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I was wondering if this question is really teasing me.

$f(x)=x^2+bx+c$, where $b=-17$ and $c=52$. What is the coefficient of $x^2$ in the function $f(f(x))$?

So basically the coefficent will be 1?

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    You can just figure out what $f(f(x))$ is. It would take a minute or two to do. Then, you can see what the answer is. I don't know what the coefficient will be, but I don't think it'll be 1. 1 would be the coefficient of $x^4$, if that's what you're thinking.2012-10-15

3 Answers 3

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No. Substitute $f(x)$ in place of $x$ and expand: $f(f(x))=f(x)^2+b\cdot f(x)+c = ...$ Since now you're interested only in the coefficient of $x^2$, count only those: something like $b^2+2c\ +b$.

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$f(f(x))=(x^2+bx+c)^2+b(x^2+bx+c)+c=x^4+...+(b^2+b+2c)x^2+...$

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    Indeed, thanx...2012-10-15
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$f(f(x)) = f(x)^2 + b f(x) + c = (x^2 + bx + c)^2 + b (x^2 + bx + c) + c$.

Now $(x^2 + bx + c)^2 = (x^2 + bx+c)(x^2 + bx + c) =$

$x^4 + 2bx^3 + (2c+b^2)x^2 + 2 b c x + c^2$.

In total, $x^2$ thus has the coefficient $2c+b^2 + b$, where the last $b$ comes from the linear term in the first line.

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    this is pretty clear, will try this out2012-10-15