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Year 10 Interactive Maths - Second Edition


Cross-Multiplication Method

The cross-multiplication method is useful to factorise (factorize) quadratic trinomials.

Place the linear factors one above the other as shown below.

The linear factors of (x+1) and (x+10) produce a middle term of 11x. But we need the linear factors that produce 7x as the middle term.

Multiply the numbers along the arms of the cross, and then add the products.  That is,
 10 + 1 = 10x+ x = 11x.  This is not the middle term.

The linear factors of (x+2) and (x+5) produce the middle term of 7x.

 5 + 2 = 5x+ 2x = 7x.  This is the middle term.

So x^2 + 7x + 10 = (x+2)(x+5)

Note:
  • The solution is read across.
  • 10 = 1 × 10 = 2 × 5 = 1 × 10 = 2 × 5
    We did not try 1 × 10 and 2 × 5 because the middle term is positive.
  • The answer can be checked using the Distributive Law.  That is:

Check the answer by expanding (x+2)(x+5).


Example 9

Factorise (factorize) the following four quadratic trinomials.

Solution:

Factors of x^2 and 14 provide the linear factors to work with.

The linear factors (x+1) and (x+14) do not produce the middle term of 9x.

 

 

 


     Multiply the numbers along the arms of the cross, and then add the products.
           14 + 1 = 14x+ x = 15x.
     We reject this pair as the middle term is 9x.

The linear factors (x+2) and (x+7) produce the middle term of 9x.

      7 + 2 = 7x+ 2x = 9x.  We accept this pair as the middle term is 9x.

So, x^2 + 9x + 14 = (x+2)(x+7).

Factors of x^2 and 12 provide the linear factors to work with.

The linear factors (x-3) and (x-4) produce the middle term of -7x.

      x× – 4 + –3 = – 4x – 3x = –7x.  We accept this pair as the middle term is –7x.

So, x^2 - 7x + 12 = (x-3)(x-4).


Note:

We did not try 1 × 12, 2 × 6 and 3 × 4 because the middle term is negative.  Also, we did not try
2 × 6 and 1 × 12 because we have already obtained the middle term by using 3 × 4.

Factors of x^2 and -7 provide the linear factors to work with.

The linear factors (x+1) and (x-7) produce the middle term of -6x.

       x × –7 + x × 1 = –7x + x = –6x.  We accept this pair as the middle term is –6x.

So, x^2 - 6x - 7 = (x+1)(x-7).

Note:

We did not try 1 × 7 because we have already obtained the middle term by using 1 × 7.


Factors of x^2 and -8 provide the linear factors to work with.

The linear factors (x-1) and (x+8) produce the middle term of 7x.

So, x^2 + 7x - 8 = (x - 1)(x + 8)


Note:

We did not try 1 × 8, 2 × 4 and 2 × 4 because we have already obtained the middle term by using 1 × 8.


Key Terms

cross-multiplication method

 

Study Another Topic in Chapter 10: Factorisation Techniques

Highest Common Factor ] Factorisation using the Common Factor ] The Difference of Two Squares ] Quadratic Trinomials ] [ Cross-Multiplication Method ] Factors of Quadratic Trinomials of the Type ax² + bx + c ] Use of Perfect Squares ] Use of Substitution ] Use of a Common Factor ] Factorisation of Four Terms ] Grouping 'Three and One' ] Real Numbers ] Completing the Square ] Problem Solving Unit ] Projects ] Symbols ] Index ]

 

Study Another Chapter
 

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