**QUADRATIC EQUATION**

Quadratic equations are equations in this form, x^{2} + 6x + 9 = 0; x^{2 } – y^{2 }= 0; 2p^{2} – 11pq + 15q^{2}; 36 – y^{2} etc., the highest power of the unknowns in these equations is 2. Generally, a quadratic **expression** is in the form **ax ^{2} + bx + c. **Notice there is no equality sign while a quadratic

**equation**has the form

**ax**

^{2 }+ bx + c = 0.** **The coefficient of the unknown with power 2 is **a **while that one with power 1 is **b** and the **constant** is **c. **When the expression is in this form:

**a**p^{2} – **b**pq + **c**q^{2}

**a**x^{2} – **b**xy + **c**y^{2}

Here, we have two unknown p and q. The coefficient of the first p^{2} is **a,** that of pq is **b **and q^{2} is **c.**

** **

** FACTORIZING QUADRATIC EXPRESSIONS **

Factorizing quadratic equation involves looking for two numbers which add up to give the sum and multiply to give the product.

The **sum** is **b **while the product is **a x c**. The signs of the factors depend on the signs of the product and sum.

We will now look at the different forms of a quadratic equation and how to factorize them

** **

**Case 1: **When both the sum and the product are **positive**.

Factorize the quadratic expressions below:

1. t^{2} + 9t + 20

2. 3p^{2 }+ 14p + 8

3. 2y^{2} + 13y + 20

4. 18 + 13d + 2d^{2}

**Solution **

Follow the steps below to factorize

**Step 1**

Write the **product **and **sum **for the equation. The product is **a** x **c** while the sum is **b**

Product = a x c

Sum = b

From t^{2} + 9t + 20,

a = 1 (coefficient of t^{2})

b = +9 (coefficient of t)

c = +20 (the constant)

**Product **= a x c

= 1 x 20

= 20

**Sum **= +9

** **

**Step 2 **

Look for the factors of the product.

20 = 1, 2, 4, 5, 10, 20

pair the factors in two so that they multiply to give 20.

** 1 and 20

** 2 and 10

** 4 and 5

There are three pairs formed

**Step 3 **

** **Pick the pairs that add up to give our sum (+9). Only 4 and 5 satisfy this. Therefore, we write:

** **Factors are 4 and 5

We will now replace the 9t in the question with 4t and 5t** **

** **

** **t^{2} + 9t + 20 *replace 9t with 4t and 5t*

t^{2} + 4t + 5t + 20 *bracket in twos*

(t^{2} + 4t) + (5t + 20) *factorize each bracket*

t(t+4) + 5(t+4)

*(t+4) is common to the two expressions, we factorize again.*

(t+4)(t+5).

The above is the result obtained when t^{2} + 9t + 20 is factorized.

*Once you grab the above steps, you will be able to factorize easily without passing through the long steps above.*

2. 3p^{2 }+ 14p + 8

** Solution**

a = 3, b = 14, c = 8

**product** = ac = 3 x 8 = 24

**sum** = 14

** factors of 24

24 = 1, 2, 3, 4, 6, 8, 12, 24

** pair the factors

(1 x 24) (2 x 12) (3 x 8) (4 x 6)

** Of the four pairs above, only 2 and 12 add to give 14

Therefore, factors are 2 and 12

** we will now replace 14p in the original expression with 2p and 12p

3p^{2 }+ 14p + 8

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

** bracket them in pair of twos

(3p^{2} + 2p) + (12p + 8)

** factorize each bracket

p(3p + 2) + 4(3p + 2)

** factorize again

(3p + 2)(p + 4)

The above is the result obtained when 3p^{2 }+ 14p + 8 is factorized.

3. 2y^{2} + 13y + 20

Solution

Product = 2 x 20 = 40

Sum = 13

1, 2, 5, 8, 20, 40

(1 x 40) (2 x 20) (5 x 8)

Factors are 5 and 8

2y^{2} + 13y + 20

2y^{2} + 5y + 8y + 20

(2y^{2} + 5y) + (8y + 20)

y is common to the first bracket and 4 to the second

y(2y + 5) + 4(2y + 5)

factorize again

(2y + 5)(y+4)

4. 18 + 13d + 2d^{2}

don't be confuse with c coming before a and b.

**Solution **

a = 2, b = 13, c = 18

Product = 2 x 18 = 36

Sum = 13

36 = 1, 2, 3, 4, 9, 12, 18,36

(1 x 36) (2 x 18) (3x12) (4 x 9)

Factors are 4 and 9

18 + 13d + 2d^{2}

18 + 4d + 9d + 2d^{2}

(18 + 4d) + (9d + 2d^{2})

2(9 + 2d) + d(9 + 2d)

(9 + 2d)(2 + d)

**Exercise A**

Factorize the following

1. t^{2} + 16t + 63

2. 3m^{2 }+ 11m + 6

3. e^{2} + 10e + 25

4. 4x^{2} + 13m + 10

5. 6 + 5f + f^{2}

6. 8 + 10v + 3v^{2}

7. 12 + 20y + 7y^{2}

**Case2: **when the sum is negative and the product is positive.

This is just like case 1, only that we now add minus (–) to the two factors

Factorize the following

1. n^{2} – 7n + 6

2. 3e^{2} – 20e + 28

3. 12a^{2} – 19a + 5

4. p^{2} – 10p – 25

**Solution **

1. n^{2} – 7n + 6

a = 1, b = - 7, c = 6

product = 6

sum = - 7

factors = 1, 2, 3, 6

(1 x 6) (2 x 3)

1 and 6 will add to give 7.

∴ Factors are – 1 and – 6.

n^{2} – 7n + 6

n^{2} – n – 6n + 6

(n^{2} – n) – (6n – 6)

Notice that + 6 has changed to – 6. This is because of the – outside the bracket.

**Anytime you open a bracket with – outside the bracket, the signs in the bracket will change, + will become – and – will change to +. **

** **

n(n – 1) – 6(n – 1)

(n – 1)(n – 6)

2. 3e^{2} – 20e + 28

**Solution **

a = 3 b = -20 c = 28

product = 84

sum = - 20

84 = 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84

(1 x 84) (2 x 42) (3 x 28) (4 x 21) (6 x 14) (7 x 12)

*6 and 14 add to give 20*

∴ factors are – 6 and – 20

3e^{2} – 20e + 28

3e^{2} – 6e – 14e + 28

(3e^{2} – 6e) – (14e – 28)

3e(e – 2) – 14(e – 2)

(e – 2)(3e – 14)

3. 12a^{2} – 19a + 5

**Solution**

a = 12 b = – 19 c = 5

product = 60

sum = – 16

60 = 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

(1 x 60) (2 x 30) (3 x 20) (4 x 15) (5 x 12) (6 x 10)

Factors are – 4 and – 15

12a^{2} – 19a + 5

12a^{2} – 4a – 15a + 5

(12a^{2} – 4a) – (15a – 5)

4a(3a – 1) – 5(3a – 1)

(3a – 1)(4a – 5)

4. p^{2} – 10p + 25

**Solution **

a = 1 b = -10 c = -25

product = 1 x 25 = 25

sum = - 10

factors of 25 = 1, 5, 25

(1 x 25) (5 x 5)

Factors are – 5 and – 5

p^{2} – 5p – 5p + 25

(p^{2} – 5p) – (5p – 25)

p(p – 5) – 5(p – 5)

(p – 5)(p – 5)

**Exercise B **

Factorize the following

1. 3p^{2} – 17p + 10

2. 5y^{2} – 18y + 9

3. m^{2} – 6m + 9

4. 9k^{2 }– 18k + 8

5. 7x^{2 }– 19x + 10

6. 2α^{2} - 13α + 18

7. t^{2 }– 18t +45

**NEXT: **Factorizing quadratic expressions continuation