<s>[INST] Unsure how to solve this quadratic formula problem and how to expand and substitute into the formula Find all values of a and b that will make $(ax+5)^2 = 16x^2 + bx + 25$ true for all $x$.
I believe that this is an equation that needs to be solved using the quadratic formula but am not sure how to expand and substitute into the formula. Any pointers/tips would be helpful.
 [/INST] The quadratic formula is not needed here
\begin{align*}
(ax+5)^2 &= 16x^2 + bx + 25\\
\implies
b &= a^2 x- 16 x + 10 a \\
\implies  \mathbf{(ax+5)^2} &= 16x^2 + \bigg( a^2 x- 16 x + 10 a \bigg)x + 25\\
&=\mathbf{(ax+5)^2} 
\end{align*}
We can see from
$\quad (ax+5)^2=(ax+5)^2\quad $
how all values of $\space x\space$ require that $\quad  b = a^2 x- 16 x + 10 a  $.
On the other hand, expanding both sides and comparing coefficients,
$$a^2 x^2 + 10 a x + 25
=16x^2 + bx + 25\\
\implies\quad a=\pm4, \quad b=10a=\pm40$$
these results are easily verified in a spreadsheet. This is also confirmed by
\begin{align*} b &= a^2 x- 16 x + 10 a\\
&= 4^2 x- 16 x + 10\cdot 4\\
 &=40
\end{align*}
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