find the complex zeros of the polynomial function.

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find the complex zeros of the polynomial function.

please help me with this!.   find the complex zeros of the polynomial function. write f in factored form.  use complex zeros to write f in factored form f(x)= x^3-6x^2+21x-26

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Algebra 2 - Finding Complex Zeros of a Polynomial Function ...


Jan 27, 2011 · How to find complex zeros of a polynomial function.

Complex Zeros and the Fundamental Theorem of Algebra


The Fundamental Theorem of Algebra states that every polynomial function of positive degree with complex coefficients has at least one complex zero. For example, the ...
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Finding complex zeros of a polynomial function - YouTube


Oct 09, 2011 · Pre-Calculus - Given complex zeros find the polynomial - Online Tutor, zeros 2,4+i - Duration: 4:51. Brian McLogan 30,721 views

SOLUTION: Find the complex zeros of the polynomial function ...


Question 697489: Find the complex zeros of the polynomial function. Write f in factored form. Use the complex zeros to write f in factored form.
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2.5 zeros of polynomial functions - Utep - Academics Portal Index


• Find rational zeros of polynomial functions. • Find conjugate pairs of complex zeros. • Find zeros of polynomials by factoring.
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Online Polynomial Roots Calculator that shows work


Online polynomial roots calculator finds the roots of any polynomial and creates a graph of the resulting ... (zeros) of the polynomial are returned. ... Complex numbers
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Complex Zeros - Cool Math


This algebra lesson explains complex zeros and shows how to find them.
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Solving Polynomials - Math is Fun


Solving Polynomials A polynomial looks like this: ... a "root" (or "zero") is where the function is equal to ... The first step in solving a polynomial is to find its ...
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Solving Polynomials: How-to | Purplemath


Solving Polynomials. ... solutions containing square roots or complex numbers, or both); these zeroes will ... Asking you to find the zeroes of a polynomial function, ...
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Suggested Questions And Answer :


find a polynomial function of the lowest degree with rational coefficients that has the given numbers as some of its zeros -3i,5

find a polynomial function of the lowest degree with rational coefficients that has the given numbers as some of its zeros -3i,5 You have a complex root, x = -3i Now the thing about complex roots is that they always come in pairs, as complex conjugates. If one complex root is (a + ib), the
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find the real and complex zeros of the following function f(x)=x^3-5x^2+8x-6

By trial and error we can see that one root is x=3 so (x-3) is a factor. Divide the polynomial by this factor and we get x^2-2x+2. Use the quadratic formula to find the remaining zeroes: x=(2+sqrt(4-8))/2=(2+sqrt(-4))/2=(2+2i)/2=1+i. So the three roots are 3, 1+i, 1-i. (I used synthetic division
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find the complex zeros of the polynomial function.


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3+i, 3 Lowest Degree

Question: 3+i, 3 Lowest Degree. Find a polynomial function of lowest degree with rational coefficients that has the given numbers as some of its zeros. When a polynomial has complex roots, they always come as a pair. So if x = 3+i is one root, then x = 3 - i is another root. (Two complex ro
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Finding a polynomial of a given degree with given zeros: Complex zeros

If the polynomial has an even degree, meaning that the highest power of the variable (for example: x) is an even number (for example: x^4), then all the zeroes could be complex. If odd, there must be at least one real zero. A complex zero is given by the complex expression: a+ib, where a and b ar
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form a polynomial function with real coefficients given the degree and zeros

f(x)=(x+5)(x-4-3i)(x+a+ib) represents the function with an added unknown complex factor. We need to find a and b to identify the third zero. We know that f(x)=91 when x=2, so we can write 91=7(-2-3i)(2+a+ib). That is: -13=(2+3i)(2+a+ib)=4+2a+2ib+6i+3ia-3b. There is no complex component in the
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complex, rational and real roots

The quintic function should have 5 roots.  The changes of sign (through Descartes) tell us the maximum number of positive roots. Since there are two changes of sign there is a maximum of 2 positive roots. To find the number of negative roots we negate the terms with odd powers and check
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(1, -20), (7, -20), (2, -40), (6, -40), (5, -50), (8, 10), (9, 46), (0, 7), (-1, 47), (-2, 89)

First put all the points (x,y) in order of the x value: -2 89 -1 47 0 7 1 -20 2 -40 5 -50 6 -40 7 -20 8 10 9 46   When x=0 y=7, so the constant term in the function is 7 (y intercept). There's a minimum (turning point) near x=5 because the values of y for x=2
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write a polynomial function in standard for with the given zeros x=-1,3,4

What this means, I believe, is that the function has zero value when the variable x is -1, 3 or 4. The polynomial must therefore contain the factors (x+1), (x-3) and (x-4). To find the function, f(x), we multiply these factors together. First multiply the first two factors to give x^2-2x-3, then mul
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find a polynomial function of degree four with -3 as a zero multiplicity 1,

(x+3)(x-3)(x-3)(x+2) = 0 (x^2 - 9)(x^2 - x - 6) = 0 x^4 - x^3 - 6x^2 - 9x^2 + 9x + 54 = 0 x^4 - x^3 - 15x^2 + 9x + 54 = 0
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