# Coefficent of an equation

I am suppose to find the coefficent of x^2 in this equation

after doing the calculation I ended up with this

but that is the wrong answer what is that I am missing?

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You only want the coefficient of $x^2$. So don't bother with the rest. And keep it letters as long as possible, it is a lot easier. –  André Nicolas Oct 17 '12 at 2:35
tks and congratutlations btw –  JackyBoi Oct 17 '12 at 2:52

$$f(f(x))=(x^2-23x+15)^2-23(x^2-23x+15)+15$$

Now instead of expanding, think about where we are getting $x^2$ from:

$\bullet$ First bracket: $-23x\cdot -23x=23^2 x^2$ and $15\cdot x^2$ twice

$\bullet$ Second bracket: $-23x^2$

So the coefficient will be $23^2+2\cdot 15-23=536$

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You have $f(x)=x^2+bx+c$, and you want $f\big(f(x)\big)$. As you say, this is $$\left(x^2+bx+c\right)^2+b\left(x^2+bx+c\right)+c\;.\tag{1}$$

Now

\begin{align*} \left(x^2+bx+c\right)^2=x^4+2bx^3+\left(b^2+2c\right)x^2+2bcx+c^2\;, \end{align*}

so $(1)$ becomes

$$x^4+2bx^3+\left(b^2+2c+b\right)x^2+\left(2bc+b^2\right)x+c^2+bc+c\;,$$

in which the coefficient of $x^2$ is $b^2+2c+b$. If $b=-23$ and $c=15$, this is

$$(-23)^2+2(15)-23=529+30-23=536\;.$$

I did it this way to follow (roughly) your approach. Note, however, that you don’t actually have to multiply everything out: you just need the $x^2$ term from $\left(x^2+bx+c\right)^2$ and the $x^2$ term from $b\left(x^2+bx+c\right)$. The latter is clearly $bx^2$. The former can only come from the terms $\left(x^2\right)(c)$, $(bx)(bx)$, and $(c)\left(x^2\right)$ in $\left(x^2+bx+c\right)^2$, so the $x^2$ term here is $\left(b^2+2c\right)x^2$. Now just add the coefficients to get $b^2+2c+b$, and do the arithmetic as before.

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Didn't quite catch what you were doing in the first few steps but the latter explanation is crystal clear on what I have to do to tackle this question , +1 for the good explanation tks. –  JackyBoi Oct 17 '12 at 3:05

There are a few places where you've doubled things instead of squaring them, in the fifth and sixth lines down.

Specifically, remember that $bx\times bx=b^2x^2,$ not $2bx^2$, $c\times c=c^2,$ not $2c$, and $b\times bx=b^2x,$ not $2bx$. Everything else looks good.

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It is actually a bit less confusing (you don't have to expand the brackets with three terms) if you write $$f(f(x))=(x^2+bx+c)^2+b(x^2+bx+c)+c=(x^2+bx+c)(x^2+bx+c+(b+c)) =x^4+bx^3+(c+b)x^2+bx^3+b^2 x^2+b(c+b)x+cx^2+bcx+c(b+c)+c$$ The it is clear tha the coefficient at $x^2$ is $c+b+b^2+c=2 \cdot 15 -23+23^2$

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