If \((ax+1)(bx+3)=6x^2+cx+3 \) for all values of \(x\), and \(a+b=5\), what are the two possible values for \(c\) ?
Expand the expression on the left side first:
$$ (ax+1)(bx+3) $$
$$ abx^2+bx+3ax+3 $$
Let's compare this with the equation on the other side, taking care to identify corresponding parts:
$$ \colorbox{aqua}{$abx^2$}+\colorbox{yellow}{$bx+3ax$}+\colorbox{chartreuse}{$3$} = \colorbox{aqua}{$6x^2$}+\colorbox{yellow}{$cx$}+\colorbox{chartreuse}{$3$} $$
From the above, we get the following equations:
From the \(x^2\) terms
$$ abx^2=6x^2 $$
$$ ab=6 $$
From the linear terms
$$ bx+3ax=cx $$
$$ b+3a=c $$
We are given that \(a+b=5\), which we can use along with \(ab=6\) to identify possible values for \(a\) and \(b\). We can solve the system using tradional methods, or use a bit of mental math.
$$ a \text{ and } b = 3 \text{ or } 2 $$
\(c\) can have two answers, based on the values we assign \(a\) and \(b\).
$$ c=b+3a $$
\(a=2, b=3\)
$$ c=3+3(2) $$
$$ c=9 $$
\(a=3, b=2\)
$$ c=2+3(3) $$
$$ c=11 $$
$$ c=\boxed{9 \text{ and } 11} $$