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# 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.