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Proof of PSF quadratic factorising 
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#1
Jul1614, 10:13 AM

P: 2

Hi! I came across the below thread where a user ('krackers') asked for a proof of the PSF factorising method for quadratic equations.
The thread is now closed so I'd like to post my proof here. (The proof considers the simplest case where there are no common factors for a,b,c in the quadratic. A proof for cases with a common factor can easily be created with the same structure as below with a common factor 'd' added.) http://www.physicsforums.com/showthread.php?t=621835 Given a polynomial ax^2 + bx + c = 0 there are rational roots if (b^2  4ac) is 0 or a positive square number The rational factors can be written as p/q and m/n (fractions in simplest form) therefore the polynomial can be written as (xp/q)(xm/n) = 0 multiplying by qn: (qxp)(nxm) = 0 which becomes nqx^2  mqx npx + mp = 0 so a = nq, c =mp and b is the sum of the middle terms (mq and np) note that a*c= mnpq and that the product of the two numbers (mq) and (np) is also mnpq so therefore if the polynomial has rational roots then there must be two numbers (mq and np) that add to give 'b' and multiply to give the product of 'a' and 'c'. These numbers can be found by trial and error (because m,n,p,q are all integers) To prove that there is only and only one pair of numbers S,T you can solve the simultaneous equations: S + T = b, ST = ac (using the quadratic formula and the condition that (b^24ac) is 0 or a positive square number) (of course there can only be one pair of numbers because otherwise there would be more than one factorisation of the polynomial resulting in there being multiple possible sets of roots). I developed this proof because I felt it was my obligation as a high school maths tutor to understand why everything that I teach works :) 


#2
Jul1714, 01:15 AM

P: 2

Addendum (I'm a newbie and couldn't see a way to edit the original post):
I wondered why S,T always turn out to be integers when you calculate them by solving simultaneous equations. The formulas for S,T are (b+root(b^24ac))/2 and (broot(b^24ac)/2). The trick is that b^24ac is only even if b is even so the numerator of each fraction consists of two even numbers added or subtracted together. In both cases this gives an even number which is halved to give an integer. Similarly b^24ac is only even if b is odd and the numerator of each fraction consists of two odd numbers added or subtracted together. This also gives an even number in both cases which is halved to give an integer. Ahh, closure. 


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