Find the real solution of the function

Question

f(x)=2x^3-3x^2-3x-5

Answer 1

2x^3 -3x^2 -3x -5

3 roots

by descartes' rule of signs max 1 positive real solution

and max 2 negative real solutions

either 1 + and 2 -,

or 1 + and 2 imaginary

f(0) = -5

f(1) = -9

f(2) = -7

f(3) = 54-27-9-5 = 13

one root is between 2 and 3

if there are other solutions they're both negative

f(-1) =-5

f(-2) = -31

f(<-2) = even more negative numbers

so there's only one real solution, and it's somewhere between 2 and 3

how about 2 1/2?

graph the cubic and where the curve intersects the x axis is that one solution

it looks like 2 1/2

2(2.5)^3 -3(2.5)^2 -3(2.5) -5

= 31.25 - 3(6.25) -7.5-5

=31.25 - 18.75 -12.5

= 31.25 -31.25

=0 Bingo

x=2.5 = 2 1/2 is the one and only real solution

divide x-2.5 into the cubic you get a quadratic with no real solution, just 2 imaginary solutions