar terms common]

= (x – 14) (x + 3) [as (x + 3) is common]

**Question 8: **Solve for factors of a^{2} + 2a – 3

**Solution:**

Given: a

^{2}+ 2a – 3To factorize the following expression we have to find two numbers a and b such that a+b = 2 and ab = -3

We know that :

3 + (-1) = 2 and 3 * (-1) = -3

Splitting the middle term i.e. 2a in the given quadratic equation we get:

a

^{2}+ 2a – 3 = a^{2}+ 3a – a – 3= a (a + 3) – 1 (a + 3) [Taking the similar terms common]

= (a – 1) (a + 3) [as (a + 3) is common]

**Question 9: **Solve for factors of a^{2} + 14a + 48

**Solution:**

Given: a

^{2}+ 14a + 48To factorize the following expression we have to find two numbers a and b such that a+b = 14 and ab = 48

We know that :

8 + 6 = 14 and 8 * 6 = 48

Splitting the middle term i.e. 14a in the given quadratic equation we get:

a

^{2}+ 14a + 48 = a^{2}+ 8a + 6a + 48= a (a + 8) + 6 (a + 8) [Taking the similar terms common]

= (a + 6) (a + 8) [as (a + 8) is common]

**Question 10: **Solve for factors of x^{2} – 4x – 21

**Solution:**

Given: x

^{2}– 4x – 21To factorize the following expression we have to find two numbers a and b such that a+b = -4 and ab = -21

We know that :

3 + (-7) = -4 and 3 * (-7) = -21

Splitting the middle term i.e. -4x in the given quadratic equation we get:

x

^{2}+ 4x – 21 = x^{2}+ 3x – 7x – 21= x (x + 3) – 7 (x + 3) [Taking the similar terms common]

= (x – 7) (x + 3) [as (x + 3) is common]

**Question 11:** Solve for factors of y^{2 }+ 5y – 36

**Solution:**

Given: y

^{2}+ 5y – 36To factorize the following expression we have to find two numbers a and b such that a+b = 5 and ab = -36

We know that :

9 + (-4) = 5 and 9 *(-4) = -36

Splitting the middle term i.e. 5y in the given quadratic equation we get:

y

^{2}+ 5y – 36 = y^{2}+ 9y – 4y – 36= y (y + 9) – 4 (y + 9) [Taking the similar terms common]

= (y – 4) (y + 9) [as (y + 9) is common]

**Question 12:** Solve for factors of (a^{2} – 5a)^{2} – 36

**Solution:**

Given: (a

^{2}– 5a)^{2}– 36(a

^{2 }– 5a)^{2}– 36 = (a^{2 }– 5a)^{2}– 6^{2}By using the formula (a

^{2}– b^{2}) = (a+b) (a-b)(a

^{2}– 5a)^{2}– 6^{2}= (a^{2}– 5a + 6) (a^{2}– 5a – 6)Now solving the second part i.e a

^{2}– 5a + 6To factorize the following expression we have to find two numbers a and b such that a+b = -5 and ab = 6

We know that :

-2 + (-3) = -5 and -2 * (-3) = 6

Splitting the middle term i.e. -5x in the given quadratic equation we get:

a

^{2}-5a + 6 = a^{2}– 2a – 3a + 6= a (a – 2) -3 (a – 2) [Taking the similar terms common]

= (a – 3) (a – 2) [as (a – 2) is common]

Now solving the first part i.e a

^{2}– 5a – 6To factorize the following expression we have to find two numbers a and b such that a+b = -5 and ab = -6

We know that :

1 + (-6) = -5 and 1 * (-6) = -6

Splitting the middle term i.e. -5x in the given quadratic equation we get:

a

^{2}-5a – 6 = a^{2}+ a – 6a – 6= a (a + 1) -6(a + 1) [Taking the similar terms common]

= (a – 6) (a + 1) [as (a + 1) is common]

Since, (a2 – 5a)

^{2}– 36 = (a^{2}– 5a + 6) (a^{2}– 5a – 6)Substituting the values of (a

^{2}– 5a + 6 ) and (a^{2}– 5a – 6 ) we get:= (a – 2) (a – 3) (a + 1) (a – 6)

**Question 13:** Solve for factors of (a + 7) (a – 10) + 16

**Solution:**

Given: (a + 7) (a – 10) + 16

(a + 7) (a – 10) + 16 = a

^{2}– 10a + 7a – 70 + 16(a + 7) (a – 10) + 16 = a

^{2}– 3a – 54To factorize the following expression we have to find two numbers a and b such that a+b = -3 and ab = -54

We know that :

6 + (-9) = -3 and 6 * (-9) = -54

Splitting the middle term i.e. -3x in the given quadratic equation we get:

a

^{2}– 3a – 54 = a^{2}+ 6a – 9a – 54= a (a + 6) -9 (a + 6) [Taking the similar terms common]

= (a – 9) (a + 6) [as (a + 6) is common]