How many complex zeros does the polynomial function have? f(x)=−3x^6−2x^4+5x+6

one way would be to factor

I can't factor it so we will have to use Descartes' Rule of Signs which is helpful for finding how many real roots you have

it goes like this:

for a polynomial with real coefients, consider .

after arranging the terms in decending order in terms of degree, count how many times the signs of the coeffients change direction and minus 2 from that number until you get to 1 or 0. that will be the number of even roots the function can have

We have (-, -, +, +). the signs changed 1 times, so it has 1 real positive root

to get the negative roots, we evaluate f(-x) and see how many times the root changes

signs are (-, -, -, +). there was 1 change in sign

so the function has 1 real negative root

a total of 2 real roots

a function of degree can have at most, roots

our function is degree 6 so it has 6 roots

if 2 are real, then the others must be complex

6-2=4 so there are 4 complex roots

you can also show that there are only 2 real roots by using a graphing utility to see that there are only 2 real roots

v = 1/3 a^2h

step-by-step explanation:

if the height is 57.7cm and the height of the slant edge is 18.2cm, then the volume is 1/3 a^2h.

$3,200 x 7.25% = $3,200 x .0725 = $232 interest monthly

$232 monthly x 12 annual = $2, 784

step-by-step explanation: