# 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 solution. Is it because there aren't any, or is it...

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 solution. Is it because there aren't any, or is it something I'm just not understanding? Please explain!

### 3 Answers | Add Yours

First graph the equation:

You can see there are 4 real solutions because the curve intersects the real x axis at 4 points. Somewhere around -5.5, -0.5, 0.5 and 1.2. You can use Newton's method to find these roots:

`x_(i+1) = x_i - (f(x_i))/(f'(x_i)) = x_i - (x_i^4+4x_i^3-8x_i^2+2)/(4x_i^3+12x_i^2-16x_i)`

trying `x_0 = -5.5` we do 4 interations to get

`x_1= -5. 4554196` , `x_2= -5. 4543853` , `x_3= -5. 4543847` ,

`x_4= -5. 4543847`

So first root is `~~` -5.4543847

trying `x_0 = -0.5` we do 4 interations to get

`x_1= -0.45833333, x_2= -0.4560571, x_3= -0.45605037`

` x_4= -0.45605037`

So our second root is ` ``~~` -0.45605037

You can work out the other two x=0.62595760, x=1. 2844775, along with x=-5.4543847 and x=-0.45605037 are the solutions to the above equation.

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` ).**

Thank you for your help! I understand it now!