The word "Factoring" means to break down something big into its small portions. In the mathematical language, it means to break a number or equation into its factors. You can get the equation or number back by multiplying the factors with each other. Let's take an example of 24, the number "24" has a lot of factors such as if you multiply 6 and 4, you can get 24. The other factors are 8 and 3, 12, and 2, and not to forget that if you multiply 24 and 1, you will also get 24. The question is what is its usefulness?

Practical use of Factorization

Factorization helps us to find unknown variables. Imagine a linear equation, that linear equation is the rate of a chemical which is reacting per unit time. For an engineer, how he/she will know the amount of chemical needed for the reaction to proceed? To find that, the engineer needs to factorize the equation and equate it a number to find the value of the unknown. Furthermore, factorization is the standard way to show an equation. An equation like this 3{ x }^{ 4 }-5{ x }^{ 3 }+8{ x }^{ 2 }-15x+33 looks messy, however, a factorized equation looks much more engaging.

The best Maths tutors available
1st lesson free!
Intasar
4.9
4.9 (23 reviews)
Intasar
£42
/h
1st lesson free!
Matthew
5
5 (17 reviews)
Matthew
£25
/h
1st lesson free!
Dr. Kritaphat
4.9
4.9 (6 reviews)
Dr. Kritaphat
£49
/h
1st lesson free!
Paolo
4.9
4.9 (11 reviews)
Paolo
£25
/h
1st lesson free!
Petar
4.9
4.9 (9 reviews)
Petar
£27
/h
1st lesson free!
Rajan
4.9
4.9 (11 reviews)
Rajan
£15
/h
1st lesson free!
Farooq
5
5 (13 reviews)
Farooq
£35
/h
1st lesson free!
Myriam
5
5 (15 reviews)
Myriam
£20
/h
1st lesson free!
Intasar
4.9
4.9 (23 reviews)
Intasar
£42
/h
1st lesson free!
Matthew
5
5 (17 reviews)
Matthew
£25
/h
1st lesson free!
Dr. Kritaphat
4.9
4.9 (6 reviews)
Dr. Kritaphat
£49
/h
1st lesson free!
Paolo
4.9
4.9 (11 reviews)
Paolo
£25
/h
1st lesson free!
Petar
4.9
4.9 (9 reviews)
Petar
£27
/h
1st lesson free!
Rajan
4.9
4.9 (11 reviews)
Rajan
£15
/h
1st lesson free!
Farooq
5
5 (13 reviews)
Farooq
£35
/h
1st lesson free!
Myriam
5
5 (15 reviews)
Myriam
£20
/h
First Lesson Free>

Factorization Methods

There are two methods of factorization and they are:

  1. Middle term breaking
  2. Quadratic Formula

The middle term breaking method is the easiest and shortest method to get factors of any equation. However, not all equations can be solved by the middle term method and that is why you can rely on the quadratic formula. You can find factors of any equation with the help of quadratic formula but using the quadratic formula is hard compared to the middle term breaking method. In addition, there is a big chance that you can make common mistakes when using the quadratic formula. That is why we recommend you use middle term breaking first but if the equation can't be factorized by the middle term breaking method then you should focus on the quadratic formula.

*Please do note that these methods are ONLY used for the quadratic equations(equations that has the power of x up to 2) only.*

Boost your learning speed and understanding with a maths tutor on Superprof.

Middle Term Breaking Method

 

Consider this equation: 2{ x }^{ 2 }-3x-2

As the method says, you need to break the middle term into two factors. In the above equation, the middle term is 3x, you can get 3 with many factors but there are some conditions when selecting the right factors otherwise this method will fail. Let's address the above term into this form a{ x }^{ 2 }-bx-c=0. So, the first step is to multiply a and c. In the above case, it will be 4. Now you have the number which you need to break into factors. The next thing you need to check is the sign of the c. If it is positive then the factors will be added and if negative then the factors will be subtracted. In the equation, 2{ x }^{ 2 }-3x-2=0, the sign of c is negative, therefore, we will be subtracting factors of 4. Let's create factors of 4, here are the possible factors 2 and 2, and 4 and 1. If we subtract 2 from 2, we will get 0, however, if we subtract 1 from 4, we will get 3 which is the value of b in the above equation. In conclusion, we will select factors 4 and 1 to factorize our equation.

 

2{ x }^{ 2 }-x(4-1)-2

So, we broke 3 into (4-1). Always double check your equation, if we subtract 1 from 4, we will get 3 which will result in the question equation. This means we are on the right path!

Multiplying the x with the factors:

2{ x }^{ 2 }-4x-(-x)-2

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

Taking common outside

2x(x-2)-1(x-2)

There is always something common, which is 1. Do you see that the variable and number in the brackets are the same? This means we are doing it correctly, it should be the same otherwise there is something wrong.

(2x-1)(x-2)

The above equations in brackets are the factors of 2{ x }^{ 2 }-x(4-1)-2. Stop reading here and try to multiply the above two factors with each other and check what did you get. You see now what happened there?

Examples:

1. 6{ x }^{ 2 }+5x-6

Break the middle term:

a \times c = 6 \times 6 = 36

Sign on the c = negative

Possible factors=1 \times 36, 6 \times 6, 2 \times 18, 3 \times 12, 4 \times 9

The middle term=9 - 4 = 5 which means we will pick 4 \times 9 factor.

 

6{ x }^{ 2 }+x(9-4)-6

6{ x }^{ 2 }+9x-4x-6

Taking common

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

(3x-2)(2x-3)

 

2. 12{ x }^{ 2 }-1=11x

Re-arrange the equation:

12{ x }^{ 2 }-11x-1=0

Break the middle term:

a \times c = 12 \times 1 = 12

Sign on the c = negative

Possible factors=1 \times 12, 6 \times 2, 3 \times 4

The middle term=12 - 1 = 11 which means we will pick 4 \times 9 factor.

 

12{ x }^{ 2 }-x(12-1)-1=0

12{ x }^{ 2 }-12x+x-1=0

12x(x-1)+1(x-1)=0

(12x+1)(x-1)=0

12x+1=0 \qquad x-1=0

12x=-1 \qquad x=1

\x=\frac { -1 }{ 12 } \qquad x=1

 

Quadratic Formula

a{ x }^{ 2 }+bx+c=0

Dividing the whole equation by a:

\frac { a{ x }^{ 2 } }{ a } +\frac { bx }{ a } +\frac { c }{ a } =\frac { 0 }{ a }

{ x }^{ 2 }+\frac { bx }{ a } +\frac { c }{ a } =0

{ x }^{ 2 }+\frac { bx }{ a }=-\frac { c }{ a }

Making the left side a perfect square:

{ x }^{ 2 }+\frac { bx }{ a }+\frac { { b }^{ 2 } }{ 4{ a }^{ 2 } }=\frac { { b }^{ 2 } }{ 4{ a }^{ 2 } }-\frac { c }{ a }

{( x +  \frac { b }{ 2a } )}^{ 2 }=\frac { { b }^{ 2 } - 4ac}{ 4{ a }^{ 2 } }

Taking root on both sides:

x + \frac { b }{ 2a } = \pm \sqrt { \frac { { b }^{ 2 } - 4ac}{ 4{ a }^{ 2 } } }

x + \frac { b }{ 2a } = \pm \frac { \sqrt { { b }^{ 2 }-4ac } }{ 2a }

x=-\frac { b }{ 2a } \pm \frac { \sqrt { { b }^{ 2 }-4ac } }{ 2a }

x=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }

 

We will solve the above examples again but this time, we will use the quadratic formula. Remember, you can use quadratic formula for any quadratic equation.

 

Examples:

1. 6{ x }^{ 2 }+5x-6

a{ x }^{ 2 }+bx+c=0

 

Comparing the both above equations:

a=6, b=5, c=-6

x=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }

 

Substituting all the values of a,b, and c in the above equation:

x=\frac { -5\pm \sqrt { { 5 }^{ 2 }-4(6)(-6) } }{ 2(6) }

x=\frac { -5\pm \sqrt { 25+144 } }{ 12 }

x=\frac { -5\pm (13) }{ 12 }

x=\frac { -5+ 13 }{ 12 } \qquad x=\frac { -5-13 }{ 12 }

x=\frac { 8 }{ 12 } \qquad x=\frac { -18 }{ 12 }

x=\frac { 2 }{ 3 } \qquad x=\frac { -3 }{ 2 }

 

2. 12{ x }^{ 2 }-1=11x

Re-arrange the equation:

12{ x }^{ 2 }-11x-1=0

a{ x }^{ 2 }+bx+c=0

 

Comparing the both above equations:

a=12, b=-11, c=-1

x=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }

 

Substituting all the values of a,b, and c in the above equation:

x=\frac { -(-11)\pm \sqrt { { -11 }^{ 2 }-4(12)(-1) } }{ 2(12) }

x=\frac { 11\pm \sqrt { 121+48 } }{ 24 }

x=\frac { 11\pm \sqrt { 169 } }{ 24 }

x=\frac { 11\pm (13) }{ 24 }

x=\frac { 11+13 }{ 24 } \qquad x=\frac { 11-13 }{ 24 }

x=\frac { 24}{ 24 } \qquad x=\frac { -2 }{ 24 }

x=1 \qquad x=\frac { -1 }{ 12 }

Find a maths tutor on Superprof today.

 

Need a Maths teacher?

Did you like the article?

1 Star2 Stars3 Stars4 Stars5 Stars 5.00/5 - 1 vote(s)
Loading...

Emma

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.