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?
 A: 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 $b\in\Bbb R$, $b^2+2b+2=(b+1)^2+1>0$.
A: If $x=1$ is a root of $P(x) = x^3 + 2bx^2 - a^2x - b^2$, then
\begin{align}
   P(1) &= 0 \\
   1+2b-a^2-b^2 &= 0 \\
   a^2 &= 1 + 2b - b^2
\end{align}
So
\begin{align}
   P(x) &= x^3 + 2bx^2 + (b^2-2b-1)x - b^2 \\
        &= (x-1)(x^2+(2b+1)x + b^2)
\end{align}
If $x=1$ is a double root of $P(x)$, then
\begin{align}
   (x^2+(2b+1)x + b^2)|_{x=1} &= 0 \\
   1 + (2b+1) + b^2 &= 0 \\
   (b+1)^2 + 1 &= 0
\end{align}
Which is impossible.

OR, if you do synthetic division by $x-1$ on $P(x)$, you get
\begin{array}{rrrrrrr}
   \phantom{1 }&|1 & 2b & -a^2 & - b^2 \\
            1  &|0 & 1 & 2b+1 & -a^2+2b+1\\
\hline
   \phantom{1 }& 1 & 2b+1 & -a^2+2b+1 & -a^2-b^2+2b+1\\
\end{array}
Which means $P(x) = (x-1)(x^2 +(2b+1)x +(-a^2+2b+1)) + (-a^2-b^2+2b+1)$
And so $P(1) = 0$ implies $-a^2-b^2+2b+1=0$
and so on
