Question

How many zeros does the function f(x)=-3x^5+2x^3+7x-5 has?

Answer 1

f(x)=-3x^5+2x^3+7x-5
0 =-3x^5+2x^3+7x-5

factor out an x
0 =x(-3x^4+2x^2+7)-5

now do you see how the expression in the ( ) is reducable into a quadratic?

let x^2 = u

Answer 2

5...tha max.power of variable gives no of zeross..oky..?

Answer 3

the solution is 3, not 5, by the way

Answer 4

I was thinking 4.. but am not absolutely sure

Answer 5

0 =x(-3x^4+2x^2+7)-5

u = x^2
0 =x (-3u^2+2u+7)-5

Answer 6

there are 3 real solutions, but also there are 2 imaginary solutions

Answer 7

but the total of 5 sol to which am talking abut.,,,

Answer 8

yes yes, you are correct zaibali.qasmi, i was wrong

Answer 9

I only considered reals solutions as possibilities for zeros