# Factoring trinomials - A complete course in algebra

Example 2. Factor x 2 x 12. Solution. ( x? x?) There are several factors of 12. Let us try 2 and 6. ( x 2 x 6) But there is no way to choose signs for 6 x and 2 x to give the middle term, which is x.
Factor these trinomials. A) 2 x 2 7 x 5 (2 x 5 x 1) b) 2 x 2 11 x 5 (2 x 1 x 5) c) 3 x 2 x 10 (3 x 5 x 2 ) d) 2 x 2 x 3 (2 x 3 x 1) e) 5 x 2 13 x 6 (5 x 3.
Place the correct signs to give the middle term. A) 2 x 2 7 x 15 (2 x 3 x 5) b) 2 x 2 7 x 15 (2 x 3 x 5) c) 2 x 2 x 15 (2 x 5 x 3) d) 2 x 2 13 x 15 (2 x 3 x 5) Note: When the constant.
Sin2 x - "sine squared x " - means (sin x )2. Problem 12. Factor each quadratic. A) x 4 x 2 2 ( x 2 2 x 2 1) b) y 6 2 y 3 8 ( y 3 4 y 3 2) c) z 8 4 z 4 3 ( z 4 1 z 4 3) d) 2.
Now continue by factoring the trinomial: 6 x 6( x 2 x 3). Problem 6. Factor completely. First remove any common factors. A) x 3 6 x 2 5 x x ( x 2 6 x 5) x ( x 5 x 1) b) x 5 4 x 4 3 x 3 x 3( x 2 4 x 3) x.
The argument is whatever is being squared. X is being squared. X is called the argument. The argument appears in the middle term. A, b, c are called constants. In this quadratic, 3 x 2 2 x 1, the constants are 3, 2, 1.
In a sense, it is the same quadratic only with a different argument. For it is the constants that distinguish a quadratic. Now, since the quadratic with argument x can be factored in this way: 3 x 2 2 x 1 (3 x 1 x 1 then the quadratic with argument x 3is factored the same way: 3 x 6.