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Think of the process as multiplying two binomials in reverse.
Prime
| Factor Constants | Product of Constants |
|---|---|
| 1 and 15 | 15 |
| - 1 and - 15 | 15 |
| 3 and 5 | 15 |
| - 3 and - 5 | 15 |
Next, let's consider the coefficient of the linear term. x^2-11x+15 For this term, we need the sum of the factors that produced the constant term to equal the coefficient of the linear term, - 11.
| Factor Constants | Product of Constants and Variables | Sum of Products |
|---|---|---|
| 1 and 15 | 1x and 15x | 16x |
| - 1 and - 15 | - 1x and - 15x | - 16x |
| 3 and 5 | 3x and 5x | 8x |
| - 3 and - 5 | - 3x and - 5x | - 8x |
We did not find a factor constants whose product is 15 and also cause the sum of the products of x and the factor constants to equal - 11x. Thus, the polynomial is not factorable.