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Think of the process as multiplying two binomials in reverse.
(y-1)(y-12)
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.
y^2-13y+ 12
In this case, we have 12. 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 12 | 12 |
| - 1 and - 12 | 12 |
| 2 and 6 | 12 |
| - 2 and - 6 | 12 |
| 3 and 4 | 12 |
| - 3 and - 4 | 12 |
Next, let's consider the coefficient of the linear term. y^2 -13y+ 12 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 12 | 13 |
| - 1 and - 12 | - 13 |
We found the factors whose product is 12 and whose sum is - 13. y^2 -13y+ 12 ⇔ (y-1)(y-12)
Distribute y-12
Distribute y
Distribute - 1
Subtract term
After applying the Distributive Property and simplifying, the result is the same as the given expression. Therefore, we can be sure our solution is correct!