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Math121 Chapter 2 Notes Lesson 2.4 – Higher Degree Polynomial Equations Example 1. 3 2 5x + 5x = 10x (First, subtract 10x from both sides.) 5x + 5x – 10x = 0 (Now, we can factor out an “x” from each term on the left side of the20.) x (5x + 5x -10) = 0 (This basically gives us a quadratic equation that we can factor out like normal. Just, this time, “x” will equal three different things, instead of just 2 like it always was in the last lesson.) x (5x – 5) (x + 2) (The “x” out by itself in the front is equal to zero. We can set the two factored sets equal to zero, also, to solve for “x”.) x = 0 x = 1 x = -2 Example 2. x + 5x + 16x – 80 = 0 (When you have a constant at the end of this kind of equation, you must look at this in pairs. x and 5x are the first pair, and 16x and -80 are the second pair. What common factors do you see within the pairs Bring 2 them outside the parentheses.) x (x + 5) and 16 (x – 5) (At this point, we’re going to look back at the pairs, and div