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(x+3)(x-5)
To factor a trinomial with a leading coefficient of 1, think of the process as multiplying two binomials in reverse. Let's start by taking a look at the constant term.
x^2-2x-15
In this case, we have - 15. This is a negative number, so for the product of the constant terms in the factors to be negative, these constants must have different signs - positive or negative.
| 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- 2x-15 For this term, we need the sum of the factors that produced the constant term to equal the coefficient of the linear term, - 2.
| Factors | Sum of Factors |
|---|---|
| 1 and - 15 | - 14 |
| -1 and 15 | 14 |
| 3 and - 5 | - 2 |
| -3 and 5 | 2 |
We found the factors whose product is - 2 and whose sum is - 15. x^2- 2x-15 ⇔ (x+3)(x-5)
Distribute (x-5)
Distribute x
Distribute 3
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!