Overview
When the a term has a coefficient of 1, polynomials can be factored into the form (a +b) (a +c) without worrying about its effect. However, if that a term has a coefficient not equal to 1, the coefficient must be considered when the polynomial is factored.
Review of FOIL
Suppose (x + 2) (x + 3) is multiplied. The First terms, x ∙x, become x2. The Outer terms, 3x, are next, then the inner terms, 2x. The last terms 2∙3, equal 6. When the polynomial is put in order, it follows the FOIL pattern of x2 + 2x + 3x + 6, or x2 + 5x + 6. (It follows the pattern of x2 + bx + c.) Because x2 is equal to 1∙ x2 in that case, the FOIL method can still be used. It just takes more attention to factor the expression.
Common Factors
Factor out the largest common factor if that a term is not equal to 1 before using FOIL to do the rest. Suppose the polynomial is 5x2 + 15x + 125. The largest common factor is 5 in all the terms. Before using FOIL, the trinomial can be partially factored as 5(x2 + 10x + 25). Then it can be factored the rest of the way as 5(x + 5) (x + 5). Suppose that the trinomial is 24x2 + 76x + 40. The common factor of 4 can be factored out to leave 4(6x2 + 19x + 10).
UnFOILing the FOIL
In the above example, 6x2 + 19x + 10, the a term is equal to 6. Therefore, the two First terms will have a product of 6x2. They could be x and 6x or 2x and 3x, as x∙ 6x = 6x2, and 2x∙3x is also 6x2. The two Last terms will have a product of 10, so they could be 5∙2; 2∙5; 1∙10; or 10∙1. Then use trial and error to determine which Outer and Inner terms will have a sum of 19, the b term. Suppose the first combination is (x +5) (6x + 2). Using FOIL, the multiplication will result in 6x2 + 2x +30x +10. The 6x2 is correct, and the 10, but the middle term, 32x, is too large. What about (x +2) (6x +5)? The only terms to try are 5x + 12x, which is 17x, too small. Similarly, (x + 1) (6x + 10) results in a middle term of 16x, smaller still; and (x + 10) (6x + 1); 61x. Next, (2x + 1) (3x + 10). The sum of the middle terms is 23x, still too large; and (2x + 10) (3x + 1); 36x. Next, (2x + 2) (3x + 5), 16x again. That leaves (2x + 5) (3x + 2); 19x.
Reading the Signs
Whatever the value is of a, the sign within the factored terms matter. Suppose the trinomial is of the form ax2 +bx +c. It will factor as two additions in (x + p) (x + q). If the trinomial is in the form ax2 –bx +c, it will factor as two subtractions (x – p) (x – q). If the last term is negative, as in ax2 +bx –c or ax2 –bx –c, it will factor as either (x + p) (x – q) or (x – p) (x + q). If the middle term is positive, then p is less than q, and (x + p) (x – q) is the correct choice. If the middle term is negative, then q is less than p and (x – p) (x + q) is the correct choice.
Figure 3: The direction of the signs matter.
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