How do i write a polynomial function of least degree that has real coeficents, the given zeros of the function?
Q. How do i write a polynomial function of least degree that has real coeficents, the given zeros of the function? Given Zeros: 2, -2, 3
Asked by Kasie - Wed Jun 3 18:40:15 2009 - - 1 Answers - 0 Comments
A. since those are zeros then you could say (x-2)(x+2)(x-3)=0 then just distribute
Answered by Paulie Walnuts - Wed Jun 3 18:44:45 2009
Q. How do i write a polynomial function of least degree that has real coeficents, the given zeros of the function? Given Zeros: 2, -2, 3
Asked by Kasie - Wed Jun 3 18:40:15 2009 - - 1 Answers - 0 Comments
A. since those are zeros then you could say (x-2)(x+2)(x-3)=0 then just distribute
Answered by Paulie Walnuts - Wed Jun 3 18:44:45 2009
How do I show that a polynomial function is continuous at a point?
Q. How do I show that a polynomial function is continuous at a point? Would this be a proof? If so how do I prove it? Any helpful websites that are easy to digest?
Asked by Josephine - Tue Oct 28 01:58:34 2008 - - 2 Answers - 0 Comments
A. Is the function F(x) = x continuous everywhere? (Yes) Is the function G(x) = 1 continuous everywhere?(Yes) If f and g are continuous everywhere, is f+g continuous? How about f*g? (Yes, Yes) Note that 3x^2 = (x*x)+(x*x)+(x*x) Show that every polynomial is the finite sum of finite products of F(x)=x and G(x)=1 and hence continuous everywhere.
Answered by Michael E - Tue Oct 28 02:16:49 2008
Q. How do I show that a polynomial function is continuous at a point? Would this be a proof? If so how do I prove it? Any helpful websites that are easy to digest?
Asked by Josephine - Tue Oct 28 01:58:34 2008 - - 2 Answers - 0 Comments
A. Is the function F(x) = x continuous everywhere? (Yes) Is the function G(x) = 1 continuous everywhere?(Yes) If f and g are continuous everywhere, is f+g continuous? How about f*g? (Yes, Yes) Note that 3x^2 = (x*x)+(x*x)+(x*x) Show that every polynomial is the finite sum of finite products of F(x)=x and G(x)=1 and hence continuous everywhere.
Answered by Michael E - Tue Oct 28 02:16:49 2008
How to write a polynomial function in standard form?
Q. Need help writing a polynomial function in standard form with real coefficients whose zeros and their multiplicities include those listed. -1 (Multiplicity 2), -2 - i (multiplicity 1)
Asked by Kevin - Mon Sep 28 02:18:25 2009 - - 1 Answers - 0 Comments
A. -1 (Multiplicity 2) x = -1 x + 1 = 0 === (x+1)^2 = 0 === x = -2 - i imaginary roots occur with conjugate [- 2 + i] is also zero === [x + 2 + i][x + 2 - i] = 0 === combine (x+1)^2 [x + 2 + i][x + 2 - i] = 0 (x+1)^2 [(x + 2)^2 + 1] = 0 f(x) = polynomial = (x^2+2x+1) [x^2 + 4x + 5] f(x) = (x^2+2x+1) [x^2 + 4x + 5] multiply and but in form f(x) = ax^4 + bx^3 + cx^2 + dx + e
Answered by anil bakshi - Wed Sep 30 02:15:27 2009
Q. Need help writing a polynomial function in standard form with real coefficients whose zeros and their multiplicities include those listed. -1 (Multiplicity 2), -2 - i (multiplicity 1)
Asked by Kevin - Mon Sep 28 02:18:25 2009 - - 1 Answers - 0 Comments
A. -1 (Multiplicity 2) x = -1 x + 1 = 0 === (x+1)^2 = 0 === x = -2 - i imaginary roots occur with conjugate [- 2 + i] is also zero === [x + 2 + i][x + 2 - i] = 0 === combine (x+1)^2 [x + 2 + i][x + 2 - i] = 0 (x+1)^2 [(x + 2)^2 + 1] = 0 f(x) = polynomial = (x^2+2x+1) [x^2 + 4x + 5] f(x) = (x^2+2x+1) [x^2 + 4x + 5] multiply and but in form f(x) = ax^4 + bx^3 + cx^2 + dx + e
Answered by anil bakshi - Wed Sep 30 02:15:27 2009
How can you tell from the factored form of a polynomial function whether the function has a repeated zero?
Q. How can you tell from the factored form of a polynomial function whether the function has a repeated zero?
Asked by bub - Tue Dec 16 20:38:38 2008 - - 2 Answers - 0 Comments
A. If any of the factors is repeated (for example, if one of the factors is (x-2) then (x-2) is repeated), then the function has a repeated 0. Another way to put it: if any of the factors has an exponent other than the implied 1, then the function has a repeated 0. _/
Answered by Cheryl B - Tue Dec 16 20:47:20 2008
Q. How can you tell from the factored form of a polynomial function whether the function has a repeated zero?
Asked by bub - Tue Dec 16 20:38:38 2008 - - 2 Answers - 0 Comments
A. If any of the factors is repeated (for example, if one of the factors is (x-2) then (x-2) is repeated), then the function has a repeated 0. Another way to put it: if any of the factors has an exponent other than the implied 1, then the function has a repeated 0. _/
Answered by Cheryl B - Tue Dec 16 20:47:20 2008
A sixth degree polynomial function has r distinct real zeros. What are the possible values of r?
Q. A sixth degree polynomial function has r distinct real zeros. WHat are the possible values of r? Please tell me how you solved the problem, not just the answer Thanks!
Asked by Steffie - Sat Sep 26 22:27:09 2009 - - 4 Answers - 0 Comments
A. Hi, The maximum number of different solutions a 6th degree polynomial can have is 6. However, since a polynomial like x + 9 = 0 has no real roots, a polynomial can have from 0 to 6 distinct real zeros. <==Answer i hope that helps!! :-)
Answered by Pi R Squared - Sat Sep 26 22:36:33 2009
Q. A sixth degree polynomial function has r distinct real zeros. WHat are the possible values of r? Please tell me how you solved the problem, not just the answer Thanks!
Asked by Steffie - Sat Sep 26 22:27:09 2009 - - 4 Answers - 0 Comments
A. Hi, The maximum number of different solutions a 6th degree polynomial can have is 6. However, since a polynomial like x + 9 = 0 has no real roots, a polynomial can have from 0 to 6 distinct real zeros. <==Answer i hope that helps!! :-)
Answered by Pi R Squared - Sat Sep 26 22:36:33 2009
What is the leading term of this polynomial function?
Q. Find the leading term of (5x + 1)(3x 1)(2x + 5)^3. Plus do you mind stating some basic facts about the domain and range of a polynomial function in general?
Asked by Ben J - Sat Mar 7 17:40:58 2009 - - 4 Answers - 0 Comments
A. the leading term will be (5x)(3x)(2x)(2x)(2x) = 120x^5 The domain of a polynomial function will always be all real numbers The range of a polynomial function where the leading term has an odd exponent (i.e. x^3 + x^2 + x + 1) will be all real numbers For polynomial functions where the leading term has an even exponent (i.e. x^2 + x + 5): if the leading term is negative, it will always be all real numbers less than a certain number if the leading term is positive, it will always be all real numbers greater than a certain number.
Answered by GunsUpTech - Sat Mar 7 17:47:31 2009
Q. Find the leading term of (5x + 1)(3x 1)(2x + 5)^3. Plus do you mind stating some basic facts about the domain and range of a polynomial function in general?
Asked by Ben J - Sat Mar 7 17:40:58 2009 - - 4 Answers - 0 Comments
A. the leading term will be (5x)(3x)(2x)(2x)(2x) = 120x^5 The domain of a polynomial function will always be all real numbers The range of a polynomial function where the leading term has an odd exponent (i.e. x^3 + x^2 + x + 1) will be all real numbers For polynomial functions where the leading term has an even exponent (i.e. x^2 + x + 5): if the leading term is negative, it will always be all real numbers less than a certain number if the leading term is positive, it will always be all real numbers greater than a certain number.
Answered by GunsUpTech - Sat Mar 7 17:47:31 2009
How do you know Zeros of a Polynomial Function?
Q. How do you know when the Zeros of a polynomial function are X or Zero positive/negative. For example: F(x) = x^4 - 13x^2 + 36 Would the answer be 2,0 neg and 2,0 pos or just 2 neg and 2 pos?
Asked by Alexa D - Mon Mar 16 21:34:34 2009 - - 4 Answers - 0 Comments
A. X = x X - 13X + 36 = 0 Delta = 169 - 4 * 36 = 169 - 144 = 25 X1 = (13 -5) / 2 = 4 X2 = (13 + 5) / 2 = 9 x = 4 <=> x = -2 or x = 2 x = 9 <=> x = -3 or x = 3 Answers : S = { -3 ; -2 ; 2 ; 3 }
Answered by John F - Mon Mar 16 21:40:54 2009
Q. How do you know when the Zeros of a polynomial function are X or Zero positive/negative. For example: F(x) = x^4 - 13x^2 + 36 Would the answer be 2,0 neg and 2,0 pos or just 2 neg and 2 pos?
Asked by Alexa D - Mon Mar 16 21:34:34 2009 - - 4 Answers - 0 Comments
A. X = x X - 13X + 36 = 0 Delta = 169 - 4 * 36 = 169 - 144 = 25 X1 = (13 -5) / 2 = 4 X2 = (13 + 5) / 2 = 9 x = 4 <=> x = -2 or x = 2 x = 9 <=> x = -3 or x = 3 Answers : S = { -3 ; -2 ; 2 ; 3 }
Answered by John F - Mon Mar 16 21:40:54 2009
What is the difference between mathematical model and algebraic model for polynomial function?
Q. I was given a table with 7 points, and estimated the polynomial function. I also calculated for first and second derivative. For the same problem: One question asks to use the mathematical model to determine the increasing and decreasing. Another one asks to use algebraic model to determine the critical numbers. What are the difference?? I'm so confused...
Asked by wildroses - Fri Jun 19 04:09:14 2009 - - 1 Answers - 0 Comments
A. They sure sound like the same to me as well. An algebraic model is clearly refering to developing a polynomial of degree six. (The degree of the polynomial that would "fit" the seven points would have to be six if they were fairly random and they were not all on a parabola or straight line of any polynomial of a lower degree. "Mathematical model" is a new phrase to me. See if that terminology is in your notes or your textbook. If "yes", there should be an example of a "mathematical model" so that you would be able to differentiate between the two (assuming, for the moment, there is a difference).
Answered by unknown - Tue Jun 23 03:03:28 2009
Q. I was given a table with 7 points, and estimated the polynomial function. I also calculated for first and second derivative. For the same problem: One question asks to use the mathematical model to determine the increasing and decreasing. Another one asks to use algebraic model to determine the critical numbers. What are the difference?? I'm so confused...
Asked by wildroses - Fri Jun 19 04:09:14 2009 - - 1 Answers - 0 Comments
A. They sure sound like the same to me as well. An algebraic model is clearly refering to developing a polynomial of degree six. (The degree of the polynomial that would "fit" the seven points would have to be six if they were fairly random and they were not all on a parabola or straight line of any polynomial of a lower degree. "Mathematical model" is a new phrase to me. See if that terminology is in your notes or your textbook. If "yes", there should be an example of a "mathematical model" so that you would be able to differentiate between the two (assuming, for the moment, there is a difference).
Answered by unknown - Tue Jun 23 03:03:28 2009
How would you write a polynomial function out of these giver zeros?
Q. how would you write a polynomial function with rational coefficients in standard form with the giver zeroes od 2+root5, 3? please take me through steps. i have a test tomorrow
Asked by the_setting_stars - Thu Dec 18 02:02:09 2008 - - 2 Answers - 0 Comments
A. Well you need factors. They gave you roots. It is easy to remedy this if you remember... (x-a)(x-b) = 0 x = a,b So all you have to do is subtract the roots from x to make factors, then just multiply to give the polynomial. (x - (2 + 5))(x - 3) x - 3x - x(2 + 5) + 3(2 + 5) x - 3x - 2x - x 5 + 6 + 3 5 x - 5x - x 5 + 6 + 3 5 x - (5 + 5)x + (6 + 3 5)
Answered by o - Thu Dec 18 02:07:54 2008
Q. how would you write a polynomial function with rational coefficients in standard form with the giver zeroes od 2+root5, 3? please take me through steps. i have a test tomorrow
Asked by the_setting_stars - Thu Dec 18 02:02:09 2008 - - 2 Answers - 0 Comments
A. Well you need factors. They gave you roots. It is easy to remedy this if you remember... (x-a)(x-b) = 0 x = a,b So all you have to do is subtract the roots from x to make factors, then just multiply to give the polynomial. (x - (2 + 5))(x - 3) x - 3x - x(2 + 5) + 3(2 + 5) x - 3x - 2x - x 5 + 6 + 3 5 x - 5x - x 5 + 6 + 3 5 x - (5 + 5)x + (6 + 3 5)
Answered by o - Thu Dec 18 02:07:54 2008
Writing a polynomial function in standard from with the giver zeros?
Q. Can someone plz give me step by stp instructions how to write this polynomial function with rational coefficients in standard from witht he given zeros. 1) x= 3+i, 1-(the square root of 5)
Asked by blonde - Tue Jan 20 14:36:51 2009 - - 1 Answers - 0 Comments
A. Although using the given zeros itself word for word is not enough, you can find the standard form of the polynomial by doing the following one of the solutions is 3+ i, and because all imaginary solution exist as conjugate pairs we automatically know another solution must be 3 - i. Note that you cannot have a polynomial where the solutions are JUST 3 + i and 1 - 5 and have real coefficients. Further your not going to get any rational coefficients with just 1 - 5 so we must also have 1 + 5 if the polynomial were in factored form it would look like this let P(x) be a polynomial function P(X) = (x - 3 - i )(x - 3 + i)(x - 1+ 5)(x - 1- 5) which equates to P(X) = x^4 - 8x^3 + 18x^2 +4x -40
Answered by Blinki - Tue Jan 20 14:59:32 2009
Q. Can someone plz give me step by stp instructions how to write this polynomial function with rational coefficients in standard from witht he given zeros. 1) x= 3+i, 1-(the square root of 5)
Asked by blonde - Tue Jan 20 14:36:51 2009 - - 1 Answers - 0 Comments
A. Although using the given zeros itself word for word is not enough, you can find the standard form of the polynomial by doing the following one of the solutions is 3+ i, and because all imaginary solution exist as conjugate pairs we automatically know another solution must be 3 - i. Note that you cannot have a polynomial where the solutions are JUST 3 + i and 1 - 5 and have real coefficients. Further your not going to get any rational coefficients with just 1 - 5 so we must also have 1 + 5 if the polynomial were in factored form it would look like this let P(x) be a polynomial function P(X) = (x - 3 - i )(x - 3 + i)(x - 1+ 5)(x - 1- 5) which equates to P(X) = x^4 - 8x^3 + 18x^2 +4x -40
Answered by Blinki - Tue Jan 20 14:59:32 2009
Find a polynomial function that has the given zeros and degree?
Q. Find a polynomial function that has the given zeros and degree. ZEROS -2, 2, 5. DEGREE: 3
Asked by Itsblackpepper - Thu Jun 25 13:51:54 2009 - - 5 Answers - 0 Comments
A. If r is a root (or zero) of a function, then (x-r) is a factor of equation f(x) = a (x+2) (x-2) (x-5) where a is a constant
Answered by mathmom28 - Thu Jun 25 13:59:06 2009
Q. Find a polynomial function that has the given zeros and degree. ZEROS -2, 2, 5. DEGREE: 3
Asked by Itsblackpepper - Thu Jun 25 13:51:54 2009 - - 5 Answers - 0 Comments
A. If r is a root (or zero) of a function, then (x-r) is a factor of equation f(x) = a (x+2) (x-2) (x-5) where a is a constant
Answered by mathmom28 - Thu Jun 25 13:59:06 2009
What is a polynomial function with the roots -1, 2, and 3?
Q. I need three polynomial functions that have those roots. What is a polynomial function with the roots at -2, -1, 0, 1, and 2? Can you write another polynomial with the same roots? What is the minimum degree of either of these polynomials? Our math teacher gave us a packet and said "figure out the rules. you will retain the knowledge better". -.-
Asked by Sabley - Mon Feb 25 22:58:12 2008 - - 4 Answers - 0 Comments
A. If -1, 2 and 3 are roots, then ( x + 1)(x - 2)(x + 3) are factors of the polynomial. Multiply it out and you will have your third degree polynomial. To find two others, simply multiply the entire polynomial by some constant, like 2 or -4. To answer the second part of your question, x(x + 2)(x - 2)(x + 1)(x - 1) is such a polynomial that has those roots. This is a fifth degree polynomial because it has 5 roots, or 5 factors. To write other such polynomials, just multiply by some constant like 2 or -4.
Answered by j g - Mon Feb 25 23:06:07 2008
Q. I need three polynomial functions that have those roots. What is a polynomial function with the roots at -2, -1, 0, 1, and 2? Can you write another polynomial with the same roots? What is the minimum degree of either of these polynomials? Our math teacher gave us a packet and said "figure out the rules. you will retain the knowledge better". -.-
Asked by Sabley - Mon Feb 25 22:58:12 2008 - - 4 Answers - 0 Comments
A. If -1, 2 and 3 are roots, then ( x + 1)(x - 2)(x + 3) are factors of the polynomial. Multiply it out and you will have your third degree polynomial. To find two others, simply multiply the entire polynomial by some constant, like 2 or -4. To answer the second part of your question, x(x + 2)(x - 2)(x + 1)(x - 1) is such a polynomial that has those roots. This is a fifth degree polynomial because it has 5 roots, or 5 factors. To write other such polynomials, just multiply by some constant like 2 or -4.
Answered by j g - Mon Feb 25 23:06:07 2008
How do you use excel to graph a polynomial function?
Q. How do you use excel to graph a polynomial function
Asked by sweetgurl8603 - Mon Oct 29 15:04:31 2007 - - 1 Answers - 0 Comments
A. It's been a long time since I plotted a polynomial function so I did a Yahoo search for "polynomial function " and got the following: mathforum.org/alejandre/p olynomial.graph.html It tells how to Get your graph. Maybe some of the other hits might work better if you need anything further.
Answered by Don R - Mon Oct 29 23:26:57 2007
Q. How do you use excel to graph a polynomial function
Asked by sweetgurl8603 - Mon Oct 29 15:04:31 2007 - - 1 Answers - 0 Comments
A. It's been a long time since I plotted a polynomial function so I did a Yahoo search for "polynomial function " and got the following: mathforum.org/alejandre/p olynomial.graph.html It tells how to Get your graph. Maybe some of the other hits might work better if you need anything further.
Answered by Don R - Mon Oct 29 23:26:57 2007
How do you write a polynomial function from 3, -3, 2-i?
Q. How do you write a polynomial function from 3, -3, 2-i? Thanks! :)
Asked by *Sweetie* - Tue Apr 17 21:38:19 2007 - - 1 Answers - 0 Comments
A. Okay.. Whenever you have a complex root like 2-i, you always have to have a conjugate of it; the conjugate of 2-i would be 2+i. Okay.. so you have 4 roots. Roots are x minus something. So.. you would have (x-3) (x+3) [x-(2-i)] [x-(2+i)]. All you need to do is factoring. To make it easy, you just do the (x+3) (x-3) first and then the other two next, basically, group by pairs. you would get (x^2)-9 for the first one. For the second one you would get.. (x^2)+5. Right?? Then you factor the two you got [(x^2)-9 ] and [(x^2)+5]. You will get x^4-4x^2-45. I hope you get it. But, if you don't please feel free to email me at mashi_cutie@hotmail.com Good luck.. ^_^
Answered by mashi_cutie1004 - Tue Apr 17 22:05:44 2007
Q. How do you write a polynomial function from 3, -3, 2-i? Thanks! :)
Asked by *Sweetie* - Tue Apr 17 21:38:19 2007 - - 1 Answers - 0 Comments
A. Okay.. Whenever you have a complex root like 2-i, you always have to have a conjugate of it; the conjugate of 2-i would be 2+i. Okay.. so you have 4 roots. Roots are x minus something. So.. you would have (x-3) (x+3) [x-(2-i)] [x-(2+i)]. All you need to do is factoring. To make it easy, you just do the (x+3) (x-3) first and then the other two next, basically, group by pairs. you would get (x^2)-9 for the first one. For the second one you would get.. (x^2)+5. Right?? Then you factor the two you got [(x^2)-9 ] and [(x^2)+5]. You will get x^4-4x^2-45. I hope you get it. But, if you don't please feel free to email me at mashi_cutie@hotmail.com Good luck.. ^_^
Answered by mashi_cutie1004 - Tue Apr 17 22:05:44 2007
How do you solve this polynomial function?
Q. My teacher said last block that polynomial functions can have real and imaginary roots ... is this true ? I'm just a little bit confused about this. And also, can you please explain how to solve this function? f(x) = 3x^4 - x^2 + x + 1 How would you find all of the roots in this function ?
Asked by Camille - Sun Mar 1 11:43:29 2009 - - 2 Answers - 0 Comments
Q. My teacher said last block that polynomial functions can have real and imaginary roots ... is this true ? I'm just a little bit confused about this. And also, can you please explain how to solve this function? f(x) = 3x^4 - x^2 + x + 1 How would you find all of the roots in this function ?
Asked by Camille - Sun Mar 1 11:43:29 2009 - - 2 Answers - 0 Comments
how to find leading coefficient of polynomial function graphs?
Q. just by looking at the graph, how do i find the coefficient of polynomial functions? I can find the factors and the sign of the coefficient but I dont understand how to find the coeffcient number.
Asked by andy - Wed Sep 10 18:03:11 2008 - - 2 Answers - 0 Comments
A. 2x^2+3x-7=y leading coefficient=2 x^3-16x+63654=y leading coefficient=1 the number that the highest degree of x is multiplied by is the leading oefficient
Answered by Mr. A - Fri Sep 12 15:29:28 2008
Q. just by looking at the graph, how do i find the coefficient of polynomial functions? I can find the factors and the sign of the coefficient but I dont understand how to find the coeffcient number.
Asked by andy - Wed Sep 10 18:03:11 2008 - - 2 Answers - 0 Comments
A. 2x^2+3x-7=y leading coefficient=2 x^3-16x+63654=y leading coefficient=1 the number that the highest degree of x is multiplied by is the leading oefficient
Answered by Mr. A - Fri Sep 12 15:29:28 2008
explain, how finite differences are related to the degree of a polynomial function. prove it with at least two
Q. explain, how finite differences are related to the degree of a polynomial function. prove it with at least two EXAMPLES?
Asked by Ali T - Mon Oct 29 00:05:24 2007 - - 1 Answers - 0 Comments
A. This is a numerical technique for estimating the value of a tabulated function. In it simplest form, you take values of f(x) at regular intervals of x. then take the differences. If the differences are all the same difference then you have a linear or 1st degree function. If you have a second order function then the regular finite differences will not be constant until you take the second difference -i.e. the differences of the differences. Here is an example let f (x) = x^2 f(1) is 1 f(2) is 4 f(3) is 9 f(4) is 16 f(5) is 25 first diffs are 3 5 7 9 second diffs are 2 2 2 2 See, the second diffs are constant so the original was a second order function. Let's try f(x) = x^2+5 f(1) is 6 f(2) is 9 f(3) is 14 f(4) is 21 f(5) is 30… [cont.]
Answered by RL612 - Mon Oct 29 21:13:51 2007
Q. explain, how finite differences are related to the degree of a polynomial function. prove it with at least two EXAMPLES?
Asked by Ali T - Mon Oct 29 00:05:24 2007 - - 1 Answers - 0 Comments
A. This is a numerical technique for estimating the value of a tabulated function. In it simplest form, you take values of f(x) at regular intervals of x. then take the differences. If the differences are all the same difference then you have a linear or 1st degree function. If you have a second order function then the regular finite differences will not be constant until you take the second difference -i.e. the differences of the differences. Here is an example let f (x) = x^2 f(1) is 1 f(2) is 4 f(3) is 9 f(4) is 16 f(5) is 25 first diffs are 3 5 7 9 second diffs are 2 2 2 2 See, the second diffs are constant so the original was a second order function. Let's try f(x) = x^2+5 f(1) is 6 f(2) is 9 f(3) is 14 f(4) is 21 f(5) is 30… [cont.]
Answered by RL612 - Mon Oct 29 21:13:51 2007
How to find all the zeros of the polynomial function? ?
Q. when you find all of the zeros in a polynomial function, do you have to use a graphing Calculator if you don't have any of the zeros? Can you find all of the zeros without a graphing calculator?
Asked by shanidesigns - Fri Jan 23 17:54:17 2009 - - 5 Answers - 0 Comments
A. In general, there isn't a method (that just uses arithmetic and radicals) to find the zeros of a chosen polynomial when the degree is 5 or more. However, there are many tricks that may be used, in special cases. For instance, if the coefficients are all rational, then we can find all the rational zeros without the use of a calculator. Also, some polynomials will factor easily, so we can find the zeros without using a calculator in this case too.
Answered by Awms A - Fri Jan 23 18:01:08 2009
Q. when you find all of the zeros in a polynomial function, do you have to use a graphing Calculator if you don't have any of the zeros? Can you find all of the zeros without a graphing calculator?
Asked by shanidesigns - Fri Jan 23 17:54:17 2009 - - 5 Answers - 0 Comments
A. In general, there isn't a method (that just uses arithmetic and radicals) to find the zeros of a chosen polynomial when the degree is 5 or more. However, there are many tricks that may be used, in special cases. For instance, if the coefficients are all rational, then we can find all the rational zeros without the use of a calculator. Also, some polynomials will factor easily, so we can find the zeros without using a calculator in this case too.
Answered by Awms A - Fri Jan 23 18:01:08 2009
write a polynomial function in standard form?
Q. with these zero's write a polynomial function in standard form: 2, -6, and 2+ 4i ??? can some1 do this for me and explain it?
Asked by John - Fri Nov 2 00:36:48 2007 - - 1 Answers - 0 Comments
A. the "zeroes" are the roots of the polynomial. That is, If p(x) is your polynonial, then p(2), p(-6), and p(2 + 4i)= 0 There is a theorem that states that r is a root of p(x), if and only if x-r is a factor of p(x). That is p(x) =(x-r)q(x) where q(x) is some new polynomial. There is another theorem that states that if a polynomial has a complex non-real root a+bi, then a-bi is also a root. For instance, if 2+i is a root, then so is 2-i. Or if 3-6i is a root, then so is 3+6i. So, since 2, -6, and 2+ 4i ***2-4i is also a root** are all roots, our polynomial is (x-2)*(x-(-6))*(x-(2+4i)) *(x-(2-4i)) = (x-2)*(x+6)*(x-(2+4i))*(x -(2-4i) Multiplying these factors out, we get p(x) = x^4 - 8x^2 + 128x - 240 The latter is your standard form. … [cont.]
Answered by guyava99 - Fri Nov 2 01:00:05 2007
Q. with these zero's write a polynomial function in standard form: 2, -6, and 2+ 4i ??? can some1 do this for me and explain it?
Asked by John - Fri Nov 2 00:36:48 2007 - - 1 Answers - 0 Comments
A. the "zeroes" are the roots of the polynomial. That is, If p(x) is your polynonial, then p(2), p(-6), and p(2 + 4i)= 0 There is a theorem that states that r is a root of p(x), if and only if x-r is a factor of p(x). That is p(x) =(x-r)q(x) where q(x) is some new polynomial. There is another theorem that states that if a polynomial has a complex non-real root a+bi, then a-bi is also a root. For instance, if 2+i is a root, then so is 2-i. Or if 3-6i is a root, then so is 3+6i. So, since 2, -6, and 2+ 4i ***2-4i is also a root** are all roots, our polynomial is (x-2)*(x-(-6))*(x-(2+4i)) *(x-(2-4i)) = (x-2)*(x+6)*(x-(2+4i))*(x -(2-4i) Multiplying these factors out, we get p(x) = x^4 - 8x^2 + 128x - 240 The latter is your standard form. … [cont.]
Answered by guyava99 - Fri Nov 2 01:00:05 2007
If a polynomial function of degree 5 with rational coefficients has 3, 7, 7 +4i as zeros.What are the?
Q. If a polynomial function of degree 5 with rational coefficients has 3, 7, 7 +4i as zeros. What are the other zeros.
Asked by TICKLES - Mon Apr 2 00:08:32 2007 - - 1 Answers - 0 Comments
A. Irrational and complex zeros come in imaginary pairs (as long as the coefficients of the polynomial are real). So you also have - 7 and 7-4i Here's a really cool way to get the polynomial that has these five zeros. Build factors. From the zero of 3, the factor is obviously (x-3). From the +- 7, we get (x+ 7)(x- 7) which equals (x^2 - 7). Now use this little "recipe" to create a quadratic factor using two of its zeros: x^2 - (sum)x + product. That is referring to the sum and product of the two zeros. 7+4i + 7 - 4i = 14 and (7+4i)(7 - 4i) = 49-16i^2 = 65. So this factor is x^2 - 14x + 65. Multiply (x-3)(x^2-7)(x^2 - 14x + 65) to get the quintic.
Answered by Kathleen K - Mon Apr 2 00:12:19 2007
Q. If a polynomial function of degree 5 with rational coefficients has 3, 7, 7 +4i as zeros. What are the other zeros.
Asked by TICKLES - Mon Apr 2 00:08:32 2007 - - 1 Answers - 0 Comments
A. Irrational and complex zeros come in imaginary pairs (as long as the coefficients of the polynomial are real). So you also have - 7 and 7-4i Here's a really cool way to get the polynomial that has these five zeros. Build factors. From the zero of 3, the factor is obviously (x-3). From the +- 7, we get (x+ 7)(x- 7) which equals (x^2 - 7). Now use this little "recipe" to create a quadratic factor using two of its zeros: x^2 - (sum)x + product. That is referring to the sum and product of the two zeros. 7+4i + 7 - 4i = 14 and (7+4i)(7 - 4i) = 49-16i^2 = 65. So this factor is x^2 - 14x + 65. Multiply (x-3)(x^2-7)(x^2 - 14x + 65) to get the quintic.
Answered by Kathleen K - Mon Apr 2 00:12:19 2007
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ue, 01 Sep 2009 07:38:54 GM
>transfer . function. of the type : > >y = a0 + a1 * x + a2 * x^2 + a3 * x^3 + a4 * x^4 + a5 * x^5 + a6 * x^6 >+ a7 * x^7 + a8 * x^8 + a9 * x^9 > >. Polynomial. semplification : > >y = a0 + x * (a1 + x * (a2 + x * (a3 + x * (a4 + x * (a5 + x ...
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