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Is there a GCF between all of the terms in the given expression? If so, you should factor out that first.
(a-4)(a-9)
To factor a trinomial with a leading coefficient of one, think of the process as multiplying two binomials in reverse. Let's start by taking a look at the constant term.
a^2-13a+36
In this case, we have 36. This is a positive number, so for the product of the constant terms in the factors to be positive, these constants must have the same sign (both positive or both negative.)
| Factor Constants | Product of Constants |
|---|---|
| 1 and 36 | 36 |
| -1 and - 36 | 36 |
| 2 and 18 | 36 |
| -2 and - 18 | 36 |
| 3 and 12 | 36 |
| -3 and - 12 | 36 |
| 4 and 9 | 36 |
| -4 and - 9 | 36 |
| 6 and 6 | 36 |
| -6 and - 6 | 36 |
Next, let's consider the coefficient of the linear term. a^2-13a+36 ⇔ a^2+(-13)a+36 For this term, we need the sum of the factors that produced the constant term to equal the coefficient of the linear term, - 13.
| Factors | Sum of Factors |
|---|---|
| 1 and 36 | 37 |
| -1 and - 36 | - 37 |
| 2 and 18 | 20 |
| -2 and - 18 | - 20 |
| 3 and 12 | 15 |
| -3 and - 12 | - 15 |
| 4 and 9 | 13 |
| -4 and - 9 | - 13 |
| 6 and 6 | 12 |
| -6 and - 6 | - 12 |
We found the factors whose product is 36 and whose sum is - 13. a^2+(-13)a+36 ⇔ (a-4)(a-9)