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Q&A #104

Teachers' Lounge Discussion: Factoring trinomials

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From: Terry Dorain <twdorain@zfree.co.nz>
To: Teacher2Teacher Public Discussion
Date: 2000100703:42:18
Subject: Re: Factoring trinomials

	I don't know whether this will help or not but I have had a good deal
of success with the following method. I have noticed that many
teachers try a 'guess and check' method that runs into trouble when
the coefficient of x^2 is greater than 1.
Starting with a trinomial with the x^2 coefficient equal to 1
e.g.  x^2 + 3x + 2
There is no common factor to all terms so the trinomial is broken into
four terms 
e.g.  x^2 + x + 2x + 2
(the break up of the middle term must be a combination of factors of
the final term, in this case 2x1)
The student then divides the expression in two i.e. x^2+x and 2x+2
and factorises each half independently x(x+1)+2(x+1)
We now have a common factor of (x+1) which is removed and giving the
final factorisation of (x+1)(x+2)
This may seem laborious but makes the factorisation of trinomials with
x^2 coefficients greater than one very easy.

e.g.  To factorise 3x^2 + 8x + 4  
We still break up the middle term but instead of using the factors of
the final term 4 to achieve this break up we must multiply the
coefficient of x^2 i.e. 3 and the constant term i.e. 4 giving 12
The break up of the middle term must use factors of 12.
This gives us 3x^2 + 6x + 2x + 4
Factorise each half independently 3x(x+2)+2(x+2)
Remove the common factor (x+2)
Giving the final answer of (x+2)(3x+2)

Hope this makes sense and helps

Regards

Terry

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