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How to Succeed in Algebra – Factoring Non-Monic Trinomials – Part II
In Part II of this article, we’ll look at a surefire way to factor non-monic trinomials. We’ll look at the evidence that gives us this powerful method. Once you complete this presentation, you will not only be able to factor any non-monic, but you will also understand why this method works.
To begin our presentation, let us take an arbitrary non-monic trinomial: to do this, we must introduce the constants a, b and c. So our generic non-monic becomes ax^2 + bx + c, where a is an integer greater than 1. Our end goal is to reverse FOIL this so we can factor it as (dx + e)* (fx + g), where d, e, f and g are integers. If you FOIL now this product, you get dfx^2 + dgx+ efx + eg, which we can write as dfx^2 + (dg + ef)x + eg, since the common factor of dg and ef is x.
If we compare this last expression with the non-monic original, we know that df must equal a; dg + ef must equal b and eg must equal c. If not, we would have a factored expression that is incorrect. Now, this is where some thinking is in order, so I urge you to think and follow closely. Feel free to proofread if necessary to make sure you’re following the argument.
Our objective is to determine d,e,f and g. To do this, we use a nifty trick. We multiply a and c together. We note that this must equal dfeg, since a = df and c = eg. Now remember that the middle coefficient, or b, is equal to dg + ef. If we decompose ac 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 out; the same if we divide ef/df = e/d. We have now solved the four unknown letters, namely d, e, f and g.
Because the last part was cryptic and probably somewhat hard to follow at first glance, I’ll make a specific example that will clarify this whole argument. Consider the non-monic trinomial 6x^2 + 17x + 5. Multiply a*c or 6*5 = 30. Now break down 30 into all of its possible pairs of factors: 1 30, 2 15, 3 10, and 5 6. Find the pair of factors which, added or subtracted, gives the average coefficient 17. This is clearly 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 the lower terms, we get 6/15 or 2/5, which tells us that 2 is d and 5 is e. So the factorization is (dx + e)*(fx + g) or (2x + 5)*(3x + 1). If you FOIL that, you will indeed see that you get the original non-monic back.
To clarify this procedure, let’s do another one. Factor the non-monic 6x^2 + 41x + 70. We multiply 6*70 = 420. The factor pairs of 420 are 1420, 2210, 3140, 4105, 5 84, 6 70, 10 42, 14 30 and 20 21. The pair that added or subtracted produces 41 is 20 21. So 20 is dg and 21 is ef. When we divide both by a which is df and reduce, we have 20/6 = 10/3 = g/f; and 21/6 = 7/2, which is e/d. So 6x^2 + 41x + 70 can be factored as (dx + e)*(fx + g) = ( 2x + 7)*(3x + 10). Now how safe is it!
Note that we have not considered here the cases of non-monics which involve negative numbers. This is the case of 6x^2 + x -70. The method, although with a slight adjustment, works just as well. Once you have the method for cases in which all numbers are positive, applying it to cases in which b or c or both are negative is academic.
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