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Zeros of Polynomial Functions

Zeros of Polynomial Functions. The Rational Zero Theorem.

lawrence-sanders
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Zeros of Polynomial Functions

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  1. Zeros of Polynomial Functions

  2. The Rational Zero Theorem • If f (x)=anxn+ an-1xn-1+…+a1x + a0 has integer coefficients and p/q (where p/q is reduced) is a rational zero, then p is a factor of the constant term a0 and q is a factor of the leading coefficient an.

  3. Find all of the possible real, rational roots of f(x) = 2x3-3x2+5 Solution: p is a factor of 5 = 1, 5 q is a factor of 2 = 1, 2 p/q = 1, 1/2, 5, 5/2 Example

  4. Properties of Polynomial Equations • If a polynomial equation is of degree n, then counting multiple roots separately, the equation has n roots • If a+bi is a root of the equation, then a-bi is also a root.

  5. Find all zeros of f(x) = x3+12x2+21x+10 p/q = 1, 2, 5, 10 f(1) = 44 f(-1) = 0 Divide out -1 to get x2+11x-10 Use the quadratic formula to find the last 2 zeros. x=-11.844 and .844 The solutions are -1, -11.844, and .844 Example

  6. 2 • 0 -6 -8 24 • 24-4-24 • 1 2 -2 -12 0 x-intercept: 2 Text Example Solve: x4-6x2- 8x + 24 = 0. SolutionThe graph of f(x) =x4-6x2- 8x + 24 is shown the figure below. Because the x-intercept is 2, we will test 2 by synthetic division and show that it is a root of the given equation. The zero remainder indicates that 2 is a root of x4-6x2- 8x + 24 = 0.

  7. (x – 2)(x3+ 2x2-2x- 12) = 0 This is the result obtainedfrom the synthetic division. Text Example cont. Solve: x4-6x2- 8x + 24 = 0. SolutionNow we can rewrite the given equation in factored form. x4-6x2+ 8x + 24 = 0 This is the given equation. x – 2 = 0 or x3+ 2x2-2x- 12 = 0Set each factor equal to zero.

  8. These are the coefficients ofx3+ 2x2-2x- 12 = 0. • 1 2 -2 -12 • 2 8 12 • 1 4 6 0 The zero remainder indicates that 2 is a root ofx3+ 2x2-2x- 12 = 0. x-intercept: 2 Text Example cont. Solve: x4-6x2- 8x + 24 = 0. SolutionWe can use the same approach to look for rational roots of the polynomial equation x3+ 2x2-2x- 12 = 0, listing all possible rational roots. However, take a second look at the figure of the graph of x4-6x2- 8x + 24 = 0. Because the graph turns around at 2, this means that 2 is a root of even multiplicity. Thus, 2 must also be a root of x3+ 2x2-2x- 12 = 0, confirmed by the following synthetic division.

  9. (x – 2)(x3+ 2x2-2x- 12) = 0 This was obtainedfrom the first synthetic division. (x – 2)(x – 2)(x2+4x+ 6) = 0 This was obtainedfrom the second synthetic division. Text Example cont. Solve: x4-6x2- 8x + 24 = 0. SolutionNow we can solve the original equation as follows. x4-6x2+ 8x + 24 = 0 This is the given equation. x – 2 = 0 or x – 2 = 0 or x2+4x+ 6 = 0Set each factor equal to zero. x= 2 x= 2 x2+4x+ 6 = 0Solve.

  10. Text Example cont. Solve: x4-6x2- 8x + 24 = 0. SolutionWe can use the quadratic formula to solve x2+4x+ 6 = 0. The solution set of the original equation is: {2, -2 - i2, -2+i2}

  11. If f (x)=anxn+ an-1xn-1+… +a2x2+a1x + a0 be a polynomial with real coefficients. 1. The number of positive real zeros of f iseither equal to the number of sign changes of f (x)or is less than that number by an even integer. If there is only one variation in sign, there is exactly one positive real zero. 2. The number of negative real zeros of f iseither equal to the number of sign changes of f (-x) or is less than that number by an even integer. If f (-x)has only one variation in sign, then f has exactly one negative real zero. Descartes’s Rule of Signs

  12. Solution 1. To find possibilities for positive real zeros, count the number of sign changes in the equation for f(x).Because all the terms are positive, there are no variations in sign. Thus, there are no positive real zeros. 2. To find possibilities for negative real zeros, count the number of sign changes in the equation for f(-x). We obtain this equation by replacing x with -x in the given function. f(x)= x3+ 2x2 + 5x + 4This is the given polynomial function. Replace xwith -x. f(-x)= (-x)3+2(-x)2+ 5(-x) + 4 =-x3+ 2x2 - 5x + 4 Text Example Determine the possible number of positive and negative real zeros of f(x)= x3+ 2x2 + 5x + 4.

  13. 1 2 3 Determine the possible number of positive and negative real zeros of f(x)= x3+ 2x2 + 5x + 4. Text Example cont. Solution Now count the sign changes. f(-x)=-x3+ 2x2 - 5x + 4 There are three variations in sign. The number of negative real zeros of fis either equal to the number of sign changes, 3, or is less than this number by an even integer. This means that there are either 3 negative real zeros or 3 - 2 = 1 negative real zero.

  14. Zeros of Polynomial Functions

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