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

In Part II of this article, we will examine a safe way to factor non-monomial trinomials. We will examine the proof that this powerful method gives us. After you’re done with this presentation, you’ll not only be able to factor any nonmonic, but you’ll understand why this method works.

To begin our presentation, we take an arbitrary nonmonic trinomial: to do so we must introduce the constants a, bi and c. So our nonmonic generic becomes ax^2 + bx + c, where a is an integer greater than 1. Our ultimate goal is to invert this into FOIL so that we can factor it into the form (dx + e)*(fx). + g), where d, e, fig are integers. If you now FOIL this product, you get dfx^2 + dgx+ efx + for example, which we can write as dfx^2 + (dg + ef)x + for example, since the common factor of dg and ef is x.

If we compare this last expression with the original non-monic, we have that df must be equal to aa; dg + ef must equal abi eg must equal c. If this were not the case, we would have a factored expression that is incorrect. Now, this is where you need to think a little, so I invite you to put your thinking head on and follow closely. Feel free to reread if necessary to make sure you’re following the plot.

Our goal is to determine d, e, fi g. To do this we use a clever trick. Let’s multiply here together. We note that this must equal dfeg, since a = df ic = p. Now remember that the average coefficient, ob, is equal to dg + ef. If we break ac into all its component factors and note which sum equals ab, then we have found dg and ef. If we then divide a, which is df by dg, we get df/dg = f/g, since d cancels; similarly if we divide ef/df = e/d. Now we have the four unknown letters solved ie d,e,fi g.

Since the last part was cryptic and probably a little hard to follow at first glance, I’ll give a concrete example that will make this whole argument clear. We take the nonmonic trinomial 6x^2 + 17x + 5. We multiply a*co 6*5 = 30. Now break 30 into all 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. 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 if it is 3; similarly if we divide 6 by 15 and reduce to the lowest terms, we get 6/15 or 2/5, which tells us that 2 is di 5 is e. So the factorization is (dx + e)*(fx + g) or (2x + 5)*(3x + 1). If you FOIL this, you will indeed see that you get the original non-monic back.

To clarify this procedure, we do another one. We factor the non-monic 6x^2 + 41x + 70. We multiply 6*70 = 420. The pairs of factors of 420 are 1 420, 2 210, 3 140, 4 105, 5 84, 6 70, 10 42, 14 30 and 20 21. The pair that adds or subtracts to produce 41 is 20 21. Thus, 20 is dg and 21 is ef. When we divide both by what 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 in the form (dx + e)*(fx + g) = (2x + 7)*(3x + 10). Now how is that safe!

Note that we have not considered here cases of nonmonics involving negative numbers. This is the case for 6x^2 + x -70. The method, although with a slight adjustment, works just as well. Once you have the method down for the cases where all the numbers are positive, applying it to the cases where boco are both negative is academic.

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