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(x-3)(x+4)
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+x- 12
In this case, we have - 12. 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 -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. x^2+1x-12 For this term, we need the sum of the factors that produced the constant term to equal the coefficient of the linear term, 1.
| Factors | Sum of Factors |
|---|---|
| 1 and -12 | -11 |
| -1 and 12 | 11 |
| 2 and - 6 | -4 |
| -2 and 6 | 4 |
| 3 and - 4 | -1 |
| -3 and 4 | 1 |
We found the factors whose product is - 12 and whose sum is 1. x^2+1n- 12 ⇔ (x-3)(x+4)
Distribute (x+4)
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!