Divide P x by x 2 1 using long division Since the remainder is 0 x 2 x 6 is a

Divide p x by x 2 1 using long division since the

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Divide P ( x ) by x 2 1 using long division. Since the remainder is 0, x 2 x 6 is a factor. P ( x ) ( x 2 1)( x 2 x 6) Factor the quotient x 2 x 6. x 2 x 6 ( x 3)( x 2) Write P ( x ) as a product of four linear factors. Check: P ( x ) is a fourth -degree polynomial and we found four zeros, two of which are complex conjugates. YOUR TURN Factor the polynomial P ( x ) x 4 3 x 3 6 x 2 12 x 8 given that x 2 i is a factor. P ( x ) = ( x - i )( x + i )( x - 3)( x + 2) P ( x ) x 2 + 1 = ( x - c )( x - d )
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4.5 Complex Zeros: The Fundamental Theorem of Algebra 439 Answer: P ( x ) [ x (1 2 i )] [ x (1 2 i )]( x 1)( x 3) Note: The zeros of P ( x ) are 1, 3, 1 2 i , and 1 2 i . # EXAMPLE 4 Factoring a Polynomial with Complex Zeros Factor the polynomial P ( x ) x 4 2 x 3 x 2 2 x 2 given that 1 i is a zero of P ( x ). Since P ( x ) is a fourth- degree polynomial, we expect four zeros. The goal in this problem is to write P ( x ) as a product of four linear factors: P ( x ) ( x a )( x b )( x c )( x d ), where a , b , c , and d are complex numbers and represent the zeros of the polynomial. Solution: S TEP 1 Write known zeros and linear factors. Since 1 i is a zero, we know that 1 i is a zero. x 1 i and x 1 i We now know two linear factors of P ( x ). [ x (1 i )] and [ x (1 i )] S TEP 2 Write P ( x ) as a product of four factors. P ( x ) [ x (1 i )][ x (1 i )]( x c )( x d ) S TEP 3 Multiply the first two terms. [ x (1 i )][ x (1 i )] First group the real parts together in each bracket. [( x 1) i ][( x 1) i ] Use the special product ( x 1) 2 i 2 ( a b )( a b ) a 2 b 2 , ( x 2 2 x 1) ( 1) where a is ( x 1) and b is i . x 2 2 x 2 S TEP 4 Rewrite the polynomial. P ( x ) ( x 2 2 x 2)( x c )( x d ) S TEP 5 Divide both sides of the equation by x 2 2 x 2, and substitute in the original polynomial ( x c )( x d ) P ( x ) x 4 2 x 3 x 2 2 x 2. S TEP 6 Divide the left side of the equation using long division. S TEP 7 Factor x 2 1. ( x 1)( x 1) S TEP 8 Write P ( x ) as a product of four linear factors. P ( x ) [ x (1 i )][ x (1 i )][ x 1][ x 1] YOUR TURN Factor the polynomial P ( x ) x 4 2 x 2 16 x 15 given that 1 2 i is a zero. P ( x ) = ( x - 1 - i )( x - 1 + i )( x - 1)( x + 1) x 4 - 2 x 3 + x 2 + 2 x - 2 x 2 - 2 x + 2 = x 2 - 1 x 4 - 2 x 3 + x 2 + 2 x - 2 x 2 - 2 x + 2 = Because an n -degree polynomial function has exactly n zeros and since complex zeros always come in conjugate pairs, if the degree of the polynomial is odd , there is guaranteed to be at least one zero that is a real number . If the degree of the polynomial is even, there is no guarantee that a zero will be real—all the zeros could be complex. Study Tip Odd-degree polynomials have at least one real zero. or
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440 CHAPTER 4 Polynomial and Rational Functions Applying Descartes’ rule of signs, we find that there are 3 or 1 positive real zeros and 2 or 0 negative real zeros. b. Because this is a fourth- degree polynomial, there are four zeros. Since complex zeros come in conjugate pairs, the table describes the possible four zeros. Applying Descartes’ rule of signs we find that there are 2 or 0 positive real zeros and 2 or 0 negative real zeros. YOUR TURN List the possible combinations of real and complex zeros for: P ( x ) = x 6 - 7 x 5 + 8 x 3 - 2 x + 1 Answer: R EAL C OMPLEX Z EROS Z EROS 0 6 2 4 4 2 6 0 R EAL Z EROS C OMPLEX Z EROS 0 4 2 2 4 0 P OSITIVE N EGATIVE C OMPLEX R EAL Z EROS R EAL Z EROS Z EROS 1 0 4 3 0 2 1 2 2 3 2 0 P OSITIVE N EGATIVE C OMPLEX R EAL Z EROS R EAL Z EROS Z EROS 0 0 4 2 0 2 0 2 2 2 2 0 Factoring Polynomials Now let’s draw on the tests discussed in this chapter to help us find all the zeros of a polynomial. Doing so will enable us to write polynomials as a product of linear factors.
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  • juan alberto
  • Complex number, real zeros, complex zeros

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