How to Factor Trinomials When A Does Not Equal One
Factoring trinomials is challenging enough, but when the first term has a leading coefficient, it becomes even more difficult. Factoring a trinomial where "a" does not equal 1 requires some guessing and checking, but there are some steps you can take to make the process a little easier before you begin. Be prepared with plenty of paper when you tackle these problems, because it can take several tries to find the right combination of factors for your answer.
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Instructions
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Put the problem in standard form, which resembles ax^2 + bx +c. In 14x + 2x^2 + 24, you need to rearrange it by moving the 2x^2 to the front of the equation. This gives you 2x^2 + 14x + 24.
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Check if you can factor out any common factors. In the first example, each term is divisible by 2. Factor out the 2 to give you 2 (x^2 + 7x + 12).
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Factor the trinomial that is inside the parentheses. Make a list of the factors of 12, then determine which combination would add together to give you 7. The factors of 12 are 1 * 12, 2 * 6, and 3 * 4. The pair that adds together to make 7 is 3 and 4, so the problem factors to 2(x + 3)(x + 4).
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Make a list of all the possible factors of the first and last terms for problems that do not contain a common factor, such as 6x^2 + x - 12. The factors of 6 are 1 * 6 and 2 * 3. The factors of -12 are 1 * -12, -1 * 12, 2 * -6, -2 * 6, 3 * -4 and -3 *4.
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Choose a pair of factors for each number and create a list of all possible factor combinations. For instance, if you use 2 * 3 and -3 * 4 as the two factors, you could have two possible combinations: (2x - 3)(3x + 4) or (2x + 4)(3x -3).
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Use the FOIL method to check if the combinations you have created give you the original trinomial. For these two combinations, (2x - 3)(3x + 4) yields 6x^2 - x - 12, and (2x + 4)(3x - 3) yields 6x^2 - 6x - 12, which are not correct.
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Continue trying combinations until you find the correct one. Start by switching the signs in the parenthesis of the two you just tried: (2x + 3)(3x -4) and (2x - 4)(3x +3). The first combination yields 6x^2 + x -12, so it is the correct way to factor this trinomial. You do not need to try the second one because there will be only one correct answer.
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Tips & Warnings
Notice that the combination (2x - 3)(3x +4) yields a trinomial that is the same as the original one, except the sign of the second term is incorrect (6x^2 - x + 12). When this occurs, the answer will be the same combination with different signs: (2x + 3)(3x - 4).
Always list all of the factors for the first and third terms before beginning. If you cannot find the correct combination after trying all of the factors, consider whether or not there is a factor that you missed. Your calculator can help you find the factors at the beginning of the problem.
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Comments
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1350ng
Oct 14, 2010
Back to the above Example. Solve: x^2 + 7x + 12 = 0. Solution. The Rule of Signs indicates that the 2 real roots are both negative. Write the factors-sets of c = 12: (-1, -12) (-2, -6) (-3, -4). The sum of the third set is -7 = -b. The 2 real roots are: -3 and -4. There are no needs of factoring. -
1350ng
Oct 14, 2010
When the constants a, and c are large numbers, the factoring method becomes confusing and consumes too much time. Guessing and checking (trial and error) the 4 terms of the two binomials (mx + n)(px + q)= 0 is a very hard work. There is a new method, called Diagonal Sum Method (Trafford 2009, that directly gives the 2 real roots without factoring. It is also a trial and error method, same as the factoring one, but it reduces the number of permutations in half by using a Rule of Signs for Real Roots.
