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

use google you idiot omg you're so annoying

step-by-step explanation:

x > 1.3

step by step:

8.5−0.9x> 9.8−1.9x

step 1: simplify both sides of the inequality.

−0.9x+8.5> −1.9x+9.8

step 2: add 1.9x to both sides.

−0.9x+8.5+1.9x> −1.9x+9.8+1.9x

1x+8.5> 9.8

step 3: subtract 8.5 from both sides.

1x+8.5−8.5> 9.8−8.5

1x> 1.3

step 4: divide both sides by 1.

1x/1 > 1.3/1