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TIC LMS

Factorisation method

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Example 1:

Using factorisation method , solve 2x² + 13x = 15


Solution

Steps to solve :

1. Move terms to the left side . 2x² + 13x - 15 = 0

2 . Multiply 2x²  by -15  = -30x2

3.  Find the factors of -30x²
   - 1x and + 30x
       lx  and  -30x
      -2x and  15x
       2x  and  -15x
      -3x and  10x
       3x  and  -10x

4 .  Add the pairs of factors . The one that adds up to +13x among the six factors above is  -2x and + 15x
-2 + 15x = +13x

5 .  Replace the  +13x in the equation given with -2x + 15x
2x² + 13x - 15 = 0
2x² -2x + 15x - 15 = 0

6. Factorize by taking the first 2 and the last 2
      2x² -2x = 2x( x - 1 )
     15x - 15 = +15 (x - 1)

7 .  Combine the factorized terms .
2x ( x - 1 ) + 15 ( x - 1 ) = 0
Note : The values in the bracket must be the same . This is proof of being correct .

8 .  Collect the terms outside the bracket
( 2x + 15 ) ( x - 1 ) = 0

9 .  Equate each  bracket to  zero
2x + 15 = 0  and  x - 1 = 0

10  Solve for x in each equation
2x + 15 = 0
2x = -15
x = -15 / 2 = -7.5
x = -7.5
x - 1 = 0
x = 1

Answer: x = 7.5 or 1

Example 2:

Factorise  a²-17a+42  and  solve for x

Solution
Steps to solve

1 .  Multiply a² and +42  = + 42a²

2.  Find the factors of  + 42a²
1a  and  42a           =  43a ( sum )
-1a  and  -42a       =  -43a ( sum )
2a  and  21 a          =  23a  ( Sum )
-2a  and  -21a        = -23a ( Sum )
За  and  14 a          =  17a ( Sum )
-3a  and  -14a        =  -17a ( sum )
6a  and  7а             = 13a ( sum )
-6a  and  -7a          =  -13a ( sum )

3. Add the pairs of factors : The one that adds up to -17a among  the eight factors above  is -3a and -14a
- 3a + ( -14a ) = -17a .

4. Replace the -17a in the equation given with -3a - 14a
a²-17a +42 = 0
a² - 3a- 14a +42 = 0

5. Factorize by taking the first two parts and the last two parts.
a - 3a = a ( a - 3 )
- 14a +42 = -14 ( a - 3 )

6 .  Combine the factorized terms .
a(a - 3 )-14 ( a - 3 ) = 0
Note : The values in the bracket must be the same

7. Collect the terms outside the brackets 
( a - 14 ) ( a - 3 ) = 0
Note : Choose one of the two ( a - 3 ) since they  are the same

8. Equate  each bracket  to zero
     a-14 = 0  and a - 3 = 0

9. Solve for x in each equation. 
     a - 14 = 0
     a = 14
     a - 3 = 0
     a = 3

Answer : x = 14 or 3

Example 3:

Factorize: 2a²+7ab-15b²
Steps to solve
1. Multiply 2a² and -15b² = - 30a²b².

2. Find the factors of  -30a²b²

-1ab  and  30ab  =  + 29ab ( sum )
1ab  and  -30ab  -29ab ( Sum )
-2ab  and  15ab  =  + 13ab ( sum )
2ab  and  -15ab  =  -13ab  ( Sum )
-3ab  and  10ab =  +7ab  ( Sum )
3ab  and  -10ab  =  -7ab  ( sum )
-5ab  and  6ab  = +1ab ( sum )
5ab  and  -6ab = -1ab ( Sum )

3. Add the pairs of factors . The one that adds up  to + 7ab among  the  eight pairs of factor above is - 3ab and + 10ab 
-3ab + 10ab = + 7ab .

4. Replace the + 7ab in the equation given with -3ab and +10ab
2a² + 7ab  -  15b²
2a² -  3ab +10ab - 15b²

5  Factorize by taking  the  first two terms  and the last two terms
2a² - 3b = a(2a - 3b)
+10ab - 15b² = + 5b ( 2a  -  3b) .

6 .  Combine the factorized terms .
a (2a - 3b ) + 5b ( 2a - 3b )
Note : The values in the bracket must be the same . 

7. Collect the terms outside the brackets
(a + 5b )( 2a - 3b )


Answer: (a + 5b )( 2a - 3b )

Example 4:

Using the factorization method , solve 7-22x + 3x²


Steps to solve

1. Multiply 7 and -3x²  = + 21x²
Note : You may first rearrange into  3x² - 22 +7 or not .

2.  Find the factors of + 21x²

+1x  and  21x  		= + 22x ( sum )
- 1x and  -21x  	=  - 22x ( sum )
+ 3x  and  + 7x  	= + 10x ( sum )
-3x  and  -7x  		= -10x ( sum )

3.  Add the pairs of factors . The  one that adds  up  to -22x among the four pairs of factors above is -1x and -21x 
- 1x + ( -21x ) = -22x

4. Replace the  -22x in the equation given with -1x-21x
7- 22x + 3x²
7 - 1x - 21x + 3x²

5. Factorize by taking the first two terms and the last two terms .
7- 1x = 1 ( 7 - x )
- 2x + 3x² = -3x ( 7 - x )

6  Combine the factorized terms
1 ( 7-x ) -3x ( 7-x ) = 0
Note : The values in the bracket  must be the same

7.  Collect the terms outside the brackets
( 1-3x ) ( 7-x ) = 0
Note : Choose one of the two ( 7 - x ) since they are the same. 

8. Equate each bracket to zero.
( 1-3x) = 0  and  7 - x = 0

9. Solve for x in each equation
1-3x = 0
1 = 3x
x = 1/3 and .

7-x = 0
7 = x
x = 7

Answer : x = 1/3  or 7

Example 5:

Solve the equation : a² - 3a = 0 

Steps to solve
1. Factorize by collecting the letter common
a² - 3a = 0
a(a - 3 ) = 0

2. Equate the letter outside the bracket to zero  and also the ones  in the bracket .
a = O  and  a - 3 = 0
3. Solve for a

a = 0  and  a = 3

Example 6:

Example 6

Solve the equation m² = 16

Method 1

Steps to  Solve
1. Take the root of both sides
√m² = √16
m = +/- 4

2. You  may separate  the answer
m = +4  or 4

Answer :  M = +4  or  -4


Method II

Steps to solve
1. Move the term '16' to  the left .
m² - 16 = 0
Note : The above equation is lacking a term consisting of m . It must be that the coefficient is 0  
i.e; m² + 0m -16 = 0
It can be re - written in this form .

2. Multiply  m² and  -16  =  -16m²

3. Find the factors of -16m²
-1m  and  16m  		= 15m ( sum )
1m  and  -16m  		= -15m ( sum )
-2m  and  8m  		= 6m ( Sum )
2m  and  -8m  		= -6m ( Sum )
-4m  and  4m  		= 0m ( Sum )

4. Add the pairs of factors . The one that adds 0m among the five pairs of factors above is -4m and 4m .

5. Replace the 0m in the equation given with -4m + 4m
m² + 0m - 16 = 0
m² - 4m + 4m - 16 = 0

6. Factorize by taking the first two terms and the last two terms 
m² - 4m = m(m-4)
+4m - 16 = +4(m-4)
Note: The values in the brackets must be the same

7. Combine the factorized terms
m(m-4) +(m-4) = 0

8. Collect the terms outside the bracket and equate them to zero separately 
(m+4)(m-4) = 0
m +4 = 0 and m-4=0

Answer: m= -4 or +4