Factor \(t^3-5t^2-3t+15\text{.}\) This example has a complication with negative signs. If we try to break up this polynomial into two groups as \(\left(t^3-5t^2\right)-(3t+15)\text{,}\) then we've made an error! In that last expression, we are subtracting a group with the term \(15\text{,}\) so overall it subtracts \(15\text{.}\) The original polynomial added \(15\text{,}\) so we are off course.

One way to handle this is to treat subtraction as addition of a negative:

\begin{align*} t^3-5t^2-3t+15\amp=t^3-5t^2+(-3t)+15\\ \amp=\left(t^3-5t^2\right)+\left(-3t+15\right)\\ \end{align*}

Now we can proceed to factor out common factors from each group. Since the second group leads with a negative coefficient, we'll factor out \(-3\text{.}\) This will result in the “\({}+15\)” becoming “\({}-5\)”.

\begin{align*} \amp=\highlight{t^2}(t-5)+\highlight{(-3)}(t-5)\\ \amp=\highlight{t^2}\attention{\overbrace{(t-5)}}\highlight{{}-3}\attention{\overbrace{(t-5)}}\\ \amp=(t-5)\highlight{\left(t^2-3\right)} \end{align*}

And remember that we can confirm this is correct by multiplying it out. If we made no mistakes, it should result in the original \(t^3-5t^2-3t+15\text{.}\)