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Free Homework Math Help: Factoring By The Trial Method

 

Trial Method

When doing factoring it is a good idea to keep in mind the general form of a trinomial:

               ax2 = bx + c

The trial method takes into account the coefficients a, b and the constant c. Factoring by the trial method involves factoring both the a coefficient and the c constant. The factors are in the form of pairs. Two groups of parenthesis are written beforehand to contain the factors. Example:

               x2 + 8x + 15

The a coefficient does not have to be factored in this case since its factors are one only. In this example, just write the variables in the first term of each group of parenthesis:

               (x      )(x      )

Next, factor the c constant which is equal to 15. The factors in pairs for 15 are: 1*15, 3*5, 5*3 and 15*1. Also, since all the terms are positive, the signs of the constants in the result will be positive.

               (x+1)(x+15)
               (x+3)(x+5)
               (x+5)(x+3)
               (x+15)(x+1)

In this particular case, the last two expressions don't have to be written since they would yield the same result as the first two. The next step is to multiply the outer terms and then the inner terms of both factors. Next, add the results to see if it equals the middle term in the polynomial.

The first expression produces x + 15x = 16x. The second produces 3x + 5x = 8x which is the same as the middle term in the original form of the polynomial. Foil can also be used to see if the result is the same as the original polynomial. The factors of x2 + 8x + 15 are (x+3)(x+5).

The following example shows how to organize factors:

The trial method can factor some quadratic equations

A different factoring example:

Factoring certain quadratic equations can be done with the trial method

A more complicated example is:

               6x2 + 25x + 14

Find the pairs of factors of the leading coefficient: 1*6, 2*3, 3*2 and 6*1.

Find the pairs of factors of the constant: 1*14, 2*7, 7*2 and 14*1.

Place the factors in parenthesis with the variable:

               (x + 1)(6x + 14)
               (x + 2)(6x + 7)
               (x + 7)(6x + 2)
               (x + 14)(6x + 1)
               (2x + 1)(3x + 14)
               (2x + 2)(3x + 7)
               (2x + 7)(3x + 2)
               (2x + 14)(3x + 1)

The second to last group of quadratic factors, (2x + 7)(3x + 2), is the answer since the sum of the inner product of terms and the outer product of terms is 2*2x + 7*3x = 4x + 21x = 25x.