Question

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

Accepted Answer

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

Answer

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

Answer

Let p(x) =4x^4+5x^3+2x^2+5x+4 = 0
Put  y= 1/x, then the equation becomes 4/y^4+5/y^3+2/y^2+5/y+4 =0 Or
4+y+y^2+5y^3+4y^4 = 0.
So P(1/x) = 0. So the given euation is areciprocal equation
Also P(-1) = 4-5+2-5+4 = 0
Therefore -1 is a root. Since also 1/-1 =-1 is also a root . As this is a reciprocal equation.
Therefore  (x- -1)^2  = (x+1)^2 is a factor.
So 4x^4+5x^3+2x^2+5x+1 = (x+1)^2 { 4x^2+kx+4}, where we chose the coefficient of x^2 and constant term = 4 in order to make  coeeficients of x^4 and constant term agree with the left.
So equating the x's on both sides:
5x = 2x*4+1*k . Or k = -3.
Therfore the other factor is 4x^2-3x+4 = 0
x = {3+or-sqrt(9-64)}/2*4 = {3+ or- sqrt(-55)}/8
So x= -1 , -1, (3+2sqrt(-14))/8 , (3-sqrt(-55))/8

Answer

We notice that it is an even reciprocal equation, where x=-1 is one of 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

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

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