How do you figure out how many real number solutions x^4+4x^3-8x^2+2 has?I am learning calculus. I can't figure out how to find out even 1...

In this case Rolle's string can be very helpful. To create
this string, you need to determine the zeroes of the 1st derivative of
function:

f'(x) = 4x^3 + 12x^2 -
16x

Now, we'll cancel it:

4x^3
+ 12x^2 - 16x = 0

We'll divide by
4:

x^3 + 3x^2 - 4x = 0

We'll
factorize by x:

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

We'll cancel each factor:

x
= 0

x^2 + 3x - 4 = 0

x^2 + 3x
- 3 - 1 = 0

(x^2 - 1) + 3(x-1) =
0

(x-1)(x+1) + 3(x-1) =
0

(x-1)(x + 1 + 3) = 0

x - 1 =
0 => x1 =1

x + 4 = 0 => x2 =
-4

The roots of the 1st dericative are -4,0 and
1.

Now, we'll create the Rolle's string. The important
elements of the string are the values of the function at the x = -4 ; x = 0 and x =
1

f(-4) = (-4)^4 + 4*(-4)^3 - 8*(-4)^2 +
2

f(-4) = 256 - 256 - 128 +
2

f(-4) = -126

f(0) =
2

f(1) = 1 + 4 - 8 + 2

f(1) =
-1

We'll arrange the zeroes of derivative on x
axis.

`-oo` -4       0    1    
+`oo`

-`oo` f(-4)  f(0)  f(1)  +
`oo`

-126     2     -1

If
there are changes in sign at two consecutive values of the function, that means that
between -126 and 2, there is a value of x for f(x) = 0. That means that the graph of
function will intersect x axis between -4 and 0.

We notice
another change of sign between 0 and 1. There are changes in sign between -`oo` and -4
and between 1 and +`oo` .

Therefore, all the
4 roots of the function are real roots and they belong to the next 4
intervals:

(-`oo` ; -4) ; (-4
; 0) ; (0 ; 1) ; (1 ; `oo` ).