Question

Solve the equation $2x^5+x^4-12x^3-12x^2+x+2=0$

Answer 1

$2x^5+x^4-12x^3-12x^2+x+2=0$

If we look at the coefficients, they are $2,1,-12,-12,1,2$

Since they are the same for $x^5$ and $x^0,x^1$ and $x^4,x^2$ and $x^3$, we conclude that $x+1$ is a factor of the polynomial.

$\therefore (x+1)(2x^4-x^3-11x^2-x+2)=0$

Now, we look at $2x^4-x^3-11x^2-x+2=0.$

Dividing throughout by $x^2$, we have $2x^2+\frac{2}{x^2}-x-\frac{1}{x}-11=0$

Substituting $x+\frac{1}{x}=a$, we have

$2(a^2-2)-a-11=0$

$\therefore 2a^2-a-15=0$

$\therefore 2a^2-6a+5a-15=0$

$\therefore (2a+5)(a-3)=0$

Resubstituting $a=x+\frac{1}{x}$ and multiplying by $x^2$, we have

$(2x^2+5x+2)(x^2-3x+1)=0$

$\Rightarrow 2x^5+x^4-12x^3-12x^2+x+2=(x+1)(2x^2+5x+2)(x^2-3x+1)=0$

$\therefore (x+1)(2x+1)(x+2)(x^2-3x+1)=0$

$\therefore x=-1,-\frac{1}{2},-2,\frac{3+\sqrt{5}}{2},\frac{3-\sqrt{5}}{2}$