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Use the following polynomial equation to do the following

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Use the following polynomial f(x)=5x^3+8x^2-4x+3

a. the fundemental theorem of algebra states that this polynomial has ____ roots.

b. find f(-x).

c. use descartes' rule of signs to complete the table:
number of positive real zeros, number of negative real zeros, number of imaginary zeros

d. use the rational root theorem to determine the possible rational roots of f(x).

e. of the possible rational roots above, which ones are roots?
asked Aug 25, 2014 in PRECALCULUS by swatttts Pupil

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Given  f(x)=5x^3+8x^2-4x+3 

(a).The fundamental theorem of algebra shows that any non-zero polynomial has a number of roots at most equal to its degree  and hence the given polynomial has 3 roots



So there is only 1 sign change in it so therfore there will always be one NEGATIVE root because you can't subract 2 from it because then there would be -1 negative roots and that's not possible. From there you can find the imaginary roots: 

If there are 2 positive and 1 negative there are 0 imaginary 

If there are 0 positive and 1 negative there are 2 imaginary 



answered Aug 25, 2014 by anonymous
selected Aug 25, 2014 by swatttts
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Identify Rational Zeros  

Usually it is not practical to test all possible zeros of a polynomial function using only synthetic substitution. The Rational Zero Theorem can be used for finding the some possible zeros to test.

d. 5x3+8x2-4x+3

If p/q is a rational zero, then p is a factor of 3 and q is a factor of 5.

The possible values of p are ± 1, ±3. The possible values for q are ± 1, ± 5.

So, possible rational roots are, p/q = ±1, ±3, ±1/5, ±3/5.

answered Aug 25, 2014 by anonymous

Rational Root Theorem, if a rational number in simplest form p/q is a root of the polynomial equation anxn + an  1xn – 1 + ... + a1x + a0 = 0, then p is a factor of a0 and q is a factor if an.

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Make a table for the synthetic division and test possible zeros.

Non of the possible rational roots giving zero remainder, so these are not the roots of the polynomial.

answered Aug 25, 2014 by anonymous
edited Aug 25, 2014

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