All the roots of $(x^2+1)^2 = x(3x^2+4x+3)$

Question

Find all the roots of the equation : $$(x^2+1)^2 = x(3x^2+4x+3)$$How do we find the roots in polynomials of degree > 2 ??

Also, In odd degree polynomials I use Descartes rule of signs to predict the number of real roots. Here, however, it does not give me any clue about the number of real/imaginary roots. Is there a better method?

Answer 1

On rearrangement we have $x^4-3x^3-2x^2-3x+1=0$

Like this, divide either sides by $x^2$ as $x\ne0$ to get $$x^2+\frac1{x^2}-3\left(x+\frac1x\right)-2=0$$

Or, $$\left(x+\frac1x\right)^2-2-3\left(x+\frac1x\right)-2=0$$

Put $x+\frac1x=u$

Answer 2

Hint: $$(x^2+1)^2-3x(x^2+1)-4x^2=0$$ $$(x^2+1-4x)(x^2+1+x)=0$$

Answer 3

Another solution:

Solve the equation $(x^2+1)^2 = 3x(x^2+1)+4x^2$ with $x^2+1$ as an unknown: $$ x^2+1=\frac{3x\pm \sqrt{9x^2+16x^2}}{2}=\frac{3x\pm 5x}{2} $$ etc.