Question

Solve the following equations which have equal roots:
$x^6-3x^5+6x^3-3x^2-3x+2=0$.

Answer 1

Give equation, $x^6-3x^5+6x^3-3x^2-3x+2=0$

Consider $f\left(x\right)=x^6-3x^5+6x^3-3x^2-3x+2$

$\therefore f^{\prime }\left(x\right)=3(2x^5-5x^4+6x^2-2x-1)$

Now, HCF of $f\left(x\right)$ and $f^{\prime }\left(x\right)$ is $(x-1)(x+1)$. Hence $x=1$ and $x=-1$ are double roots of $f\left(x\right)=0$

$f\left(x\right)$ can be factored as $(x-1{)}^2(x+1{)}^2(x^2-3x+2)\text{or}(x-1{)}^3(x+1{)}^2(x-2)$

$\therefore$ Roots of the given equation are $-1,-1,1,1,1,2$.