QUADRATIC EQUATION

Quadratic equations are equations in this form, x2 + 6x + 9 = 0; x2  – y2 = 0; 2p2 – 11pq + 15q2; 36 – y2 etc., the highest power of the unknowns in these equations is 2. Generally, a quadratic expression is in the form ax2 + bx + c. Notice there is no equality sign while a quadratic equation has the form ax2 + 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:

ap2bpq + cq2

ax2bxy + cy2

Here, we have two unknown p and q. The coefficient of the first p2 is a, that of pq is b and q2 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.     t2 + 9t + 20

      2.     3p2 + 14p + 8

      3.     2y2 + 13y + 20

      4.     18 + 13d + 2d2

 

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 t2 + 9t + 20,

a = 1 (coefficient of t2)

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          

 

            t2 + 9t + 20                               replace 9t with 4t and 5t

            t2 + 4t + 5t + 20                       bracket in twos

            (t2 + 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 t2 + 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.    3p2 + 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

           

            3p2 + 14p + 8

            3p2 + 2p + 12p + 8

** bracket them in pair of twos

            (3p2 + 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 3p2 + 14p + 8 is factorized.

 

       3.    2y2 + 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

            2y2 + 13y + 20

            2y2 + 5y + 8y + 20

            (2y2 + 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 + 2d2

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 + 2d2

18 + 4d + 9d + 2d2

(18 + 4d) + (9d + 2d2)

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

(9 + 2d)(2 + d)

 

            Exercise A

Factorize the following

      1.      t2 + 16t + 63

      2.      3m2 + 11m + 6

      3.      e2 + 10e + 25

      4.      4x2 + 13m + 10

      5.      6 + 5f + f2

      6.      8 + 10v + 3v2

      7.      12 + 20y + 7y2

 

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.   n2 – 7n + 6

    2.   3e2 – 20e + 28

    3.   12a2 – 19a + 5

    4.   p2 – 10p – 25

 

Solution

     1.   n2 – 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.

n2 – 7n + 6

n2 – n – 6n + 6

(n2 – 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.   3e2 – 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

3e2 – 20e + 28

3e2 – 6e – 14e + 28

(3e2 – 6e) – (14e – 28)

3e(e – 2) – 14(e – 2)

(e – 2)(3e – 14)

 

    3.   12a2 – 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

12a2 – 19a + 5

12a2 – 4a – 15a + 5

(12a2 – 4a) – (15a – 5)

4a(3a – 1) – 5(3a – 1)

(3a – 1)(4a – 5)

 

    4.    p2 – 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

            p2 – 5p – 5p + 25

            (p2 – 5p) – (5p – 25)   

            p(p – 5) – 5(p – 5)

            (p – 5)(p – 5) 

 

            Exercise B

Factorize the following

   1.   3p2 – 17p + 10

   2.   5y2 – 18y + 9

   3.   m2 – 6m + 9

   4.   9k2 – 18k + 8

   5.   7x2 – 19x + 10

   6.   2α2 - 13α + 18

   7.   t2 – 18t +45  

 

NEXT: Factorizing quadratic expressions continuation

 

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