**General Trinomials**

*x*

^{2}+ (

*a + b*)

*x + ab*= (

*x + a*)(

*x + b*)

The expression *x*^{2} +(*a + b*)*x* +* ab* tells us that we need two numbers whose product is the last term and whose sum is the coefficient of the middle term. Consider the trinomial *x*^{2} + 5*x* + 6 . Now, two factors of 6 are 1 and 6, but 1 + 6 â‰ 5. However, 2 and 3 are also factors of 6, and 2 + 3 = 5. Hence, *x*^{2} + 5*x* + 6 = (*x* + 2)(*x* + 3).

**Example:**

**Column A **

*x *

**Column B**

7

where *x*^{2} - 7*x* - 18 = 0

*, x*

^{2}- 7

*x*- 18 =(

*x*+ 2)(

*x*- 9)= 0 . Setting each factor equal to zero yields

*x*+ 2 = 0 and

*x Â–*â€“ 9 = 0. Solving these equations yields

*x*= Â–â€“2 and 9. If

*x*= Â–â€“2, then Column B is larger. However, if

*x*= 9, then Column A is larger. This is a double case, and the answer is (D).

**Complete Factoring**

When factoring an expression, first check for a common factor, then check for a difference of squares, then for a perfect square trinomial, and then for a general trinomial.

**Example:**

Factor the expression 2*x*^{3} - 2*x* ^{2} -12*x* completely.

**Solution**:

First check for a common factor: 2x is common to each term. Factoring 2x out of each term yields 2*x*(*x*^{2} - *x* - 6). Next, there is no difference of squares, and *x*^{2} - *x* - 6 is not a perfect square trinomial since x does not equal twice the product of the square roots of *x*^{2} and 6. Now, Â–â€“3 and 2 are factors of Â–â€“6 whose sum is Â–â€“1. Hence, 2*x*(*x*^{2} -* x* - 6) factors into 2*x*(*x* Â–â€“ 3)(*x* + 2).