# How to solve the equation 4x^4 + 5x^3 + 2x^2 + 5x + 4=0?

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When you graph the equation, it hits -1 twice on the x axis; therefore, -1 is a root.

-1 / 4 5 2 5 4

4 1 1 4 0 use synthetic division

-1/ 4 1 1 4

4 -3 4 0 you get 4x^2-3x + 4 now use the quadratic equation

3 +- square root of 9 - 4(4)(4) all over 2(4)

3+- square root of -55 all over 8

Final answer: 3 +- i squareroot 55 all over 8

Being an even reciprocal equation, x=-1 is one if it's four roots.

We'll apply the following algorithm, to find out the other roots. The calculus will be made with the equation's coefficients:

(-1)*4+5=**1**

(-1)***1**+2=**1**

(-1)***1**+5=**4**

(-1)***4**+4=**0, when the final result is 0, that certifies "-1" as root.**

We've obtained an equation of third degree:

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

We'll apply again the same algorithm:

(-1)*4+1=**-3**

(-1)***-3**+1=**4**

(-1)***4+**4**=0, the second root will be again "-1"**

The equation obtained with Horner's algorithm is:

4x^2-3x+4=0

In order to find out the following 2 roots, we'll calculate delta:

delta=(-3)^2-4*4*4=9-64=55

**x3=(-(-3)+sqrt 55)/2*4=(3+sqrt 55)/8**

**x4=(3-sqrt 55)/8**