Q&A #104

Teachers' Lounge Discussion: Factoring trinomials

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From: Terry Dorain

To: Teacher2Teacher Public Discussion
Date: 2000100704:42:18
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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