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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), then the other complex root is (a - ib) since you have x = -3i, then you must also have x = 3i. Your lowest polynomial will have three roots, so three factors, (x - 3i), (x + 3i) and (x - 5), giving a cubic function. The function then is. (x - 3i)(x + 3i)( x - 5) = (x^2 - (3i)^2)(x - 5 = (x^2 + 9)(x - 5) = x^3 + 9x - 5x^2 - 45 = f(x) = x^3 - 5x^2 + 9x - 45
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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 to divide the known factor into the polynomial, but algebraic division could also have been used.)
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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 roots will always be of the form a + ib, a - ib. ) So we have at least three roots: x = 3, x = 3 + i, x = 3 - i. Creating the polynomial from these three roots. (x - 3)(x - (3 + i))(x - (3 - i)) = 0 (x - 3)(x^2 - 3x + 3i - 3x - 3i + 9 + 1) = 0 (x - 3)(x^2 - 6x + 10) = 0 x^3 - 6x^2 + 10x - 3x^2 + 18x - 30 = 0 x^3 - 9x^2 + 28x - 30 = 0
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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 are real, and i=sqrt(-1). If the polynomial has real coefficients (all the numbers in at are, or represent, real numbers), as is usually the case, then the imaginary part of the complex zeroes must combine to make a real number. For example, if one complex zero is identified as a+ib, then there will be another zero a-ib such that (x-a-ib)(x-a+ib)=x^2-2ax+a^2+b^2. (Note that, although a and b are real, they are not necessarily rational numbers.) There are no imaginary components in this expansion. Example: x^4+x^3-x-1 is degree 4 polynomial. It has real zeroes 1 and -1 so that it factorises to (x-1)(x+1)(x^2+x+1). The zeroes of x^2+x+1 are found by setting the quadratic to zero and solving using the formula; so x=(-1+sqrt(1-4))/2=(-1+sqrt(-3))/2=(-1+isqrt(3))/2. The two complex roots are -1/2-isqrt(3)/2 and -1/2+isqrt(3)/2. In this case a=-1/2 and b=sqrt(3)/2. Note that a^2+b^2=1. Example: you're told that a degree 5 polynomial has a real zero 1, and two complex zeroes 1+i and -2-i. Find all zeroes and the polynomial. The two missing zeroes will be 1-i and -2+i. The polynomial is (x-1)(x-1-i)(x-1+i)(x+2+i)(x+2-i)=(x-1)(x^2-2x+2)(x^2+4x+5)=x^5+x^4-3x^3-x^2+12x-10. Strictly speaking, the polynomial is a(x^5+x^4-3x^3-x^2+12x-10), where a is a real number, because multiplying a polynomial by a constant doesn't affect its zeroes.
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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 number 91, so we can write 4+2a-3b=-13 and 2b+6+3a=0. The first of these equations gives b in terms of a: b=(4+2a+13)/3, and this can be substituted into the second equation: (8+4a+26)/3+6+3a=0, or 4a+34+18+9a=0. So 13a=-52, making a=-4 and b=3 by substituting for a in either equation. f(x) then becomes the cubic (n, the order, is 3): f(x)=(x+5)(x-4-3i)(x-4+3i)=(x+5)((x-4)^2+9)=(x+5)(x^2-8x+25). So, f(x)=x^3-3x^2-15x+125.
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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 for sign changes: that means that there is at most one negative root, because there's only one sign change between the first and second terms. Complex roots always come in matching or complementary pairs, so that means in this case 2 or 4.  Put x=-1: function is positive; for x=-2 function is negative, so there is a real root between -1 and -2, because the x axis must be intercepted between -1 and -2. This fulfils the maximum for negative roots. That leaves 4 more roots. They could be all complex; there could be two complex and two positive roots. Therefore there are at least two complex roots. To go deeper we can look at calculus and a graph of the function: The gradient of the function is 10x^4-5. When this is zero there is a turning point: x^4=1/2. If we differentiate again we get 40x^3. When x is negative this value is negative so the turning point is a maximum when we take the negative fourth root of 1/2; at the positive fourth root of 1/2 the turning point is a minimum, and these are irrational numbers. The value of the function at these turning points is positive. The fourth root of 1/2 is about 0.84. and once the graph has crossed the x axis between -1 and -2, it stays positive, so all other roots must be complex. The function is 12 when x=0, so we can now see the behaviour: from negative values of x, the function intersects the x axis between -1 and -2 (real root); at x=4th root of 1/2 (-0.84) it reaches a local maximum, intercepts the vertical axis at 12 until it reaches 4th root of 1/2 (0.84) and a local minimum, after which it ascends rapidly at a steep gradient.  [Incidentally, one way to find the real root is to rearrange the equation: x^5=2.5x-6=-6(1-5x/12); x=-(6(1-5x/12))^(1/5)=-6^(1/5)(1-5x/12)^(1/5) We can now use an iterative process to find x. We start with x=0, so x0, the first approximation of the solution, is the fifth root of 6 negated=-1.430969 approx. To find the next solution x1, we put x=x0 and repeat the process to get -1.571266. We keep repeating the process until we get the accuracy we need, or the calculator reaches a fixed value. After just a few iterations, my calculator gave me -1.583590704.]
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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 and 6 (-40) are more positive than -50. The gradient (dy/dx) gets positively steeper as x increases after x=5, and negatively steeper between x=-2 and +2. The degree of the polynomial is even because the function is positive for larger magnitude positive and negative values of x. The zeroes are between x=0 and 1 and between x=7 and 8. The function can be split into two parts: (1) y for -2 Read More: ...

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 multiply by the third factor, x^3-2x^2-3x-4x^2+8x+12. Gather similar terms together and we get x^3-6x^2+5x+12. Check the answer by substituting the three values of x for which f(x)=0. Other polynomials are possible. The answer given is the simplest. By multiplying the function by a constant and/or repeats of the discovered factors, more polynomials can be created with the same number of zeroes.
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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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