How do you write $4(x +1)^2 + 1$ in the form $(ax +b)^2 + c\;$? We can write $4x^2+8x+5$ in the form $a(x+b)^2+c$ as $4(x+1)^2+1$. However, the question I am doing asks me to write it in the form $(ax+b)^2+c$. How do I change it to that form?

You can also solve the equation $(ax+b)^2+c\equiv 4x^2+8x+15$ (the $\equiv$ sign means "true for all values of $x$"—the coefficients on both sides of the equation have to be the same). If we expand the left-hand side, we get
$$
(ax+b)^2+c\equiv a^2x^2+2abx+b^2+c\equiv 4x^2+8x+5 \, .
$$
This means that $a^2=4$ and $2ab=8$ and $b^2+c=15$. If $a^2=4$ then $a=2$ or $a=-2$. Suppose that $a=2$. Then, $4b=8$, and so $b=2$. Finally, $4+c=5$, and so $c=1$. This gives us the solution
$$
(2x+2)^2+1\equiv 4x^2+8x+5 \, .
$$
If $a=-2$, then $-4b=8$, and so $b=-2$. Finally, $4+c=5$, and so $c=1$. This gives us the solution
$$
(-2x-2)^2+1 \equiv  4x^2+8x+5 \, .
$$
However, the first solution is arguably simpler because it doesn't have any minus signs in it.
