Algebra - Ch. 7: Factoring Practice Problems (5 of 21) Example 4

In this video I will use 3 different methods (trial/error/guessing, product-sum, and foil method) to factor 8a^3-25a^2+3a. Example 4

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[Lglandorf]

When factoring 8a^2 - 25a + 3 have you ever seen the method where you put the first coefficient in each set of parentheses: (8a - )(8a - )? Next you find 2 factors of 24 (8 x 3) that add (since 3 is positive) up to 25. Choices are 1 x 24, 2 x 12, 3 x 8, 4 x 6. 1 and 24 work so we have (8a - 1)(8a - 24). Now see if you can divide by a common factor (common to just the terms in the same parentheses) from either or both sets of parentheses. (8a - 1) stays the same, but each term in (8a - 24) can be divided by 8, leaving (a - 3). Therefore the answer is (8a - 1)(a - 3). Another example: 6x^2 + 5x - 6 gives us (6x + )(6x - ). Finding two factors of 36 (6 x 6) that subtract (3rd term is negative) to get 5 would be 4 x 9 with the larger digit going with the plus sign (second term is positive). This gives us (6x + 9)(6x - 4). Now divide by a common factor in each set of parentheses. The first parentheses has a common factor of 3 and the second parentheses has a common factor of 2. Dividing them by 3 and 2 respectively, gives the answer of (2x + 3)(3x + 2). A student of mine showed me this years ago. :)