EXAMPLE 3 Find zeros when the leading coefficient is not 1

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Presentation transcript:

EXAMPLE 3 Find zeros when the leading coefficient is not 1 Find all real zeros of f (x) =10x4 – 11x3 – 42x2 + 7x + 12. SOLUTION STEP 1 6 3 2 1 12 + , + , + , + , + , + , + , + , + 4 5 10 + , + , + , + , + , + , , + List the possible rational zeros of f :

EXAMPLE 3 Find zeros when the leading coefficient is not 1 STEP 2 x = – , x = , x = , and x = 3 2 12 5 1 are reasonable based on the graph shown at the right. Choose reasonable values from the list above to check using the graph of the function. For f , the values

↑ EXAMPLE 3 Find zeros when the leading coefficient is not 1 STEP 3 Check the values using synthetic division until a zero is found. 2 3 – 10 –11 –42 7 12 –15 39 – 9 69 4 10 – 26 – 3 – 23 21 1 is a zero. 2 – ↑ 10 – 11 – 42 7 12 – 5 8 17 –12 10 – 16 – 34 24 0 –1

Find zeros when the leading coefficient is not 1 EXAMPLE 3 Find zeros when the leading coefficient is not 1 STEP 4 Factor out a binomial using the result of the synthetic division. f (x) = x + (10x3 – 16x2 – 34x + 24) 1 2 Write as a product of factors. = x + (2)(5x3 – 8x2 – 17x + 12) 1 2 Factor 2 out of the second factor. = (2x +1)(5x3 – 8x2 – 17x +12) Multiply the first factor by 2.

EXAMPLE 3 Find zeros when the leading coefficient is not 1 Repeat the steps above for g(x) = 5x3 – 8x2 – 17x + 12. Any zero of g will also be a zero of f. The possible rational zeros of g are: x = + 1, + 2, + 3, + 4, + 6, + 12, + , + , + , + , + , + 1 5 2 3 4 6 12 STEP 5 The graph of g shows that may be a zero. Synthetic division shows that is a zero and 3 5 g (x) = x – (5x2 – 5x – 20) 3 5 = (5x – 3)(x2 – x – 4). f (x) = (2x + 1) g (x) It follows that: = (2x + 1)(5x – 3)(x2 – x – 4)

Find zeros when the leading coefficient is not 1 EXAMPLE 3 Find zeros when the leading coefficient is not 1 STEP 6 Find the remaining zeros of f by solving x2 – x – 4 = 0. Substitute 1 for a, –1 for b, and –4 for c in the quadratic formula. x = –(–1) + √ (–1)2 – 4(1)(–4) 2(1) x = 1 + √17 2 Simplify. ANSWER 1 3 The real zeros of f are , , 1 + √17, and 1 – √17. – 2 5 2 2

GUIDED PRACTICE for Example 3 Find all real zeros of the function. 5. f (x) = 48x3+ 4x2 – 20x + 3 1 2 3 4 6 , , ANSWER – 6. f (x) = 2x4 + 5x3 – 18x2 – 19x + 42 3 2 2, , 1, +2 √2 ANSWER –