SECTION 4.3 The Real Zeros of a Polynomial Function 221 Figure 29 f x x x x x x 4 16 37 84 84 6 5 3 2 ( ) = + − − − − Step 3 Figure 29 shows the graph of f using Geogebra.The graph has the characteristics expected of the given polynomial function of degree 6: no more than five turning points, y-intercept 84 − , and it behaves like y x6 = for large x . Step 4 From Figure 29, 2− appears to be a zero. Additionally, 2− is a potential rational zero. Evaluate f 2 ( ) − and find that f 2 0 ( ) − = . By the Factor Theorem, x 2 + is a factor of f . We use synthetic division to factor f. ) − − − − − − − − − − − 214 0 16 37 84 84 2 4 816 42 84 12 4 8 2142 0 Factor f as f x x x x x x x 2 2 4 8 21 42 5 4 3 2 ( ) ( ) ( ) = + + − − − − Now work with the first depressed equation: q x x x x x x 2 4 8 21 42 0 1 5 4 3 2 ( ) = + − − − − = Repeat Step 4 Since 2− might be a zero of even multiplicity we check the potential rational zero 2− again. Since q 2 0 1( ) − = , then x 2 + is a factor. Use synthetic division to factor ( ) q x 1 . ) − − − − − − − − 2 1 2 4 8 21 42 2 0 8 0 42 1 0 4 0 21 0 So, ( ) ( ) ( ) = + − − q x x x x 2 4 21 1 4 2 and f x x x x x 2 2 4 21 4 2 ( ) ( ) ( )( ) = + + − − Repeat Step 4 The depressed equation ( ) = − − = q x x x4 21 0 2 4 2 can be factored. ( )( ) − − = − + = − = + = = = ± x x x x x x x x 4 21 7 3 0 7 0 or 3 0 7 7 4 2 2 2 2 2 2 Since x 3 0 2 + = has no real solutions, the real zeros of f are 7, 7 − , and 2− , with 2− being a zero of multiplicity 2. The factored form of f is ( )( )( ) ( ) ( ) = + − − − − = + + − + f x x x x x x x x x x 4 16 37 84 84 2 7 7 3 6 5 3 2 2 2 Note that one positive real zero and three negative real zeros agrees with our result from Step 1. Now Work PROBLEM 45
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