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.
On
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*}
If its true for all $x$, then its true for $x=1$ and $x=-1$. Evaluate those values of $x$, then sum the resulting equations and you will be able to solve for $a$. Then substitute the value of $a$ to solve for $b$.