Register to reply

Proof of PSF quadratic factorising

by bairdos
Tags: factorising, proof, quadratic
Share this thread:
Jul16-14, 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.)

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 (x-p/q)(x-m/n) = 0
multiplying by qn: (qx-p)(nx-m) = 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^2-4ac) 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 :)
Phys.Org News Partner Mathematics news on
Math journal puts Rauzy fractcal image on the cover
Heat distributions help researchers to understand curved space
Professor quantifies how 'one thing leads to another'
Jul17-14, 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^2-4ac))/2 and (-b-root(b^2-4ac)/2).
The trick is that b^2-4ac 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^2-4ac 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.

Register to reply

Related Discussions
Factorising a quadratic? General Math 19
Quadratic Proof Precalculus Mathematics Homework 16
Factorising a complicated Quadratic Calculus 7
Factorising quadratic Precalculus Mathematics Homework 10
Factorising a Quadratic I still struggle.. Introductory Physics Homework 10