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Let us consider a quadratic that is of the form a x 2 + b x . We can see here that x is a common factor of both terms. Therefore, a x 2 + b x factorises to x ( a x + b ) . For example, 8 y 2 + 4 y factorises to 4 y ( 2 y + 1 ) .

Another type of quadratic is made up of the difference of squares. We know that:

( a + b ) ( a - b ) = a 2 - b 2 .

This is true for any values of a and b , and more importantly since it is an equality, we can also write:

a 2 - b 2 = ( a + b ) ( a - b ) .

This means that if we ever come across a quadratic that is made up of a difference of squares, we can immediately write down what the factors are.

Find the factors of 9 x 2 - 25 .

  1. We see that the quadratic is a difference of squares because:

    ( 3 x ) 2 = 9 x 2

    and

    5 2 = 25 .
  2. 9 x 2 - 25 = ( 3 x ) 2 - 5 2
  3. ( 3 x ) 2 - 5 2 = ( 3 x - 5 ) ( 3 x + 5 )
  4. The factors of 9 x 2 - 25 are ( 3 x - 5 ) ( 3 x + 5 ) .

These types of quadratics are very simple to factorise. However, many quadratics do not fall into these categories and we need a more general method to factorise quadratics like x 2 - x - 2 ?

We can learn about how to factorise quadratics by looking at how two binomials are multiplied to get a quadratic. For example, ( x + 2 ) ( x + 3 ) is multiplied out as:

( x + 2 ) ( x + 3 ) = x ( x + 3 ) + 2 ( x + 3 ) = ( x ) ( x ) + 3 x + 2 x + ( 2 ) ( 3 ) = x 2 + 5 x + 6 .

We see that the x 2 term in the quadratic is the product of the x -terms in each bracket. Similarly, the 6 in the quadratic is the product of the 2 and 3 in the brackets. Finally, the middle term is the sum of two terms.

So, how do we use this information to factorise the quadratic?

Let us start with factorising x 2 + 5 x + 6 and see if we can decide upon some general rules. Firstly, write down two brackets with an x in each bracket and space for the remaining terms.

( x ) ( x )

Next, decide upon the factors of 6. Since the 6 is positive, these are:

Factors of 6
1 6
2 3
-1 -6
-2 -3

Therefore, we have four possibilities:

Option 1 Option 2 Option 3 Option 4
( x + 1 ) ( x + 6 ) ( x - 1 ) ( x - 6 ) ( x + 2 ) ( x + 3 ) ( x - 2 ) ( x - 3 )

Next, we expand each set of brackets to see which option gives us the correct middle term.

Option 1 Option 2 Option 3 Option 4
( x + 1 ) ( x + 6 ) ( x - 1 ) ( x - 6 ) ( x + 2 ) ( x + 3 ) ( x - 2 ) ( x - 3 )
x 2 + 7 x + 6 x 2 - 7 x + 6 x 2 + 5 x + 6 x 2 - 5 x + 6

We see that Option 3 (x+2)(x+3) is the correct solution. As you have seen that the process of factorising a quadratic is mostly trial and error, there is some information that can be used to simplify the process.

Method: factorising a quadratic

  1. First, divide the entire equation by any common factor of the coefficients so as to obtain an equation of the form a x 2 + b x + c = 0 where a , b and c have no common factors and a is positive.
  2. Write down two brackets with an x in each bracket and space for the remaining terms.
    ( x ) ( x )
  3. Write down a set of factors for a and c .
  4. Write down a set of options for the possible factors for the quadratic using the factors of a and c .
  5. Expand all options to see which one gives you the correct answer.

There are some tips that you can keep in mind:

  • If c is positive, then the factors of c must be either both positive or both negative. The factors are both negative if b is negative, and are both positive if b is positive. If c is negative, it means only one of the factors of c is negative, the other one being positive.
  • Once you get an answer, multiply out your brackets again just to make sure it really works.

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Source:  OpenStax, Siyavula textbooks: grade 10 maths [ncs]. OpenStax CNX. Aug 05, 2011 Download for free at http://cnx.org/content/col11239/1.2
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