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