Problem #25 in the calculator-allowed section of Practice Test #4, please.

This is a fun question! First, note that the polynomials provided have a little something in common:

    \begin{align*}f(x)&=2x^3+6x^2+4x\\g(x)&=x^2+3x+2\end{align*}

Do you see it? How about now:

    \begin{align*}f(x)&=2x(x^2+3x+2)\\g(x)&=x^2+3x+2\\f(x)&=2x[g(x)]\end{align*}

Takeaway lesson from that: when a question at first looks like it might take a very long time to do (in this case, when it looks at first like you’re going to have to do long division possibly four times) slow down for a minute and think about whether there’s a shortcut built into the problem.

You’re looking for a way to combine f(x) and g(x) so that they’re divisible by 2x+3. Once you see that you can pull a 2x (as in, the same 2x that’s part of 2x+3) out of f(x) and get g(x), you know to use that as your base of operations. Let’s just see what happens if we multiply g(x) by 2x+3:

    \begin{align*}&(2x+3)g(x)\\=&2x[g(x)]+3g(x)\end{align*}

Ready for the last step? All we need to do is substitute because we know f(x)=2x\times g(x):

    \begin{align*}&(2x+3)g(x)\\=&2x[g(x)]+3g(x)\\=&f(x)+3g(x)\end{align*}

That matches choice B, and we know that it’s divisible by 2x+3 because we made it by multiplying by 2x+3!

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