HOW TO FACTOR. Next you reason that if it factors, then the factorization will look something like,

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1 HOW TO FACTOR ax bx I now want to talk a bit about how to fator ax bx where all the oeffiients a, b, and are integers. The method that most people are taught these days in high shool (assuming you go to a high shool where the basis of algebra are still onsidered important) is the trial-and-error method. In this method, you start with a trinomial suh as, 5x Next you reason that if it fators, then the fatorization will look something like, (_ x _)(_ x _) From here it is easy to see that the numbers that go into the red blanks (the oeffiients of x) must multiply to 3, and the numbers that go into the blue blanks (the onstant terms) must multiply to. Thus, you proeed by finding all pairs of integers that multiply to 3 and all pairs that multiply to, and then by trial-and-error, you begin plugging those numbers into the blanks until you find a ombination that gives you the right middle term. For example, ( )( x ) yields the orret middle term of 5x, but ( )( x ) results in a middle term of 7x whih is not what we are looking for. The advantage of the trial and error method is that it is oneptually easy to understand, but the downside is that if our onstant term or our oeffiient for x have a lot of fators, then it an be very diffiult and time onsuming to find just the right one that works. Thus, I want to demonstrate and alternative method that is easier to implement and that takes all the guess work out of fatoring, and then I ll show you why the method works. First, though, reall how we went about fatoring a trinomial like the following where the oeffiient of x is. x 5x 6 Our method in this ase is to find integers u and v suh that u v= 5 and uv = 6. Clearly, u = and v = 3 do the trik, and one we ve found u and v, we an immediately omplete our fatorization as follows. x 5x 6 = ( x u)( x v) = ( x )( x 3) For a trinomial of the form partiular: ax bx, we are now going to follow similar steps. In. Find integers u and v suh that u v= b and uv = a.. Write ax bx as ax ux vx = ( ax ux) ( vx ).

2 3. Fator by grouping. Here is an illustration using our original trinomial, 5x We now seek integers u and v suh that u v= 5 and uv = 3 = 6, and one again, u = and v = 3 fit the bill. Thus, we rewrite our polynomial replaing 5x with x, and then we fator by grouping. 5x x ( x) ( ) x ( ) ( ) ( x )( ) And we re done! Notie in the above example all our oeffiients were positive integers. If some of our oeffiients are, on the other hand, negative, then the method still works, but you may have to fator out a negative sign at some point in order to omplete the fatoring by grouping. For instane, onsider the following example. 5x x ( x) ( ) x ( ) ( ) ( x )( ) And now we want to understand why this method works, and to keep things simple, I will assume that all the oeffiients are positive integers. And as we saw in the last example, if we do have some negative oeffiients, then the only adjustment that might have to be made to the method is to fator out a negative rather than a positive ommon fator when doing the fatoring by grouping. Hene, let s begin with ax bx where all the oeffiients are assumed to be positive integers and the greatest ommon fator of a, b, and is. The next step is to find integers u and v suh that, u v= b and uv = a

3 We an usually find suh a pair by trial-and-error, but by treating this as a system of two equations in two unknowns, u and v, we an easily find a solution by using the quadrati formula. For example, v= b u and u v= b and uv = a uv = u( b u) = ub u = a 0 = u bu a b± b 4a u = This last formula tells us several things. First, b 4a must be a perfet square sine, otherwise, u won t be an integer. And seond, we get two solutions to our equation, and b b 4a if we let one of them be u =, then the other solution an be denoted by b b 4a v =. However, it is quiker to get v by just using the equation above that v= b u. The third thing we notie is that if our system of equations, u v= b and uv = a, has a solution, then there is only one pair of integers that we an represent as u and v. It doesn t really matter, though, whih number we designate as u and whih as v, but when onsidered as a pair, there is only one solution to the system. This will be important to us! Next, we want to show that if a pair of integers, u and v, exist suh that u v= b and uv = a, then the trinomial is definitely fatorable. To see how, just follow the following hain of logi. ax bx = ax ux vx uv x ux vx = ( ) = uv x u x v x ( ) ( ) = uv x u x v x = [ ux( vx ) ( vx ) ] = [ ( ux )( vx ) ]

4 At this point, notie that sine the oeffiients of our original trinomial are integers, the must at some point anel out. Hene, we must be able to write as = mm where m divides u and m divides v. In other words, there exist m and m suh that = mm, u = r m, and v= s m. Hene, we an ontinue our derivation above as, [ ( ux )( vx ) ] = ( rm x mm )( sm x mm ) mm = ( rm x mm ) ( sm x mm ) m m rm m m sm m m = x x m m m m = ( rx m )( sx m ) We have now overed two steps of what I onsider a three-step argument. In the first step, we showed that if integers u and v exists suh that u v= b and uv = a, then that pair is unique. We an always swith between whih integer we all u and whih integer we all v, but the pair itself is one of a kind. In the seond step, we showed that if our integers u and v exist, then the trinomial an indeed be fatored. And in our final third step, we are going to show that if the integers u and v exist, then we an fator the trinomial using the partiular method of fatoring by grouping that we illustrated at the beginning. Let s get into it! Now if our trinomial an be fatored into a produt of two binomials with integer oeffiients, then that means we an write, ax bx = ( p x p )( p x p ) 3 4 If we multiply out the right hand side of this equation a bit, then we an rewrite it as, 3 4 = ( p x p )( p x p ) p p x p p x p p x p p But notie that in this instane p p 3 p p 4 = b and pp3 pp4 = pp3 pp4 = a. Hene, we an set u = pp3 and v= pp4, and remember that this is the only pair of integers whih will now solve the equations u v= b and uv = a. Again, it doesn t matter whih of these produts we all u and whih one we all v. It will still be the same pair in one order or the other, and we an proeed as follows.

5 ax bx = ax ux vx = p px ppx ppx pp = ( p px ppx) ( ppx pp) = px( px p ) p ( px p ) 3 4 = ( px p)( px p) 3 4 And we re done!