Example  4
Factor \(2x^2 - x - 15 \) using inspection.

Step 1: Look for factors of \(a\) and \(c\).

Factors of \(a\):

\(2 = 1(2)\)
\(2 = -1(-2)\)

Because \(a\) is positive, you can assume both first terms of the binomials are also positive, so the factored form will look like

\(\left( {2x + \_\_\_\_} \right)\left( {x + \_\_\_\_} \right)\)

Factors of \(c\):

\(-15 = 1(-15) \)
\(-15 = -1(15) \)
\(-15 = 3(-5) \)
\(-15 = -3(5) \)

Step 2: Using trial and error, look for the combination that will give the correct polynomial if expanded. You only need to find the middle term, in other words multiply the Outer terms and Inner terms of FOIL, and add them together.

\(\begin{align}
 &\left( {2x + 1} \right)\left( {x + \left( { -15} \right)} \right) \to - 30x + x = -29x &{\rm{No}} \\
 &\left( {2x + \left( { -1} \right)} \right)\left( {x + 15} \right) \to 30x - x = 29x &{\rm{No}} \\
 &\left( {2x + 15} \right)\left( {x + \left( { -1} \right)} \right) \to - 2x + 15x = 13x &{\rm{No}} \\
 &\left( {2x + \left( { -15} \right)} \right)\left( {x + 1} \right) \to 2x - 15x = -13x &{\rm{No}} \\
 &\left( {2x + 3} \right)\left( {x + \left( { -5} \right)} \right) \to - 10x + 3x = -7x &{\rm{No}} \\
 &\left( {2x + \left( { -3} \right)} \right)\left( {x + 5} \right) \to 10x - 3x = 7x &{\rm{No}} \\
 &\left( {2x + 5} \right)\left( {x + \left( { -3} \right)} \right) \to - 6x + 5x = -1x &{\rm{YES!}} \\
 &\left( {2x + \left( { -5} \right)} \right)\left( {x + 3} \right) \to 6x - 5x = 1x &{\rm{No}}  \end{align}\)

Therefore, the factored form of \(2x^2 - x - 15 \) is \((2x + 5)(x - 3)\).
That method seems like a lot of work, but with practice, you can try out various combinations in your head, and it will eventually get easier and faster.