Example 3
Completion requirements
| Example 3 |
Factor \(x^2 - x - 6 \).
The \(b\) value is \(-1 \) and the \(c\) value is \(-6 \). This means you need \(r + s = -1\) and \(rs = -6\). Sometimes you can determine these values quickly by inspection and sometimes a table is more helpful. Start with \(r \) and \(s \) values that will multiply to give \(-6\), and then check to see which pairing gives \(r + s = -1\).
In the last line of the table, \(r + s = -1\) and \(rs = -6\). Use \(r = 2\) and \(s = -3\) to write the trinomial in factored form.
\(x^2 - x - 6 = \left( {x {\color{blue} + 2}} \right)\left( {x {\color{blue} - 3}} \right)\)
You can check by expanding.
\(\begin{align}
\left( {x + 2} \right)\left( {x - 3} \right) &= x^2 - 3x + 2x - 6 \\
&= x^2 - x - 6 \\
\end{align}\)
| \(r \) | \(s \) |
\(rs\) | \(r + s\) |
Works? |
| \(-2\) | \(3\) |
\(-2(3) = -6 \) |
\(-2 + 3 = 1\) |
No |
| \(-1\) | \(6\) | \(-1(6) = -6 \) |
\(-1 + 6 = 5\) |
No |
| \(1\) |
\(-6\) | \(1(-6) = -6 \) |
\(1 - 6 = -5 \) |
No |
| \(2\) |
\(-3\) | \(2(-3) = -6 \) |
\(2 - 3 = -1 \) |
Yes! |
In the last line of the table, \(r + s = -1\) and \(rs = -6\). Use \(r = 2\) and \(s = -3\) to write the trinomial in factored form.
\(x^2 - x - 6 = \left( {x {\color{blue} + 2}} \right)\left( {x {\color{blue} - 3}} \right)\)
You can check by expanding.
\(\begin{align}
\left( {x + 2} \right)\left( {x - 3} \right) &= x^2 - 3x + 2x - 6 \\
&= x^2 - x - 6 \\
\end{align}\)