Find a polynomial with integer coefficients and a leading coefficient of one that satisfies the given?
Q. Find a polynomial with integer coefficients and a leading coefficient of one that satisfies the given conditions. P has degree 2, and zeros 1 + i 5 and 1 - i 5. P(x) =?
Asked by Math - Tue Sep 22 16:54:24 2009 - - 1 Answers - 0 Comments
A. If x = a is a zero, then x - a is a factor. Since x = 1 + i 5 and x = 1 - i 5 are solutions, we have the following equation for P(x): P(x) = [x - (1 + i 5)][x - (1 - 5)] ==> P(x) = (x - 1 - i 5)(x - 1 + i 5) You can re-write this and get: P(x) = [(x - 1) + i 5][(x - 1) - i 5] This is a difference of squares, so we can get: P(x) = (x - 1) - (i 5) ==> P(x) = x - 2x + 1 - i (5) ==> P(x) = x - 2x + 1 - (-1)(5) ==> P(x) = x - 2x + 1 + 5 ==> P(x) = x - 2x + 6 Answer Verification: I hope that helps!
Answered by Brian - Tue Sep 22 20:24:30 2009
Q. Find a polynomial with integer coefficients and a leading coefficient of one that satisfies the given conditions. P has degree 2, and zeros 1 + i 5 and 1 - i 5. P(x) =?
Asked by Math - Tue Sep 22 16:54:24 2009 - - 1 Answers - 0 Comments
A. If x = a is a zero, then x - a is a factor. Since x = 1 + i 5 and x = 1 - i 5 are solutions, we have the following equation for P(x): P(x) = [x - (1 + i 5)][x - (1 - 5)] ==> P(x) = (x - 1 - i 5)(x - 1 + i 5) You can re-write this and get: P(x) = [(x - 1) + i 5][(x - 1) - i 5] This is a difference of squares, so we can get: P(x) = (x - 1) - (i 5) ==> P(x) = x - 2x + 1 - i (5) ==> P(x) = x - 2x + 1 - (-1)(5) ==> P(x) = x - 2x + 1 + 5 ==> P(x) = x - 2x + 6 Answer Verification: I hope that helps!
Answered by Brian - Tue Sep 22 20:24:30 2009
How do you find a polynomial with zeros given?
Q. I've been given 2, -3, and 1-2i as the zeros and I need to find a polynomial containing those zeros with real coefficients. How would I do that?
Asked by confused guy - Wed Dec 17 20:31:34 2008 - - 2 Answers - 0 Comments
A. Since 2 is a zero, (x - 2) is a factor Since -3 is a zero, (x + 3) is a factor Since 1 - 2i is a zero, its conjugate 1 + 2i is also a root. Find trinomial with the above complex numbers as roots. Sum of roots = (1 - 2i) + (1 + 2i) = 2 = -b; b = -2 Product = (1 - 2i)(1 + 2i) = 1 + 4 = 5 = c (x^2 - 2x + 5) is a factor polynomial f(x) = (x - 2)(x + 3)(x^2 - 2x + 5) f(x) = (x^2 + x - 6)(x^2 - 2x + 5) f(x) = x^4 - x^3 - 3x^2 + 17x - 30
Answered by Jerome J - Wed Dec 17 20:40:48 2008
Q. I've been given 2, -3, and 1-2i as the zeros and I need to find a polynomial containing those zeros with real coefficients. How would I do that?
Asked by confused guy - Wed Dec 17 20:31:34 2008 - - 2 Answers - 0 Comments
A. Since 2 is a zero, (x - 2) is a factor Since -3 is a zero, (x + 3) is a factor Since 1 - 2i is a zero, its conjugate 1 + 2i is also a root. Find trinomial with the above complex numbers as roots. Sum of roots = (1 - 2i) + (1 + 2i) = 2 = -b; b = -2 Product = (1 - 2i)(1 + 2i) = 1 + 4 = 5 = c (x^2 - 2x + 5) is a factor polynomial f(x) = (x - 2)(x + 3)(x^2 - 2x + 5) f(x) = (x^2 + x - 6)(x^2 - 2x + 5) f(x) = x^4 - x^3 - 3x^2 + 17x - 30
Answered by Jerome J - Wed Dec 17 20:40:48 2008
How do you find a polynomial of minimum degree that have the given zeros?
Q. Find a polynomial of minimum degree (there are many) that have the given zeros. 1) -2, 0, 5, 6 Make the leading coefficient equal to 1: f (x)= 2) 1 - square root of 5 and 1 + square root of 5 Make the leading coefficient equal to 1 and eliminate the parentheses: f (x) = Can you explain how to solve these? Thanks.
Asked by mshealerless - Sun Nov 5 19:59:22 2006 - - 3 Answers - 0 Comments
A. if a is a zero x-a is a factor 1) so factors are (x+2) ,x (x-5)(x-6) multiply them and you get f(x) = x(x+2)(x-5)(x-6) multiply them and you get leading coefficient 1 2) factors are (x-(1-sqrt(5)) and (x-(1+sqrt(5)) so (x-1-sqrt(5))(x-1+ sqrt(5)) (a+b a-b form ) = (x-1)^2 - 5 = x^2-2x-4
Answered by mein Hoon na - Sun Nov 5 20:34:43 2006
Q. Find a polynomial of minimum degree (there are many) that have the given zeros. 1) -2, 0, 5, 6 Make the leading coefficient equal to 1: f (x)= 2) 1 - square root of 5 and 1 + square root of 5 Make the leading coefficient equal to 1 and eliminate the parentheses: f (x) = Can you explain how to solve these? Thanks.
Asked by mshealerless - Sun Nov 5 19:59:22 2006 - - 3 Answers - 0 Comments
A. if a is a zero x-a is a factor 1) so factors are (x+2) ,x (x-5)(x-6) multiply them and you get f(x) = x(x+2)(x-5)(x-6) multiply them and you get leading coefficient 1 2) factors are (x-(1-sqrt(5)) and (x-(1+sqrt(5)) so (x-1-sqrt(5))(x-1+ sqrt(5)) (a+b a-b form ) = (x-1)^2 - 5 = x^2-2x-4
Answered by mein Hoon na - Sun Nov 5 20:34:43 2006
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 sarah - 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 sarah - 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
What is the flow for finding zeros of a polynomial function?
Q. I have the following directions and equation: Find all the zeros of the polynomial function and write the polynomial as a product of its leading co-efficient and its linear factor (hint: first determine the rational zeros). P(x)=x^4+x^3-2x^2+4x-24 I am not getting the order and answers I think I need. Please help. Thanks.
Asked by __A_YAHOO_USER__ - Mon Sep 7 08:49:50 2009 - - 1 Answers - 0 Comments
A. Use the rational root theorem. Where p is the constant term and q is the leading coefficient. factors of p: 24, 12, 8, 6, 4, 3, 2, 1 factors of q: 1 So p/q: 24, 12, 8, 6, 4, 3, 2, 1 Plug those values into your given function and see which ones give you a zero. You will see that the x = -3 and x = 2 give you zeros. Use synthetic division with one of your zeros to find the others -3 1 1 -2 4 -24 ___-3__6_-12_24___ 1 -2 4 -8 0 So now use x^3 -2x^2 + 4x -8 and x = 2 and synthetic division to find the others... 2 1 -2 4 -8 ___2__0__8_ 1 0 4 0 So now you have x^2 + 4 = 0 You can easily find the other roots by solving for x x^2 + 4 = 0 x^2 = -4 x = 2i So your roots… [cont.]
Answered by unknown - Mon Sep 7 09:04:21 2009
Q. I have the following directions and equation: Find all the zeros of the polynomial function and write the polynomial as a product of its leading co-efficient and its linear factor (hint: first determine the rational zeros). P(x)=x^4+x^3-2x^2+4x-24 I am not getting the order and answers I think I need. Please help. Thanks.
Asked by __A_YAHOO_USER__ - Mon Sep 7 08:49:50 2009 - - 1 Answers - 0 Comments
A. Use the rational root theorem. Where p is the constant term and q is the leading coefficient. factors of p: 24, 12, 8, 6, 4, 3, 2, 1 factors of q: 1 So p/q: 24, 12, 8, 6, 4, 3, 2, 1 Plug those values into your given function and see which ones give you a zero. You will see that the x = -3 and x = 2 give you zeros. Use synthetic division with one of your zeros to find the others -3 1 1 -2 4 -24 ___-3__6_-12_24___ 1 -2 4 -8 0 So now use x^3 -2x^2 + 4x -8 and x = 2 and synthetic division to find the others... 2 1 -2 4 -8 ___2__0__8_ 1 0 4 0 So now you have x^2 + 4 = 0 You can easily find the other roots by solving for x x^2 + 4 = 0 x^2 = -4 x = 2i So your roots… [cont.]
Answered by unknown - Mon Sep 7 09:04:21 2009
How is dividing a polynomial by a binomial similar to or different from the long division you learned in elem.?
Q. How is dividing a polynomial by a binomial similar to or different from the long division you learned in elementary school? Can understanding how to do one kind of division help you with understanding the other kind? What are some examples from real life in which you might use polynomial division?
Asked by Daddy's Girl - Sun Dec 20 01:09:21 2009 - - 3 Answers - 0 Comments
A. In synthetic division (polynomials) expressions are grouped into like terms. In integer division, numbers are grouped into multiples of powers of ten. That is the only difference that comes to mind, so I suppose you could say that they are similar in every other way. Real life? I have seen synthetic division only in theoretical problems, but I would have to say that most people who use it still consider their lives to be real.
Answered by Pope - Sun Dec 20 01:29:14 2009
Q. How is dividing a polynomial by a binomial similar to or different from the long division you learned in elementary school? Can understanding how to do one kind of division help you with understanding the other kind? What are some examples from real life in which you might use polynomial division?
Asked by Daddy's Girl - Sun Dec 20 01:09:21 2009 - - 3 Answers - 0 Comments
A. In synthetic division (polynomials) expressions are grouped into like terms. In integer division, numbers are grouped into multiples of powers of ten. That is the only difference that comes to mind, so I suppose you could say that they are similar in every other way. Real life? I have seen synthetic division only in theoretical problems, but I would have to say that most people who use it still consider their lives to be real.
Answered by Pope - Sun Dec 20 01:29:14 2009
How do I find remaining factors of a polynomial based on this equation?
Q. Given the binomial, x-2, is a factor of x^3-6x^2+11x-6, find the remaining factors of the polynomial.
Asked by tinkerbellwantabe - Tue Mar 11 21:29:24 2008 - - 2 Answers - 0 Comments
A. divide x^3 - 6x^2 + 11x - 6 x goes into x^3 this many times: x^2 x^2 times (x-2) = x^3 - 2x^2 subtract that from x^3 - 6x^2 + 11x - 6 = -4x^2 + 11x - 6 x goes into -4x^2 this many times: -4x -4x times (x-2) = -4x^2 + 8x subtract that from -4x^2 + 11x - 6 = 3x - 6 x goes into 3x, 3 times 3 times (x-2) = 3x-6 subtract that, there is no remainder so (x^3 - 6x^2 + 11x - 6) / (x-2) = x^2 - 4x + 3 Now factor that binominial into (x-3)(x-1) Final answer: (x-1)(x-2)(x-3)
Answered by Steve A - Tue Mar 11 21:37:53 2008
Q. Given the binomial, x-2, is a factor of x^3-6x^2+11x-6, find the remaining factors of the polynomial.
Asked by tinkerbellwantabe - Tue Mar 11 21:29:24 2008 - - 2 Answers - 0 Comments
A. divide x^3 - 6x^2 + 11x - 6 x goes into x^3 this many times: x^2 x^2 times (x-2) = x^3 - 2x^2 subtract that from x^3 - 6x^2 + 11x - 6 = -4x^2 + 11x - 6 x goes into -4x^2 this many times: -4x -4x times (x-2) = -4x^2 + 8x subtract that from -4x^2 + 11x - 6 = 3x - 6 x goes into 3x, 3 times 3 times (x-2) = 3x-6 subtract that, there is no remainder so (x^3 - 6x^2 + 11x - 6) / (x-2) = x^2 - 4x + 3 Now factor that binominial into (x-3)(x-1) Final answer: (x-1)(x-2)(x-3)
Answered by Steve A - Tue Mar 11 21:37:53 2008
How do I determine whether or not a polynomial is a perfect square trinomial?
Q. How do I determine whether or not a polynomial is a perfect square trinomial?
Asked by purpuhlrowboht - Mon Nov 16 20:00:41 2009 - - 1 Answers - 0 Comments
A. It must fit this pattern: A 2AB + B and it will equal (A B) respectively for + or sign in middle term. 4x 12x + 9 fits the pattern with A = 2x and B=3 . . . LOOK: (2x) 2(2x)(3) + 3 = (2x + 3)
Answered by Mark - Mon Nov 16 20:05:57 2009
Q. How do I determine whether or not a polynomial is a perfect square trinomial?
Asked by purpuhlrowboht - Mon Nov 16 20:00:41 2009 - - 1 Answers - 0 Comments
A. It must fit this pattern: A 2AB + B and it will equal (A B) respectively for + or sign in middle term. 4x 12x + 9 fits the pattern with A = 2x and B=3 . . . LOOK: (2x) 2(2x)(3) + 3 = (2x + 3)
Answered by Mark - Mon Nov 16 20:05:57 2009
Find a polynomial of degree 4 with real coefficients and with zeros ?
Q. Find a polynomial of degree 4 with real coefficients and with zeros 3 + i, 0, 2. You may leave your result as a product of linear and irreducible quadratic factors.
Asked by Simm - Fri Dec 5 16:46:00 2008 - - 1 Answers - 0 Comments
A. x(x-2)(x - 6x + 8)
Answered by Track P - Fri Dec 5 16:54:32 2008
Q. Find a polynomial of degree 4 with real coefficients and with zeros 3 + i, 0, 2. You may leave your result as a product of linear and irreducible quadratic factors.
Asked by Simm - Fri Dec 5 16:46:00 2008 - - 1 Answers - 0 Comments
A. x(x-2)(x - 6x + 8)
Answered by Track P - Fri Dec 5 16:54:32 2008
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 write a polynomial function with rational coeffcients in standard form, with the given zeros?
Q. I need to write a polynomial function with rational coefficents in standard form. with the zeros 2i, and suare root of 3 I need the full function.
Asked by joeblake15 - Sun Jan 6 02:05:05 2008 - - 4 Answers - 0 Comments
A. f(x) = (x - 2i)(x + 2i)(x + 3)(x - 3) = (x - 4i )(x - 3) = (x + 4)(x - 3) = x^4 - 3 x + 4 x - 12 = x^4 + x -12 => f(x) = x^4 + x -12
Answered by piano - Sun Jan 6 02:25:06 2008
Q. I need to write a polynomial function with rational coefficents in standard form. with the zeros 2i, and suare root of 3 I need the full function.
Asked by joeblake15 - Sun Jan 6 02:05:05 2008 - - 4 Answers - 0 Comments
A. f(x) = (x - 2i)(x + 2i)(x + 3)(x - 3) = (x - 4i )(x - 3) = (x + 4)(x - 3) = x^4 - 3 x + 4 x - 12 = x^4 + x -12 => f(x) = x^4 + x -12
Answered by piano - Sun Jan 6 02:25:06 2008
How do you write a polynomial function with: Degree 3, -2 multiplicity, and 3 zeros?
Q. How do you write a polynomial function with: Degree 3, -2 multiplicity, and 3 zeros? Thank you.
Asked by ADH - Sun Jan 24 17:37:34 2010 - - 1 Answers - 0 Comments
A. Assuming you mean " 1 zero with multiplicity 2" and "3 different zeros" Then it is not possible. A polynomial fuction (complex or real) with 3 different zeros has the form: f = A (x-a) (x-b) x-c) with 3 different (complex or real) numbers a,b,c; but to have a zero with multiplicity 2 2 of those numbers have do be equal. In a polinomial with degree 3 there cannot be both. The degree would neeed to be 4 (or higher), than it could be: f = A (x-a) (x-b) (x-c)^2 with a,b,c different numbers and c being a zero with multiplicity 2.
Answered by hopefully - Tue Jan 26 06:24:19 2010
Q. How do you write a polynomial function with: Degree 3, -2 multiplicity, and 3 zeros? Thank you.
Asked by ADH - Sun Jan 24 17:37:34 2010 - - 1 Answers - 0 Comments
A. Assuming you mean " 1 zero with multiplicity 2" and "3 different zeros" Then it is not possible. A polynomial fuction (complex or real) with 3 different zeros has the form: f = A (x-a) (x-b) x-c) with 3 different (complex or real) numbers a,b,c; but to have a zero with multiplicity 2 2 of those numbers have do be equal. In a polinomial with degree 3 there cannot be both. The degree would neeed to be 4 (or higher), than it could be: f = A (x-a) (x-b) (x-c)^2 with a,b,c different numbers and c being a zero with multiplicity 2.
Answered by hopefully - Tue Jan 26 06:24:19 2010
Given the roots of a 4th degree polynomial and one other point, how do you find the leading coefficient?
Q. I have been given a project where I need to create an application in Excel that has the user input the roots of a 4th degree polynomial and one other point and has the program output the formula of the polynomial, a graph of the polynomial, and the y-intercept and leading coefficient of the polynomial. I have everything solved except for the leading coefficient. Can anyone tell me how to do this?
Asked by Harry - Tue Dec 1 14:22:07 2009 - - 1 Answers - 0 Comments
A. With roots of r1, r2, r3, and r4, your factored polynomial is C(x-r1)(x-r2)(x-r3)(x-r4) , where C is the leading coefficient. To find out what C is, given an x,y pair, evaluate (x-r1)(x-r2)(x-r3)(x-r4) and divide that into the y value you were given. For example, let's take roots of 1, 2, 3, and 4, and a point (0, 240): (0-1)(0-2)(0-3)(0-4) = (-1)(-2)(-3)(-4) = 24 240/24 = 10 So our leading coefficient, C, is 10.
Answered by Conan - Tue Dec 1 14:31:01 2009
Q. I have been given a project where I need to create an application in Excel that has the user input the roots of a 4th degree polynomial and one other point and has the program output the formula of the polynomial, a graph of the polynomial, and the y-intercept and leading coefficient of the polynomial. I have everything solved except for the leading coefficient. Can anyone tell me how to do this?
Asked by Harry - Tue Dec 1 14:22:07 2009 - - 1 Answers - 0 Comments
A. With roots of r1, r2, r3, and r4, your factored polynomial is C(x-r1)(x-r2)(x-r3)(x-r4) , where C is the leading coefficient. To find out what C is, given an x,y pair, evaluate (x-r1)(x-r2)(x-r3)(x-r4) and divide that into the y value you were given. For example, let's take roots of 1, 2, 3, and 4, and a point (0, 240): (0-1)(0-2)(0-3)(0-4) = (-1)(-2)(-3)(-4) = 24 240/24 = 10 So our leading coefficient, C, is 10.
Answered by Conan - Tue Dec 1 14:31:01 2009
How do you determine if a polynomial is the difference of two squares?
Q. the answer needs to be in 50 words. i understand what a polynomial is but what i'm getting stuck on is what kind of answer they are looking for, for i have like three answers down, could anyone answer this so i can compare notes and understand this assignment.
Asked by Quinnzel - Sat Aug 8 16:02:43 2009 - - 1 Answers - 0 Comments
A. You determine if a polynomial is the difference of two squares by saying that: The numbers is(are) perfect number(s). The exponent(s) is(are) even. If the numbers are not perfect square, and/or the exponent(s) is(are) not even, then a polynomial is not the difference of two squares. Formula: a -b = (a-b)(a+b) Ex. x -4 = (x-2)(x+2)
Answered by su u uo p - Sat Aug 8 16:11:56 2009
Q. the answer needs to be in 50 words. i understand what a polynomial is but what i'm getting stuck on is what kind of answer they are looking for, for i have like three answers down, could anyone answer this so i can compare notes and understand this assignment.
Asked by Quinnzel - Sat Aug 8 16:02:43 2009 - - 1 Answers - 0 Comments
A. You determine if a polynomial is the difference of two squares by saying that: The numbers is(are) perfect number(s). The exponent(s) is(are) even. If the numbers are not perfect square, and/or the exponent(s) is(are) not even, then a polynomial is not the difference of two squares. Formula: a -b = (a-b)(a+b) Ex. x -4 = (x-2)(x+2)
Answered by su u uo p - Sat Aug 8 16:11:56 2009
What do complex solutions do to polynomial functions?
Q. How do the complex numbers (a-bi/a+bi) affect the graph of a given polynomial function if they are imaginary? What do they do exactly?
Asked by Anonymous - Mon Mar 8 18:58:47 2010 - - 0 Answers - 0 Comments
Q. How do the complex numbers (a-bi/a+bi) affect the graph of a given polynomial function if they are imaginary? What do they do exactly?
Asked by Anonymous - Mon Mar 8 18:58:47 2010 - - 0 Answers - 0 Comments
What polynomial represents the new area?
Q. Joseph's basketball court is square, with one side being "x" feet. He decides to lengthen one side by 8 feet, and shorten the other side by 2 feet. What polynomial represents the new area?
Asked by bigsexy_velvet - Mon Nov 3 17:46:47 2008 - - 1 Answers - 0 Comments
A. (x + 8)(x-2) = x^2 + 6x - 16
Answered by Coolio - Mon Nov 3 17:50:32 2008
Q. Joseph's basketball court is square, with one side being "x" feet. He decides to lengthen one side by 8 feet, and shorten the other side by 2 feet. What polynomial represents the new area?
Asked by bigsexy_velvet - Mon Nov 3 17:46:47 2008 - - 1 Answers - 0 Comments
A. (x + 8)(x-2) = x^2 + 6x - 16
Answered by Coolio - Mon Nov 3 17:50:32 2008
Is a binomial a factor of a polynomial if it leaves a remainder when divided?
Q. Is a binomial a factor of a polynomial if it leaves a remainder when divided? Thanks! you saved my ass, your Mcawesome!
Asked by SuperPi - Tue Dec 23 13:35:50 2008 - - 6 Answers - 0 Comments
A. In many books, this is called "The remainder Theorem"... it goes like this: If a polynomial p(x) is divided by (x-r) then the remainder will be p(r). Since you claim that "there is a remainder" you really mean that the remainder is non-zero. Having p(r) not equal to zero means that p(x) cannot be rewriten as (x-r)q(x) for any polynomial q.
Answered by Mike Robertson - Tue Dec 23 13:46:05 2008
Q. Is a binomial a factor of a polynomial if it leaves a remainder when divided? Thanks! you saved my ass, your Mcawesome!
Asked by SuperPi - Tue Dec 23 13:35:50 2008 - - 6 Answers - 0 Comments
A. In many books, this is called "The remainder Theorem"... it goes like this: If a polynomial p(x) is divided by (x-r) then the remainder will be p(r). Since you claim that "there is a remainder" you really mean that the remainder is non-zero. Having p(r) not equal to zero means that p(x) cannot be rewriten as (x-r)q(x) for any polynomial q.
Answered by Mike Robertson - Tue Dec 23 13:46:05 2008
What is the best algorithm for solving polynomial roots?
Q. I have been looking at several algorithms for solving roots of polynomial and some of them are quite reasonable. But which is the best algorithm? I would like to find one that is: easy to write in program language, can solve complex roots unlike the bisection method, and convergence is fast and almost guaranteed. Thanks
Asked by John221 - Mon Dec 22 21:40:43 2008 - - 3 Answers - 0 Comments
A. Are there any conditions on the polynomials? Particularly: 1. the nature of the coefficients 2. the degree? In any case try the wikipedia link below. If you have been tinkering, you've probably already seen it, but just in case you haven't, it contains numerous links. You might also try reverse engineering some of the freeware graphers out there... Keep us posted on your results. bye for now. --ADDED-- Newton's method (previous post) does not meet the criteria you have cited of fast and guaranteed convergence. --Adding again-- A link to a bound theorem for real roots (related to Sturm's) that may be helpful for initial values if using Newton.
Answered by Toddio - Mon Dec 22 22:00:17 2008
Q. I have been looking at several algorithms for solving roots of polynomial and some of them are quite reasonable. But which is the best algorithm? I would like to find one that is: easy to write in program language, can solve complex roots unlike the bisection method, and convergence is fast and almost guaranteed. Thanks
Asked by John221 - Mon Dec 22 21:40:43 2008 - - 3 Answers - 0 Comments
A. Are there any conditions on the polynomials? Particularly: 1. the nature of the coefficients 2. the degree? In any case try the wikipedia link below. If you have been tinkering, you've probably already seen it, but just in case you haven't, it contains numerous links. You might also try reverse engineering some of the freeware graphers out there... Keep us posted on your results. bye for now. --ADDED-- Newton's method (previous post) does not meet the criteria you have cited of fast and guaranteed convergence. --Adding again-- A link to a bound theorem for real roots (related to Sturm's) that may be helpful for initial values if using Newton.
Answered by Toddio - Mon Dec 22 22:00:17 2008
How do you find a polynomial by its zeros?
Q. If a polynomial has only two zeros, its degree is going to be 2 right? and also, how do you find a polynomial by its zeros? Also, how do you find two polynomials that share the same two zeros?
Asked by scdesperado15 - Sat Jun 6 16:05:29 2009 - - 4 Answers - 0 Comments
A. Suppose f(x) = 0 for x = a and x = b. Then f(x) = (x - a)(x - b) = x - ax - bx + ab Example (zeroes at x = 3 and x = -2) f(x) = x -3x - (-2)x + 3(-2) = x - x - 6
Answered by Marlon JD - Sat Jun 6 16:09:37 2009
Q. If a polynomial has only two zeros, its degree is going to be 2 right? and also, how do you find a polynomial by its zeros? Also, how do you find two polynomials that share the same two zeros?
Asked by scdesperado15 - Sat Jun 6 16:05:29 2009 - - 4 Answers - 0 Comments
A. Suppose f(x) = 0 for x = a and x = b. Then f(x) = (x - a)(x - b) = x - ax - bx + ab Example (zeroes at x = 3 and x = -2) f(x) = x -3x - (-2)x + 3(-2) = x - x - 6
Answered by Marlon JD - Sat Jun 6 16:09:37 2009
How to find the increasing/decreasing interval of a polynomial?
Q. Given the polynomial f(x) = 2x^3-3x^2-3x+2 state the intervals where it is increasing and decreasing. Am I right in thinking that it increases when its derivative is decreasing, and vice versa? I graphed this on the computer, and it does not appear to be true. Any help is very appreciated, thanks !
Asked by Will - Sat Jan 10 20:05:11 2009 - - 1 Answers - 0 Comments
A. use the first derivative test. first, find the points at which the slope is zero, i.e. where the graph is a max/min/inflection point. then, find a random number between 2 adjacent zero-slope points. if the derivative is positive, then the function is increasing. if the derivative is negative, the function is decreasing.
Answered by ohSnap - Sat Jan 10 20:10:33 2009
Q. Given the polynomial f(x) = 2x^3-3x^2-3x+2 state the intervals where it is increasing and decreasing. Am I right in thinking that it increases when its derivative is decreasing, and vice versa? I graphed this on the computer, and it does not appear to be true. Any help is very appreciated, thanks !
Asked by Will - Sat Jan 10 20:05:11 2009 - - 1 Answers - 0 Comments
A. use the first derivative test. first, find the points at which the slope is zero, i.e. where the graph is a max/min/inflection point. then, find a random number between 2 adjacent zero-slope points. if the derivative is positive, then the function is increasing. if the derivative is negative, the function is decreasing.
Answered by ohSnap - Sat Jan 10 20:10:33 2009
From Yahoo Answer Search: 'Polynomial'
Thu Mar 11 15:46:08 2010 [ refresh local cache ]
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Inside GNSS
... from the surface to a given altitude, by integrating through the profile vertically and then fitting a polynomial to the resulting data. ...
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Roots and Turning Points The degree of a polynomial tells you even more about it than the limiting behavior Specifically an nth degree polynomial can have at most n real roots
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joeyjoejoe
ue, 12 Jan 2010 08:00:00 GM
Just took a qualifying exam (and did pretty well on it I think), but got stumped by one of the questions. Show that if a . polynomial. has all real roots.
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