Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.

Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.

Like this presentation? Why not share!

- Module 2 Lesson 2 Notes by toni dimella 2910 views
- 3 2 Polynomial Functions And Their ... by silvia 15638 views
- MATH GRADE 10 LEARNER'S MODULE by PRINTDESK by Dan 378049 views
- Csr2011 june18 14_00_sudan by CSR2011 418 views
- Day 5 examples by jchartiersjsd 244 views
- Advanced functions ppt (Chapter 1) ... by Tan Yuhang 986 views

1,203 views

Published on

No Downloads

Total views

1,203

On SlideShare

0

From Embeds

0

Number of Embeds

13

Shares

0

Downloads

50

Comments

0

Likes

2

No embeds

No notes for slide

- 1. Polynomial Functions and Equations<br />Continuation<br />
- 2. Polynomial Functions<br />A polynomial function is a function of the form:<br />n must be a positive integer<br />an ≠ 0<br />All of these coefficients are real numbers<br />The degree of the polynomial is the largest power on any x term in the polynomial.<br />
- 3. Given f(x) =an xn + an-1 xn-1+…+a1 x +a0, <br /> where an ≠ 0 and r is a real number, <br />the following statements are equivalent:<br /><ul><li>x = r is a zero of the function f.
- 4. x = r is a solution of the polynomial equation f(x) = 0.
- 5. (x – r ) is a factorof the polynomial f(x).
- 6. (r , 0) is an x-interceptof the graph of f. </li></li></ul><li>Rational zeros of polynomial functions<br />
- 7. Rational Zeros of Polynomial Functions<br />The zeros of a polynomial function are the solutions that can be found when each of the factors of the polynomial is set equal to zero and the value of the variable is solved.<br />
- 8. Fundamental Theorem of Algebra<br />Every polynomial function whose defining equation is f(x) =an xn + an-1 xn-1+…+ a1 x + a0, where an ≠ 0 and n ≥ 1 has at least one complex zero.<br />
- 9. Fundamental Theorem of Algebra<br />Every polynomial function whose defining equation is f(x) =an xn + an-1 xn-1+…+ a1 x + a0, where an ≠ 0 and n ≥ 1 has at least one complex zero.<br />Note that a complex zero of the polynomial can either be a real or imaginary number.<br />
- 10. Rational Zeros of Polynomial Functions<br />The factored form of the polynomial<br /> f(x) =an xn + an-1 xn-1+…+ a1 x +a0<br /> is<br /> f(x) =an (x – r1) (x – r2)… (x – rn)<br /> where r1, r2, … rnare the zeros of the polynomial. <br />Ifa factor (x – r) occurs ktimes, <br />then ris called a zero of multiplicity k.<br />
- 11. Example 1a.<br />Find the zeros of the given function and determine the multiplicity of each zero.<br />Solution:<br />-1 is a zero of multiplicity 3 and <br />1 is a zero of multiplicity 1.<br />
- 12. Example 1b.<br />Find the zeros of the given function and determine the multiplicity of each zero.<br />Solution:<br />1 is a zero of multiplicity 2, <br />-3 is a zero of multiplicity 3, and <br />2 is a zero of multiplicity 1.<br />
- 13. Example 1c.<br />Find the zeros of the given function and determine the multiplicity of each zero.<br />Solution:<br />½ and -½ are zeros of multiplicity 1, <br />3/2 is a zero of multiplicity 4, and <br />-3 is a zero with multiplicity 5.<br />
- 14. Example 2a.<br />Find a polynomial equation of least possible degree which satisfies the given condition: <br />zeros are 2, 1, and ¼<br />Solution:<br />If f(r ) = 0, then (x – r) is a factor of f(x). <br />Therefore, the desired polynomial is <br />f(x) = (x – 2)(x – 1)( x – ¼).<br />
- 15. Example 2b.<br />Find a polynomial equation of least possible degree which satisfies the given condition: <br />zeros are -1, 3 and 1 of multiplicity 2<br />Solution:<br />If f(r ) = 0, then (x – r) is a factor of f(x). <br />Therefore, the desired polynomial is <br />f(x) = (x +1)(x – 3)(x – 1)2<br />
- 16. Example 2c.<br />Find a polynomial equation of least possible degree which satisfies the given condition: <br />f(1) = f(2) = f(3) = f(0) = 0, f(4) = 24<br />Solution:<br />By the Factor Theorem, f(x) = an (x – r1) (x – r2)… (x – rn)<br />Since f(1) = f(2) = f(3) = f(0) = 0, then (x – 0), <br /> (x – 1), (x – 2), and (x – 3) are factors of the polynomial f(x). <br />If f(4) = 24, then<br /> 24 = an (4 – 0)(4 – 1)(4 – 2)(4 – 3) <br />Therefore, the desired polynomial is <br />f(x) = x(x – 1)(x – 2)(x – 3) <br />
- 17. Descartes’s Rule of Signs<br />If f(x) =an xn + an-1 xn-1+…+ a1 x +a0 is a polynomial with real number coefficients, then the following holds true:<br />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.<br /> 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. <br />
- 18. Example 3a.<br />Determine the maximum number of positive roots and the maximum number of negative roots of the given polynomial function:<br />Solution:<br />Apply the Descartes’s Rule of Signs<br />Since f(x) is a polynomial of 3rd , there are 3 zeros of the function.<br />There is no variation of signs in f(x) so there no positive roots.<br />This implies that all roots are negative.<br />Now, <br />There are 3 variations of signs in f(x) so there are 3 negative roots.<br />Therefore, all 3 roots of the function are negative.<br />
- 19. Example 3b.<br />Determine the maximum number of positive roots and the maximum number of negative roots of the given polynomial function:<br />Solution:<br />Apply the Descartes’s Rule of Signs<br />Since f(x) is a polynomial of 3rd degree, there are 3 zeros of the function.<br />There is only 1 variation of signs in f(x) so there are either 1 positive root or none at all.<br />Now, <br />There are 2 variations of signs in f(x) so there are either 2 negative roots or none at all.<br />Therefore, there are 1 positive and 2 negative roots of the function.<br />
- 20. Example 3c.<br />Determine the maximum number of positive roots and the maximum number of negative roots of the given polynomial function:<br />Solution:<br />Apply the Descartes’s Rule of Signs<br />Since f(x) is a polynomial of 4th degree, there are 4 zeros of the function.<br />There are 2 variations of signs in f(x) so there are either 2 positive roots or none at all.<br />Now, <br />There are 2 variations of signs in f(x) so there are either 2 negative roots or none at all.<br />Therefore, there are 2 positive and 2 negative roots of the function.<br />
- 21. Example 3d.<br />Determine the maximum number of positive roots and the maximum number of negative roots of the given polynomial function:<br />Solution:<br />Apply the Descartes’s Rule of Signs<br />Since f(x) is a polynomial of 3rd degree, there are 3 zeros of the function.<br />There are 2 variation of signs in f(x) so there are either 2 positive roots or none at all.<br />Now, . There is only 1 variation of sign in f(x) so there is exactly 1 negative root.<br />Therefore, there are 2 positive roots and 1 negative root.<br />
- 22. Example 3e.<br />Determine the maximum number of positive roots and the maximum number of negative roots of the given polynomial function:<br />Solution:<br />Apply the Descartes’s Rule of Signs<br />Since f(x) is a polynomial of 5th degree, there are 5 zeros of the function.<br />There are 5 variations of signs in f(x) so there are either 5 or 3 positive roots.<br />Now, <br />There is no variation of sign in f(x) so there is no negative root.<br />Therefore, there are 5 positive roots of the function.<br />
- 23. The Rational Zero Theorem<br />If the rational number p/q , a fraction in lowest terms, is a root of the equation <br />f(x) =an xn + an-1 xn-1+…+ a1 x + a0, <br /> where each a1 , a2 ,…, an is an integral coefficient, <br /> then p is an exact divisor of an and <br /> q is an exact divisor of a0 . <br />
- 24. Example 4a.<br />Determine the zeros of the given polynomial function:<br />Solution:<br />Applying the Descartes’s Rule of Signs (see Example 3a), there are 3 negative roots. <br />By the Rational Zero Theorem, <br />p must be an exact divisor of 1 ; p = 1, -1<br />q must be an exact divisor of 1; q = 1, -1<br />the set of possible rational zeros p/q of f(x) is {1, -1}.<br />However, since there are 3 negative roots, we will only try x = -1. Perform synthetic division.<br />
- 25. -1<br />-1<br />1<br />1<br />1 3 3 1 <br />-2<br />-1<br />-1<br />1<br />0<br />2<br />1 2 1 <br />-1<br />-1<br />1<br />0<br />Therefore, the zero of f(x) is 1 of multiplicity 3.<br />
- 26. Example 4b.<br />Determine the zeros of the given polynomial function:<br />Solution:<br />Applying the Descartes’s Rule of Signs (see Example 3b), there are 1 positive and 2 negative roots. <br />By the Rational Zero Theorem, <br />p must be an exact divisor of -1 ; p = 1, -1<br />q must be an exact divisor of 1; q = 1, -1<br />the set of possible rational zeros p/q of f(x) is {1, -1}.<br />
- 27. Example 4b.<br />Determine the zeros of the given polynomial function:<br />Solution:<br />Now use the values in the set of possible rational zeros to find the roots of the function.<br />Suppose x = 1. <br />Perform synthetic division.<br />
- 28. 1<br />-1<br />1<br />1<br />1 1 -1 -1 <br />2<br />1<br />1<br />1<br />0<br />2<br />1 2 1 <br />-1<br />-1<br />1<br />0<br />Therefore, the zeros of f(x) are -1, of multiplicity 2, and 1.<br />
- 29. Example 4c.<br />Determine the zeros of the given polynomial function:<br />Solution:<br />Applying the Descartes’s Rule of Signs, there are 2 positive and 2 negative roots. <br />By the Rational Zero Theorem, <br />p must be an exact divisor of 4 ; p = 1, -1, 2, -2, 4, -4<br />q must be an exact divisor of 1; q = 1, -1<br />the set of possible rational zeros p/q of f(x) is<br /> {1, 2, 4}.<br />
- 30. 1<br />-1<br />1<br />1<br />1 0 -5 0 4 <br />-4<br />1<br />-4<br />1<br />0<br />1<br />-4<br />-4<br />1 1 -4 -4 <br />4<br />-1<br />0<br />-4<br />0<br />0<br />Note that <br /> So, <br />Therefore, the zeros of f(x) are 2, -2, 1, and -1. <br />
- 31. Exercises:<br />1.Find the zeros of the given functions and determine the multiplicity of each zero.<br />1.1<br />1.2<br />1.3<br />2. Find a polynomial equation of least possible degree which satisfies the given conditions: <br />2.1 zeros are 1, -1, and ¾<br />2.2 Zeros are 4 of multiplicity 3, -1 of multiplicity 2, and 3<br />
- 32. Exercises:<br />3. Determine the maximum number of positive roots and the maximum number of negative roots of the given polynomial function:<br />3.1<br />3.2<br />3.3<br />4. Determine the zeros of the given polynomial functions:<br />4.1 <br />4.2<br />4.3<br />

No public clipboards found for this slide

×
### Save the most important slides with Clipping

Clipping is a handy way to collect and organize the most important slides from a presentation. You can keep your great finds in clipboards organized around topics.

Be the first to comment