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Calculus1 9 Online
OpenStudy (elenathehomeschooler):

For the polynomial F(x)= x^5-2x^4+8x^2-13x+6 answer the following: a.) How many zeroes does the function have over the set of complex numbers? b.) What is the maximum number of local extrema (maxima or minima) the graph of the function can have? c.) List the possible rational zeroes of this function. D. ) Factor this polynomail completely over the set of complex numbers.

OpenStudy (anonymous):

The number of zeroes is the same as the degree. Do you know what the degree is for that polynomial?

OpenStudy (elenathehomeschooler):

is it 5?

OpenStudy (anonymous):

yes.

OpenStudy (anonymous):

The maximum number of extrema is one less than the degree.

OpenStudy (elenathehomeschooler):

so it would be 4

OpenStudy (anonymous):

yes. for the possible rational zeros, divide the factors of the constant by the factors of the leading coefficient. The constant is 6 and the leading coefficient is one, so you need to do \[\frac{ factors~of~6 }{ factors~of~1}\]

OpenStudy (elenathehomeschooler):

so would it be \[\pm1 \pm2 \pm3 \pm6\]

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

do you know how to factor it?

OpenStudy (elenathehomeschooler):

no

OpenStudy (anonymous):

you have to test the possible zeros you found in part c using either long or synthetic division. That should give you the real zeros. You can graph it to pick them out instead of guessing, but I'd imagine you have to show your work.

OpenStudy (elenathehomeschooler):

ok

OpenStudy (anonymous):

do you have any of the real zeros?

OpenStudy (elenathehomeschooler):

hang on i messed up

OpenStudy (whpalmer4):

another way to test a possible root is by evaluating the polynomial at that value. if \(a\) is a root of \(P(x)\), then \(P(a) = 0\) You can use Horner's rule to evaluate polynomials quickly.

OpenStudy (anonymous):

@whpalmer4 I completely forgot about that test.

OpenStudy (elenathehomeschooler):

-2

OpenStudy (anonymous):

yes, so one of your factors is (x + 2).

OpenStudy (elenathehomeschooler):

1 as well

OpenStudy (anonymous):

yes. and and 1 is a double root, so factor so far are (x - 1)²(x + 2). Use the quadratic formula on what's left from the division to get the complex roots.

OpenStudy (elenathehomeschooler):

can you show me how to do that?

OpenStudy (anonymous):

ok. what did you have left after all the division?

OpenStudy (elenathehomeschooler):

what do yuo mean?

OpenStudy (anonymous):

Every time you test the polynomial and find a root, you divide by the factor. Then divide the result by the next factor, and so on |dw:1452897033359:dw|

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