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Factorising a Quadratic

Figure 1
Khan Academy video on factorising a quadratic.

Factorisation can be seen as the reverse of calculating the product of factors. In order to factorise a quadratic, we need to find the factors which when multiplied together equal the original quadratic.

Let us consider a quadratic that is of the form ax2+bxax2+bx
. We can see here that xx is a common factor of both terms. Therefore,
ax2+bxax2+bx
factorises to x(ax+b)x(ax+b). For example, 8y2+4y8y2+4y
factorises to
4y(2y+1)4y(2y+1).

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

( a + b ) ( a - b ) = a 2 - b 2 . ( a + b ) ( a - b ) = a 2 - b 2 .
(1)

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

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

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.

Exercise 1: Difference of Squares

Find the factors of 9x2-259x2-25.

Solution

  1. Step 1. Examine the quadratic :

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

    ( 3 x ) 2 = 9 x 2 ( 3 x ) 2 = 9 x 2
    (3)

    and

    5 2 = 25 . 5 2 = 25 .
    (4)
  2. Step 2. Write the quadratic as the difference of squares :
    9 x 2 - 25 = ( 3 x ) 2 - 5 2 9 x 2 - 25 = ( 3 x ) 2 - 5 2
    (5)
  3. Step 3. Write the factors :
    ( 3 x ) 2 - 5 2 = ( 3 x - 5 ) ( 3 x + 5 ) ( 3 x ) 2 - 5 2 = ( 3 x - 5 ) ( 3 x + 5 )
    (6)
  4. Step 4. Write the final answer :

    The factors of 9x2-259x2-25
    are (3x-5)(3x+5)(3x-5)(3x+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 x2-x-2x2-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)(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 . ( x + 2 ) ( x + 3 ) = x ( x + 3 ) + 2 ( x + 3 ) = ( x ) ( x ) + 3 x + 2 x + ( 2 ) ( 3 ) = x 2 + 5 x + 6 .
(7)

We see that the x2x2
term in the quadratic is the product of the xx-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 x2+5x+6x2+5x+6
and see if we can decide upon some general rules. Firstly, write down two brackets with an xx in each bracket and space for the remaining terms.

( x ) ( x ) ( x ) ( x )
(8)

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

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

Therefore, we have four possibilities:

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

Table 3
Option 1 Option 2 Option 3 Option 4
( x + 1 ) ( x + 6 ) ( x + 1 ) ( x + 6 ) ( x - 1 ) ( x - 6 ) ( x - 1 ) ( x - 6 ) ( x + 2 ) ( x + 3 ) ( x + 2 ) ( x + 3 ) ( x - 2 ) ( x - 3 ) ( x - 2 ) ( x - 3 )
x 2 + 7 x + 6 x 2 + 7 x + 6 x 2 - 7 x + 6 x 2 - 7 x + 6 x 2 + 5 x + 6 x 2 + 5 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 ax2+bx+c=0ax2+bx+c=0
    where aa, bb and cc have no common factors and aa is positive.
  2. Write down two brackets with an xx in each bracket and space for the remaining terms.
    (x)(x)(x)(x)
    (9)
  3. Write down a set of factors for aa and cc.
  4. Write down a set of options for the possible factors for the quadratic using the factors of aa and cc.
  5. Expand all options to see which one gives you the correct answer.

There are some tips that you can keep in mind:

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

Exercise 2: Factorising a Quadratic

Find the factors of 3x2+2x-13x2+2x-1.

Solution
  1. Step 1. Check whether the quadratic is in the form ax2+bx+cax2+bx+c
    with aa positive. :

    The quadratic is in the required form.

  2. Step 2. Write down two brackets with an xx
    in each bracket and space for the remaining terms. :
    ( x ) ( x ) ( x ) ( x )
    (10)

    Write down a set of factors for aa and cc. The possible factors for aa are: (1,3). The possible factors for cc are: (-1,1) or (1,-1).

    Write down a set of options for the possible factors of the quadratic using the factors of aa and cc. Therefore, there are two possible options.

    Table 4
    Option 1 Option 2
    ( x - 1 ) ( 3 x + 1 ) ( x - 1 ) ( 3 x + 1 ) ( x + 1 ) ( 3 x - 1 ) ( x + 1 ) ( 3 x - 1 )
    3 x 2 - 2 x - 1 3 x 2 - 2 x - 1 3 x 2 + 2 x - 1 3 x 2 + 2 x - 1
  3. Step 3. Check your answer :
    ( x + 1 ) ( 3 x - 1 ) = x ( 3 x - 1 ) + 1 ( 3 x - 1 ) = ( x ) ( 3 x ) + ( x ) ( - 1 ) + ( 1 ) ( 3 x ) + ( 1 ) ( - 1 ) = 3 x 2 - x + 3 x - 1 = x 2 + 2 x - 1 . ( x + 1 ) ( 3 x - 1 ) = x ( 3 x - 1 ) + 1 ( 3 x - 1 ) = ( x ) ( 3 x ) + ( x ) ( - 1 ) + ( 1 ) ( 3 x ) + ( 1 ) ( - 1 ) = 3 x 2 - x + 3 x - 1 = x 2 + 2 x - 1 .
    (11)
  4. Step 4. Write the final answer :

    The factors of 3x2+2x-13x2+2x-1
    are (x+1)(x+1) and (3x-1)(3x-1).

Factorising a Trinomial

  1. Factorise the following:
    Table 5
    (a) x2+8x+15x2+8x+15
    		(b) x2+10x+24x2+10x+24
    		(c) x2+9x+8x2+9x+8
    (d) x2+9x+14x2+9x+14
    		(e) x2+15x+36x2+15x+36
    		(f) x2+12x+36x2+12x+36
    Click here for the solution
  2. Factorise the following:
    1. x2-2x-15x2-2x-15
    2. x2+2x-3x2+2x-3
    3. x2+2x-8x2+2x-8
    4. x2+x-20x2+x-20
    5. x2-x-20x2-x-20

      Click here for the solution
  3. Find the factors for the following trinomial expressions:
    1. 2x2+11x+52x2+11x+5
    2. 3x2+19x+63x2+19x+6
    3. 6x2+7x+26x2+7x+2
    4. 12x2+8x+112x2+8x+1
    5. 8x2+6x+18x2+6x+1

      Click here for the solution
  4. Find the factors for the following trinomials:
    1. 3x2+17x-63x2+17x-6
    2. 7x2-6x-17x2-6x-1
    3. 8x2-6x+18x2-6x+1
    4. 2x2-5x-32x2-5x-3

      Click here for the solution

Factorisation by Grouping

One other method of factorisation involves the use of common factors. We know that the factors of 3x+33x+3
are 3 and (x+1)(x+1). Similarly, the factors of 2x2+2x2x2+2x
are 2x2x
and (x+1)(x+1). Therefore, if we have an expression:

2 x 2 + 2 x + 3 x + 3 2 x 2 + 2 x + 3 x + 3
(12)

then we can factorise as:

2 x ( x + 1 ) + 3 ( x + 1 ) . 2 x ( x + 1 ) + 3 ( x + 1 ) .
(13)

You can see that there is another common factor: x+1x+1. Therefore, we can now write:

( x + 1 ) ( 2 x + 3 ) . ( x + 1 ) ( 2 x + 3 ) .
(14)

We get this by taking out the x+1x+1 and seeing what is left over. We have a +2x+2x
from the first term and a +3+3 from the second term. This is called factorisation by grouping.

Exercise 3: Factorisation by Grouping

Find the factors of 7x+14y+bx+2by7x+14y+bx+2by
by grouping

Solution

  1. Step 1. Determine if there are common factors to all terms :

    There are no factors that are common to all terms.

  2. Step 2. Determine if there are factors in common between some terms :

    7 is a common factor of the first two terms and bb is a common factor of the second two terms.

  3. Step 3. Re-write expression taking the factors into account :
    7 x + 14 y + b x + 2 b y = 7 ( x + 2 y ) + b ( x + 2 y ) 7 x + 14 y + b x + 2 b y = 7 ( x + 2 y ) + b ( x + 2 y )
    (15)
  4. Step 4. Determine if there are more common factors :

    x+2yx+2y
    is a common factor.

  5. Step 5. Re-write expression taking the factors into account :
    7 ( x + 2 y ) + b ( x + 2 y ) = ( x + 2 y ) ( 7 + b ) 7 ( x + 2 y ) + b ( x + 2 y ) = ( x + 2 y ) ( 7 + b )
    (16)
  6. Step 6. Write the final answer :

    The factors of 7x+14y+bx+2by7x+14y+bx+2by
    are (7+b)(7+b) and (x+2y)(x+2y).

Figure 2
Khan Academy video on factorising a trinomial by grouping.

Factorisation by Grouping

  1. Factorise by grouping: 6x+a+2ax+36x+a+2ax+3
    Click here for the solution
  2. Factorise by grouping: x2-6x+5x-30x2-6x+5x-30
     Click here for the solution
  3. Factorise by grouping: 5x+10y-ax-2ay5x+10y-ax-2ay
     Click here for the solution
  4. Factorise by grouping: a2-2a-ax+2xa2-2a-ax+2x
     Click here for the solution
  5. Factorise by grouping: 5xy-3y+10x-65xy-3y+10x-6
     Click here for the solution

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