1. Solving Quadratic Equations by Factoring
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Try to rewrite the middle term, bx, as two terms.
(x+5)(x+8)
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.
x^2+13x+40
In this case, we have 40. 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 40 | 40 |
| -1 and - 40 | 40 |
| 2 and 20 | 40 |
| - 2 and - 20 | 40 |
| 4 and 10 | 40 |
| - 4 and - 10 | 40 |
| 5 and 8 | 40 |
| - 5 and - 8 | 40 |
Next, let's consider the coefficient of the linear term. x^2+13x+40 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 40 | 41 |
| -1 and - 40 | - 41 |
| 2 and 20 | 22 |
| - 2 and - 20 | - 22 |
| 4 and 10 | 14 |
| - 4 and - 10 | - 14 |
| 5 and 8 | 13 |
| - 5 and - 8 | - 13 |
We found the factors whose product is 40 and whose sum is 13. x^2+13x+40 ⇔ (x+5)(x+8)