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Step 3: In the table, look for a pair of factors whose sum is 5. For this example, the two factors for - 24 that sum to 5 are - 3 and 8. These numbers repre sent the two coefficients for x : - 3 for – 3 x and 8 for 8x that is used to replace the constant value b , in bx, which is 5 x .

Using the quadratic formula, the roots are:

The trinomial expression has the following factors:

Step 4: Regroup and factor the expression.

We demonstrate this method using the same exam ple as above, 6 x 2 + 5 x – 4.

Step 1: Use the Quadratic Formula to solve for the roots of the following equation 6 x 2 + 5 x – 4 = 0.

As we did earlier, we identify the coefficients for each of the terms in the trinomial. In this example, a = 6, b = 5, and c = - 4. Using these coefficients, we insert them into the formula and find:

Since students already know how to factor integers, listing the possible factors in a table allow students to see the possible options, which helps them to identify the correct factors. Method 1 Extension: This is similar to Method 1. The difference is that, after finding the values for m and n, instead of separating b into m + n, we can write these terms in the form ( ax + m ) ( ax + n ) di rectly. This allows us to factor out any common constants. For example, using the same problem we identify the coefficients for each term as before, 6 x 2 + 5 x – 4; a = 6, b = 5, c = - 4. Then, using the table of factors, we let m = - 3, n = 8 and ac = 6 . We write it out in the following way, ( ax + m ) ( ax + n ) = (6 x – 3) (6 x + 8). Then, we factor out the common factor for each group of terms, 3(2 x – 1) × 2(3 x + 4). At this point, we divide by the common factors 2 and 3, or 6, because we essentially multi plied by these factors twice in the previous step. As a result, our factored trinomial is, 6 x 2 + 5 x – 4 = (3 x + 4) (2 x – 1). Another way to view this, is since the factors are the zeros of the polynomial function and our poly nomial has no common constant factor, the gener ated constants, 2 and 3, can be deleted without changing the roots of the polynomial function. Since ( ax + m ) ( ax + n ) = 0, solving for the roots by equating (3 x + 4) = 0; (2 x – 1) = 0, is the same as (6 x + 8) = 0; (6 x – 3) = 0. The roots will not change. This same perspective is used when ex plaining the quadratic formula method and the gra phing method shared in this article.

Step 2: Place the roots into the factor group form:

Step 3: Eliminate the fractions by multiplying the entire expression by a common denominator, 6. The result is 6 x 2 + 5 x – 4 = (3 x + 4) (2 x – 1). This method is especially useful when the product of a and c is large and Method 1 results with too many combinations to guess and check.

Method 3: Factoring by Tic Tac Toe

For this method, we place the polynomial into standard form on the first line of the Tic Tac Toe table. Next, we identify possible factors for the leading term, a, and the constant term, c. These are placed in the corresponding vertical columns on the Tic Tac Toe table. Then, we cross multiply the factor of one of the leading terms with one of the factors of the constant. We write the product in the middle of the second and third row and add them together. When the sum equals the middle term, b of the polynomial, we write out the factors in the second line and third line into two separate paren theses. These are the factors we were seeking. The details for the Tic Tac Toe method using our exam ple follows.

Step 1: Draw a Tic Tac Toe table and write the pol ynomial in standard form (see Table 1). Place term

Method 2: Factoring by the Quadratic Formula

Virginia Mathematics Teacher vol. 46, no. 2

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