How do you find all zeros of a function using the rational zero theorem?
Q. I know how to find all possible rational zeros, but how do you use that information to find the real zeros. One of the problems is: f(x)=x^3-5x^2-22x+56.
Asked by -Matt- - Mon Jan 28 18:40:40 2008 - - 1 Answers - 0 Comments

A. First: find all possible zeroes... Second: Start testing! Find out which of these possibilities actually ARE factors. To see if "2" is a root (to see if the the expression is divisible by x-2, just divide (synthetically). 2| 1 -5 -22 56 2 -6 -56 --- 1 -3 -28 0 Since the rightmost number is 0, the expression is divisible by x-2. Thus, we've converted the original function TO: f(x)=(x-2)(x^2-3x-28) Now we can factor the second part: f(x)=(x-2)(x-7)(x-4) Now, set f(x)=0 0=(x-2)(x-7)(x-4) x=2 x=7 x=4 :)
Answered by SaintPretz59 - Mon Jan 28 18:51:45 2008

How do you find the zeros of a quadratic function without having to graph it?
Q. -the graph is the shape of a parabola -in algebra -zeros are the point(s) that cross the x-axis on a coordinate plane -how would u find them without having to graph the function?
Asked by little indian - Tue Feb 19 18:07:57 2008 - - 2 Answers - 0 Comments

A. I believe you can either do the quadratic formula with the equation/function, or you can factor it and then solve for x. Either way, you'll end up with values for x, which should be the zeros.
Answered by shoomy1030 - Tue Feb 19 18:16:57 2008

How do you do long or synthetic division to find the other zeros?
Q. How do you write the complete factored form given that -4 is a zero? How would i use long or synthetic division to find the other zeros?
Asked by Emily - Thu Sep 24 03:06:47 2009 - - 1 Answers - 0 Comments

A. If x=-4 is a zero of the polynomial, then (x-4) will be factor of the polyniklial and it can be written as (x-4)* g(x) where g(x) will be obtained by dividing polynomial by (x-4).
Answered by DOVE - Thu Sep 24 03:21:50 2009

How do you determine whether zeros of a polynomial are complex conjugates?
Q. What is a complex conjugates? And how do you determine whether the zeros of the polynomial are complex conjugates?
Asked by Desiree D - Sun Feb 3 17:49:23 2008 - - 2 Answers - 0 Comments

A. If all coefficients of the polynomial are real, all non-real roots will be complex conjugates. This follows from the fact that all polynomials with real coefficients can be factored into first and second degree polynomials with real coefficients, and all quadratic equations with real coefficients but no real roots have distinct roots that are complex conjugates.
Answered by jpschnei - Sun Feb 3 18:01:06 2008

How to count evens odds and zeros in a given integer by JAVA?
Q. Please help me. I am not so familiar with programming in JAVA. User enters an arbitrary integer. Task is to print the number of zeros, odd , and even digits in this input. Forexample: Enter: 1098 Output: Evens:1 Odds:2 Zeros:1 Any ideas please?
Asked by Name N - Sat Sep 13 12:38:06 2008 - - 1 Answers - 0 Comments

A. Process the input as a String. Use String.toCharArray() to get a char[] and loop through its elements and count by the three categories of your interest.. For each element, test if ch=='0' else, use Integer.intValue() to convert it into an int val. if val/2*2 == val, it is even. Note in int domain, 1/2=0, 2/2=1, 3/2=1, 4/2=2
Answered by Yuman - Sun Sep 14 22:22:16 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 Dancing Beauty - 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

FInd the zeros of cubic equation with out using a calculator , 3x^3 -9x + 1 = 0, How can I maually find them?
Q. I am trying to find the zeros of this cubic equation manually , But I have no Idea where to start ? Please help .
Asked by krishna - Sat Nov 14 02:31:08 2009 - - 2 Answers - 0 Comments

How to find a polynomial functions with just the zeros?
Q. I was given the zeros -3, radical 5 and radical -5 how do i find it?
Asked by confused guy - Tue Jan 27 18:17:43 2009 - - 3 Answers - 0 Comments

A. The factor theorem states: If 'a' is a zero of a polynomial P(x) then x - 'a' is a factor of P(x). Multiply (x + 3)(x - rad 5)(x + rad 5) (I'm presuming you mean -rad 5)
Answered by Martin F - Tue Jan 27 18:27:04 2009

How do I find the zeros in a 3rd degree polynomial?
Q. In the equation 2x^3-3x^2-16x+10 I found one zero (-5/2) using the rational zeros theorem, but there are two more and I can't remember how to find them. Thanks everyone! I get it now. I was making it far more difficult than it was. I can finally go to bed. :-)
Asked by Amber M - Mon Mar 19 02:09:55 2007 - - 3 Answers - 0 Comments

A. Let p(x) = 2x^3 - 3x^2 - 16x + 10 Since you found one zero, it's pretty much downhill from here, because if (-5/2) is a zero, it follows that [x - (-5/2)] is a factor, or [x + (5/2)] is a factor. BUT, this also means any multiple of this is a factor, even a multiple of 2. 2(x + (5/2)) = [2x + 5] is a factor. Your next step would be to do synthetic long division; (2x + 5) INTO (2x^3 - 3x^2 - 16x + 10) Since synthetic long division is difficult to show on here, I'll tell you right now that the quotient should be x^2 - 4x + 2, so 2x^3 - 3x^2 - 16x + 10 = (2x + 5)(x^2 - 4x + 2) Now that you have a quadratic as a factor, you can use the quadratic formula on x^2 - 4x + 2.
Answered by Puggy - Mon Mar 19 02:15:33 2007

How does binary work besides being zeros and ones?
Q. Is there a key that converts the alphabet where each letter is represented by zeros and ones? If so could you provide what this key is for each letter?
Asked by sylvester - Sat May 20 08:05:53 2006 - - 5 Answers - 1 Comments

A. The key for the conversion and reconversion of the data in Binary ie the machine language and the real world characters is the ASCII code. You can see the list of ASCII Code in Its a long table where u can see all the characters which you see on the keyboard has a corresponding binary equivalent of the Binary code which is actually processed or stored
Answered by vamsich_gem - Sat May 20 08:53:28 2006

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

How to find real zeros in polynomial function?
Q. The question asks "What are the real zeros for f(x) = x^3 + 2x^2 -4? (Approximate the zeros to the nearest tenth) Any help is appreciated. By the way the answer is supposed to be 1.1...I don't understand why though.
Asked by The Questioner - Thu Jun 12 18:02:09 2008 - - 1 Answers - 0 Comments

A. By going to 1.13039543476728
Answered by unknown - Thu Jun 12 18:07:13 2008

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

How can I find the zeros of a fourth degree polynomial without graphing it?
Q. Whilst attempting to find the solutions to c-x = 1/(sqrt(1-x^2/c^2)), where c is the speed of light (3.00 * 10^8 m/s), I ran into a fourth degree polynomial. I'm more interested in knowing of any theorem that gives all zeros of a fourth degree polynomial. Also, I'm curious to know if a theorem exists that gives all zeros of an n-th degree polynomial.
Asked by Bob S - Tue Jul 17 12:34:06 2007 - - 3 Answers - 0 Comments

A. A previous answerer gave you the general solution to any 4th degree polynomial so I won't reiterate his solution. I will add that there is NO general solution to the general n-th degree polynomial for n >= 5. (Of course, for special types of n-th degree polynomials there are easily found solutions (e.g., x^n - 1 = 0). This is a rather famous theorem in Group Theory (Abel's Theorem) proved in the 1800s I believe. Math Rules!
Answered by Math Chick - Tue Jul 17 12:57:42 2007

What is the relationship between the degree of a polynomial and the maximum number of zeros it can have?
Q. a. The number of zeros is on less than the degree of the polynomial b.The degree if the polynomial is one less than the number of zeros. c. They are the same value d. The number of zeros is the square of the degree of the polynomal.
Asked by trulyurs87 - Sat Jun 14 10:49:32 2008 - - 1 Answers - 0 Comments

A. A polynomial of degree n has exactly n zeros. (Note they may be repeated, but there are exactly n of them.) The polynomial can be written as a constant times the product of n linear factors of the form (x-r) where r is a root. So the answer is choice c.
Answered by Duke - Sat Jun 14 10:55:17 2008

How do I find the complex zeros of the polynomial P(x) = x^4 + 2x^3 - 2x^2 + 2x - 3?
Q. I know that the real zeros are 1 and -3, but how do I determine the complex zeros? Help
Asked by Pride of Troy - Mon Oct 22 15:04:49 2007 - - 3 Answers - 0 Comments

A. Assume the complex zeroes as A+iB and C+iD. You already have the real zeroes. So - do this (x-A-iB)(x-iC-D)(x-1)(x+3 ) and then equate the exponents of what you get to the original P(x) . Remember that there will be no imaginary component so take any 'i' common and equate whatvever inside the brackets to zero. Solving these equations in A,B,C,D you would easily get the other two zeroes. Good luck!
Answered by sumita4u - Mon Oct 22 15:12:55 2007

How do I find the complex zeros of the polynomial p(x)=2x^3-13x^2+36x-40?
Q. I know that the real zero is 2.5, but I am having trouble getting the complex zeros. I am still confused, anymore help anyone?
Asked by Stephanie H - Tue Oct 23 18:06:56 2007 - - 1 Answers - 0 Comments

A. Since you know the real zero is 2.5, the factor theorem (look up on Wikipedia) tells you that (x-2.5) or (2x-5) is a factor of the polynomial. Doing the long division of the cubic p(x) by (2x-5) should give you the quadratic x^2-4x+8. You can then easily find the complex zeros by using the quadratic formulae.
Answered by Kenny Boy - Tue Oct 23 18:31:13 2007

Find a polynomial of degree 6 with real coefficients and with zeros ?
Q. Find a polynomial of degree 6 with real coefficients and with zeros 3 + 4i, 2 5i , 0, 2, and such that P( 1) = 14688. You may leave your result as a product of linear and irreducible quadratic factors.
Asked by Simm - Fri Dec 5 16:44:02 2008 - - 2 Answers - 0 Comments

A. Complex roots always occur in conjugate pairs. So, if (3 + 4i) is a root, then (3 - 4i) is a root. Also, if (2 - 5i) is a root, then (2 + 5i) is a root. The six roots are: 3 + 4i, 3 - 4i, 2 - 5i, 2 + 5i, 0, and 2 The polynomial can be written as a product of the roots as follows: (x - (3 + 4i)) * (x - (3 - 4i)) * (x - (2 - 5i)) * (x - (2 + 5i)) * (x - 0) * (x - 2) Simplifying: P(x) = (x - 3 - 4i) * (x - 3 + 4i) * (x - 2 + 5i) * (x - 2 - 5i) * (x) * (x - 2) However, when you solve for P(-1): P(-1) = (-4 - 4i) * (-4 + 4i) * (-3 + 5i) * (-3 - 5i) * (-1) * (-3) = (16 + 16) * (9 + 25) * 3 = 32 * 34 * 3 P(-1) = 3264 This doesn't equal -14668. That's because there needs to be a multiplier for the entire equation, which can be… [cont.]
Answered by Math Wiz - Fri Dec 5 17:16:07 2008

Trying to find the zeros of a quadratic?
Q. I have this quadratic: -2x^2-19x-24 I need to find its zeros and factor it but I havent had any luck. I also tried the quadratic formula but it produces negative roots?
Asked by MissF-18 - Wed Sep 16 22:03:38 2009 - - 5 Answers - 0 Comments

A. So whats wrong with negative roots? The zeros for this function are at x=-8 and x = -3/2
Answered by Chief_Iron_Cock - Wed Sep 16 22:12:35 2009

How do I find the zeros of this polynomial?
Q. Find all zeros (rational, irrational, and imaginary) exactly. P(x) = 2x^3 - 9x^2 - 2x + 30 Any help would be great, thanks.
Asked by Zak - Wed Apr 1 05:21:55 2009 - - 3 Answers - 0 Comments

A. You factor it P(x) = 2x 9x 2x + 30 P(x) = (2x 5)(x 2x 6) To find the zeroes, equate ea of the factors to zero and solve (2x 5) = 0 2x = 5 x = 5/2 (x 2x 6) = 0 x = {2 [2 - (4)(1)(-6)]}/2(1) x = (2 28)/2 x = (2 2 7)/2 x = 1 7 x = {5/2, 1 7, 1 + 7}
Answered by Jerome J - Wed Apr 1 05:36:16 2009