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So I got the question $a^2(y-3)+a(3-y)+7(y-3)$ on my homework but I'm not sure how to start to solve it. Do I group the two $(y-3)s$ together or factor a $-1$ out of $(3-7)$ or something entirely different? Some pointers in the right direction would be greatly appreciated.

Thank you in advance!

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    Take out common factor $y-3$ to get $$a^2-a+7$$ – 2017-02-15
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    But what about the (3−y) left over? – 2017-02-15
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    Just replace $(3-y)$ by $-(y-3)$. See my answer – 2017-02-15

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Note that $3-y=-(y-3)$ and so $$\begin{align}a^2(y-3)+a(3-y)+7(y-3)&=a^2(y-3)-a(y-3)+7(y-3)\\ &=(a^2-a+7)(y-3) \end{align}$$

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    Thank you so much! This makes a lot more sense now. – 2017-02-15