The document discusses the upper and lower bound theorem for real roots of polynomial functions. The theorem states that if the remainder of synthetic division of a polynomial f(x) by x-b has only non-negative numbers, b is an upper bound for the real roots of f(x)=0. If the remainder alternates in sign when dividing f(x) by x-a, a is a lower bound. An example demonstrates finding the bounds between -3 and 2 for a 4th degree polynomial. The intermediate value theorem and fundamental theorem of algebra are also summarized.
* Find zeros of polynomial functions
* Use the Fundamental Theorem of Algebra to find a function that satisfies given conditions
* Find all zeros of a polynomial function
* Find zeros of polynomial functions
* Use the Fundamental Theorem of Algebra to find a function that satisfies given conditions
* Find all zeros of a polynomial function
* Evaluate a polynomial using the Remainder Theorem.
* Use the Factor Theorem to solve a polynomial equation.
* Use the Rational Zero Theorem to find rational zeros.
* Find zeros of a polynomial function.
* Use the Linear Factorization Theorem to find polynomials with given zeros.
* Use Descartes’ Rule of Signs.
Create a polynomial function that meets the following conditions- Expl.docxmrichard5
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4. Degree 3, 2 positive real zeros, 1 negative real zero, 0 complex zeros.
Polynomial:
Solution
Polynomial may be : f(x) = -x^3 + x^2 + x - 1 Reasoning : the number of positive roots of the polynomial is either equal to the number of sign differences between consecutive nonzero coefficients. so here has two sign change(the sequence of pairs of successive signs is -+, ++, +-). So it has 2 positive roots. for negative root we change the signs of the coefficients of the terms with odd exponents, i.e., apply Descartes\' rule of signs to the polynomial f(-x), to obtain a second polynomial f(-x) = x^3 + x^2 - x - 1 it has 1 sign change so f(-x) has one positive root so f(x) has 1 negative root.
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The fundamental thorem of algebra
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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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