Roots of polynomials with repeated roots Let $a$ and $b$ be real numbers. Consider the cubic equation $$x^3+2bx^2-ax^2-b^2=0$$
(i) Show that if $x=1$ is a solution of the cubic then $$ -1+\sqrt{2}\leq b\leq1+\sqrt{2} $$
(ii) Show that there is no value of $b$ for which $x=1$ is a repeated root of the cubic
I need some assistance on (ii). In order to show that there is no value of $b$ for which $x=1$ is a repeated root of the cubic. I attempted to factorise the cubic. If $x=1$ is a solution, then the cubic may be expressed as follows with no remainder. $x^3+2bx^2-a^2x-b^2 \equiv (x-1)(Ax^2+Bx+C)$. Comparing coefficients, $A = 1, C = b^2, B = a^2+b^2$ $\therefore (x-1)(x^2+(a^2+b^2)x+b^2)$. Finally, if we sub $x=1$ into the resulting quadratic, we get $1+a^2+b^2+b^2 > 0$ for all real values of $a,b$. Therefore, $x=1$ cannot be a repeated root.
I have been told that $x^2+(a^2+b^2)x+b^2$ is not the quadratic equation for this and in fact it's $x^2+(2b+1)x+b^2$
$(x-1)(x^2+(2b+1)x+b^2) = x^3+2bx^2+(b^2-2b-1)x-b^2$. The coefficient of the $x$ term is correct, but the $b^2-2b-1 \not= -a^2$ therefore this cannot be right aswell.
My solution:
$(x-1)(x^2+(a^2+b^2)x+b^2) = x^3+(a^2+b^2-1)x^2-a^2x-b^2$
So my solution seems to get the correct coefficient for $x$ term, but the incorrect $a^2+b^2-1$ coefficient for the $x^2$ term.
What is going on here? Has math been broken?

By polynomial long division, you get that$$x^3+2bx^2-ax^2-b^2=(x-1)^2(x+2b+2)+(-a^2+4b+3)x-(b^2+2b+2).$$So, if $1$ was a repeated root of $x^3+2bx^2-ax^2-b^2$, that is, if$$(x-1)^2\mid x^3+2bx^2-ax^2-b^2,$$the polynomial $(-a^2+4b+3)x-(b^2+2b+2)$ would be the null polynomial. But, for every $b\in\Bbb R$, $b^2+2b+2=(b+1)^2+1>0$.
