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Where could I possibly have gone wrong?!

Mathematics
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Consider all monic quadratic polynomials \(f(x)\) satisfying the following relation:\[f(f(-1)) = f(f(0)) = 0\]Find the sum of \(f(3)\) across all such polynomials. Here's my foolproof solution:
OK, so let \(f(x) = x^2 + bx + c \) for a second. Now \(f(0) = c \Rightarrow f(f(0)) = f(c) = 0\). This means that \(c\) is a root of this quadratic! Now the other root must be \(1\) so that the product of roots becomes \(c\). Therefore, so far, \(f(x) = (x-1)(x-c)\). We use the second relation to find \(c\). As you can clearly see, \(f(-1) = 2c + 2\) meaning that \(f(2c + 2) = 0\). Thus, \(2c+2\) is also a root of the polynomial - but \(2c+2\) must either be \(c\) or \(1\). We consider both possibilities: 1. \(2c + 2 = 1\Rightarrow c = -1/2\) 2. \(2c + 2 = c\Rightarrow c = -2\) Thus, the two possible quadratics are \((x-1)(x+1/2)\) and \((x-1)(x+2)\).

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The answer comes out to be \(2\cdot 3.5 + 2\cdot 5 = \boxed{17}\). Which isn't the answer.
it looks very good to me! not able to find any mistakes
\[f(x) = x^2 + x\]also seems to satisfy this.
I didn't consider the case \(c=0\) :)
If \(c=0\), then the polynomial is in the form \(f(x) = x^2 + bx\) so \(f(0) = 0\) and so \(f(f(-1))=0 \Rightarrow f(1-b) = 0 \Rightarrow (1-b)^2 + b(1 - b) = 0 \).\[\Rightarrow b^2 - 2b + 1 + b - b^2 = 0\]\[\Rightarrow b = 1\]
\[f(3) = 9 + 3 = 12 \]so\[s = 29\]
Thanks guys - I'm done.
good job :)
Nice!

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