Upper and Lower Bounds for Roots 3.5: More on Zeros of Polynomial Functions The Upper  and  Lower  Bound  Theorem  helps u...
EXAMPLE:   Finding Bounds for the Roots Show that all the real roots of the equation 8 x 3      10 x 2    39 x  + 9   ...
EXAMPLE:   Finding Bounds for the Roots Show that all the real roots of the equation 8 x 3      10 x 2    39 x  + 9   ...
The Intermediate Value Theorem   The Intermediate Value Theorem for Polynomials Let  f   ( x ) be a polynomial function wi...
EXAMPLE : Approximating a Real Zero a.   Show that the polynomial function  f   ( x )     x 3      2 x     5   has a re...
EXAMPLE : Approximating a Real Zero b.   A numerical approach is to evaluate  f   at successive tenths between 2 and  3, l...
EXAMPLE : Approximating a Real Zero b.   We now follow a similar procedure to locate the real zero between  successive hun...
The Fundamental Theorem of Algebra 3.5: More on Zeros of Polynomial Functions We have seen that if a polynomial equation i...
The Linear Factorization Theorem   The Linear Factorization Theorem  If  f   ( x )      a n x n     a n  1 x n  1    ...
3.5: More on Zeros of Polynomial Functions EXAMPLE: Finding  a Polynomial  Function with Given Zeros   Find a fourth-degre...
3.5: More on Zeros of Polynomial Functions EXAMPLE: Finding  a Polynomial  Function with Given Zeros   Find a fourth-degre...
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The fundamental thorem of algebra

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The fundamental thorem of algebra

  1. 1. Upper and Lower Bounds for Roots 3.5: More on Zeros of Polynomial Functions The Upper and Lower Bound Theorem helps us rule out many of a polynomial equation's possible rational roots. The Upper and Lower Bound Theorem Let f ( x ) be a polynomial with real coefficients and a positive leading coefficient, and let a and b be nonzero real numbers. 1. Divide f ( x ) by x  b (where b  0) using synthetic division. If the last row containing the quotient and remainder has no negative numbers, then b is an upper bound for the real roots of f ( x )  0. 2. Divide f ( x ) by x  a (where a  0) using synthetic division. If the last row containing the quotient and remainder has numbers that alternate in sign (zero entries count as positive or negative), then a is a lower bound for the real roots of f ( x )  0.
  2. 2. EXAMPLE: Finding Bounds for the Roots Show that all the real roots of the equation 8 x 3  10 x 2  39 x + 9  0 lie between –3 and 2. Solution We begin by showing that 2 is an upper bound. Divide the polynomial by x  2. If all the numbers in the bottom row of the synthetic division are non­negative, then 2 is an upper bound . All numbers in this row are nonnegative. 3.5: More on Zeros of Polynomial Functions 35 13 26 8 26 52 16 9  39 10 8 2 more more
  3. 3. EXAMPLE: Finding Bounds for the Roots Show that all the real roots of the equation 8 x 3  10 x 2  39 x + 9  0 lie between –3 and 2. Solution The nonnegative entries in the last row verify that 2 is an upper bound. Next, we show that  3 is a lower bound. Divide the polynomial by x  (  3), or x  3. If the numbers in the bottom row of the synthetic division alternate in sign, then  3 is a lower bound. Remember that the number zero can be considered positive or negative. Counting zero as negative, the signs alternate:  ,  ,  ,  . By the Upper and Lower Bound Theorem, the alternating signs in the last row indicate that  3 is a lower bound for the roots. (The zero remainder indicates that  3 is also a root.) 3.5: More on Zeros of Polynomial Functions 35 13 26 8  9 42  24 9  39 10 8  3
  4. 4. The Intermediate Value Theorem The Intermediate Value Theorem for Polynomials Let f ( x ) be a polynomial function with real coefficients. If f ( a ) and f ( b ) have opposite signs, then there is at least one value of c between a and b for which f ( c ) = 0. Equivalently, the equation f ( x )  0 has at least one real root between a and b . 3.5: More on Zeros of Polynomial Functions
  5. 5. EXAMPLE : Approximating a Real Zero a. Show that the polynomial function f ( x )  x 3  2 x  5 has a real zero between 2 and 3. b. Use the Intermediate Value Theorem to find an approximation for this real zero to the nearest tenth 3.5: More on Zeros of Polynomial Functions a. Let us evaluate f ( x ) at 2 and 3. If f (2) and f (3) have opposite signs, then there is a real zero between 2 and 3. Using f ( x )  x 3  2 x  5, we obtain Solution This sign change shows that the polynomial function has a real zero between 2 and 3. and f (3)  3 3  2  3  5  27  6  5  16. f (3) is positive. f (2)  2 3  2  2  5  8  4  5  1 f (2) is negative.
  6. 6. EXAMPLE : Approximating a Real Zero b. A numerical approach is to evaluate f at successive tenths between 2 and 3, looking for a sign change. This sign change will place the real zero between a pair of successive tenths. Solution a. Show that the polynomial function f ( x )  x 3  2 x  5 has a real zero between 2 and 3. b. Use the Intermediate Value Theorem to find an approximation for this real zero to the nearest tenth The sign change indicates that f has a real zero between 2 and 2.1. Sign change Sign change 3.5: More on Zeros of Polynomial Functions f (2.1)  (2.1) 3  2(2.1)  5  0.061 2.1 f (2)  2 3  2(2)  5   1 2 f ( x )  x 3  2 x  5 x more more
  7. 7. EXAMPLE : Approximating a Real Zero b. We now follow a similar procedure to locate the real zero between successive hundredths. We divide the interval [2, 2.1] into ten equal sub- intervals. Then we evaluate f at each endpoint and look for a sign change. Solution a. Show that the polynomial function f ( x )  x 3  2 x  5 has a real zero between 2 and 3. b. Use the Intermediate Value Theorem to find an approximation for this real zero to the nearest tenth The sign change indicates that f has a real zero between 2.09 and 2.1. Correct to the nearest tenth, the zero is 2.1. Sign change 3.5: More on Zeros of Polynomial Functions f (2.07)  0.270257 f (2.03)  0.694573 f (2.1)  0.061 f (2.06)  0.378184 f (2.02)  0.797592 f (2.09)  0.050671 f (2.05)  0.484875 f (2.01)  0.899399 f (2.08)  0.161088 f (2.04)  0.590336 f (2.00)  1
  8. 8. The Fundamental Theorem of Algebra 3.5: More on Zeros of Polynomial Functions We have seen that if a polynomial equation is of degree n, then counting multiple roots separately, the equation has n roots. This result is called the Fundamental Theorem of Algebra. The Fundamental Theorem of Algebra If f ( x ) is a polynomial of degree n, where n  1, then the equation f ( x )  0 has at least one complex root.
  9. 9. The Linear Factorization Theorem The Linear Factorization Theorem If f ( x )  a n x n  a n  1 x n  1   …  a 1 x  a 0 b, where n  1 and a n  0 , then f ( x )  a n ( x  c 1 ) ( x  c 2 ) … ( x  c n ) where c 1 , c 2 ,…, c n are complex numbers (possibly real and not necessarily distinct). In words: An n th-degree polynomial can be expressed as the product of n linear factors. Just as an n th-degree polynomial equation has n roots, an n th-degree polynomial has n linear factors. This is formally stated as the Linear Factorization Theorem. 3.5: More on Zeros of Polynomial Functions
  10. 10. 3.5: More on Zeros of Polynomial Functions EXAMPLE: Finding a Polynomial Function with Given Zeros Find a fourth-degree polynomial function f ( x ) with real coefficients that has  2, and i as zeros and such that f (3)  150. Solution Because i is a zero and the polynomial has real coefficients, the conjugate must also be a zero. We can now use the Linear Factorization Theorem.  a n ( x  2)( x  2)( x  i )( x  i ) Use the given zeros: c 1  2, c 2  2, c 3  i , and, from above, c 4  i . f ( x )  a n ( x  c 1 )( x  c 2 )( x  c 3 )( x  c 4 ) This is the linear factorization for a fourth-degree polynomial.  a n ( x 2  4)( x 2  i ) Multiply f ( x )  a n ( x 4  3 x 2  4) Complete the multiplication more more
  11. 11. 3.5: More on Zeros of Polynomial Functions EXAMPLE: Finding a Polynomial Function with Given Zeros Find a fourth-degree polynomial function f ( x ) with real coefficients that has  2, and i as zeros and such that f (3)  150. Substituting  3 for a n in the formula for f ( x ) , we obtain f ( x )  3( x 4  3 x 2  4) . Equivalently, f ( x )  3 x 4  9 x 2  12. Solution f (3)  a n (3 4  3  3 2  4)  150 To find a n , use the fact that f (3)  150. a n (81  27  4)  150 Solve for a n . 50 a n  150 a n  3
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