How to Succeed in Algebra – Factoring Non-Monic Trinomials – Part II

In Part II of this article, we are going to examine a sure-fire way to factor non-monic trinomials. We are going to examine the proof which gives us this powerful method. After you are done with this presentation, you will not only be able to factor any non-monic, you will understand the reason why this method works.

To begin our presentation, let us take an arbitrary non-monic trinomial: to do this we need to introduce the constants a, b, and c. Thus our generic non-monic becomes ax^2 + bx + c, in which a is an integer greater than 1. Our end aim is to reverse FOIL this so that we can factor it into the form (dx + e)*(fx + g), where d, e, f, and g are integers. If you now FOIL this product, you obtain dfx^2 + dgx+ efx + eg, which we can write as dfx^2 + (dg + ef)x + eg, since the common factor of both dg and ef is x.

If we compare this last expression with the original non-monic, we that df must be equal to a; dg + ef must equal b and eg must equal c. If this were not the case, we would have a factored expression which is incorrect. Now this is where some thought is required, so I enjoin you to put your thinking cap on and follow closely. Feel free to reread as necessary to make sure you are following the argument.

Our goal is to determine d,e,f, and g. To do this we use a clever trick. We multiply a and c together. We note that this must equate to dfeg, since a = df and c = eg. Now remember that the middle coefficient, or b, is equal to dg + ef. If we break ac down into all its component factors, and observe which sum is equal to b, then we have found dg and ef. If we then divide a, which is df by dg, we get df/dg = f/g, since the d’s cancel; similarly if we divide ef/df = e/d. We now have all four unknown letters solved, namely d,e,f, and g.

Because the last part was cryptic and probably somewhat hard to follow at first blush, I am going to do a specific example which will make this whole argument clear. Let us take the non-monic trinomial 6x^2 + 17x + 5. Let us multiply a*c or 6*5 = 30. Now break down 30 into all of its possible factor pairs: 1 30, 2 15, 3 10, and 5 6. Find the factor pair which when added or subtracted yields the middle coefficient 17. Clearly this is the pair 2 15. Since b = dg + ef, we know that 2 is dg and 15 is ef. If we divide 2 by 6 (which is dg by df) we get 1/3, which tells us that g is 1 and f is 3; similarly if we divide 6 by 15 and reduce to lowest terms, we get 6/15 or 2/5, which tells us that 2 is d and 5 is e. Thus the factorization is (dx + e)*(fx + g) or (2x + 5)*(3x + 1). If you FOIL this, you will see indeed that you get back the original non-monic.

To make this procedure clear, let us do another. Let us factor the non-monic 6x^2 + 41x + 70. We multiply 6*70 = 420. The factor pairs of 420 are 1 420, 2 210, 3 140, 4 105, 5 84, 6 70, 10 42, 14 30, and 20 21. The pair which added or subtracted produces 41 is 20 21. Thus 20 is dg and 21 is ef. When we divide both of these by a which is df and reduce, we have 20/6 = 10/3 = g/f; and 21/6 = 7/2, which is e/d. Thus 6x^2 + 41x + 70 can be factored into the form (dx + e)*(fx + g) = ( 2x + 7)*(3x + 10). Now how’s that for sure-fire!

Note that we have not considered here cases of non-monics which involve negative numbers. Such is the case with 6x^2 + x -70. The method, albeit with a slight adjustment, works just as well. Once you have the method down for the cases in which all the numbers are positive, applying it to cases in which either b or c or both are negative, is academic.