Example 1
Completion requirements
| Example 1 |
Factor \(2x^2 - x - 15 \) using grouping by decomposition.
Step 1: If possible, identify the GCF.
There is no GCF.
Step 2: Find a pair of numbers with a sum of \(b\) (\(-1\) in this case) and a product of \(ac\) (\(-15(2) = -30\) in this case).
One strategy is to list numbers that multiply to \(-30\), and then look within those pairs for a sum of \(-1\), much like in Example 3 in the previous section.
Step 3: The numbers \(5\) and \(–6\) add to \(–1\) and multiply to \(–30\). Rewrite the middle term (\(b\)-term) as two terms with coefficients \(5\) and \(–6\). This step “decomposes” the middle term into two usable terms.
\(2x^2 - x - 15 = 2x^2 + {\color{red}5x - 6x} - 15\)
Step 4: Group the first two terms and the last two terms.
\(2x^2 + {\color{red}5x - 6x} - 15 = (2x^2 + {\color{red}5x}) + ({\color{red}-6x} - 15)\)
Step 5: Remove the greatest common factor from each of the two groups.
Note that the goal is to be left with identical factors in each set of brackets. As such, sometimes this means factoring a negative number from one of the two groups, which technically wouldn’t make it the “Greatest” common factor.
Step 6: The result is a common factor, \(2x + 5 \), in each group. Remove the common factor and simplify.
Note that this is similar to collecting like terms, \(xa - 3a = a(x - 3) \). Instead of \(a\) being the common term, \((2x + 5)\) is common.
There is no GCF.
Step 2: Find a pair of numbers with a sum of \(b\) (\(-1\) in this case) and a product of \(ac\) (\(-15(2) = -30\) in this case).
One strategy is to list numbers that multiply to \(-30\), and then look within those pairs for a sum of \(-1\), much like in Example 3 in the previous section.
| First number | Second number | Product | Sum | Works? |
| \(1\) | \(–30\) | \(1(–30) = –30\) | \(1 - 30 = –29\) |
No
|
| \(–1\) | \(30\) | \(–1(30) = –30\) | \(–1 + 30 = 29\) |
No
|
| \(2\) | \(–15\) | \(2(–15) = –30\) | \(2 - 15 = –13\) |
No
|
| \(–2\) | \(15\) | \(–2(15) = –30\) | \(–2 + 15 = 13\) |
No
|
| \(3\) | \(–10\) | \(3(–10) = –30\) | \(3 - 10 = –7\) |
No
|
| \(–3\) | \(10\) | \(–3(10) = –30\) | \(–3 + 10 = 7\) |
No
|
| \(5\) | \(–6\) | \(5(–6) = –30\) | \(5 - 6 = –1\) |
YES!
|
| \(–5\) | \(6\) | \(–5(6) = –30\) | \(–5 + 6 = 1\) |
No
|
Step 3: The numbers \(5\) and \(–6\) add to \(–1\) and multiply to \(–30\). Rewrite the middle term (\(b\)-term) as two terms with coefficients \(5\) and \(–6\). This step “decomposes” the middle term into two usable terms.
\(2x^2 - x - 15 = 2x^2 + {\color{red}5x - 6x} - 15\)
Step 4: Group the first two terms and the last two terms.
\(2x^2 + {\color{red}5x - 6x} - 15 = (2x^2 + {\color{red}5x}) + ({\color{red}-6x} - 15)\)
Step 5: Remove the greatest common factor from each of the two groups.
\((2x^2 + {\color{red} 5x}) + ({\color{red} - 6x} - 15) = x({\color{red} 2x + 5}) - 3({\color{red}2x + 5})\)
Note that the goal is to be left with identical factors in each set of brackets. As such, sometimes this means factoring a negative number from one of the two groups, which technically wouldn’t make it the “Greatest” common factor.
Step 6: The result is a common factor, \(2x + 5 \), in each group. Remove the common factor and simplify.
\(x({\color{red}2x + 5}) - 3({\color{red}2x + 5}) = ({\color{red}2x + 5})(x - 3)\)
Verify by expanding.
\(\begin{align}
\left( {2x + 5} \right)\left( {x - 3} \right) &= 2x^2 - 6x + 5x - 15 \\
&= 2x^2 - x - 15 \\
\end{align}\)
Verify by expanding.
\(\begin{align}
\left( {2x + 5} \right)\left( {x - 3} \right) &= 2x^2 - 6x + 5x - 15 \\
&= 2x^2 - x - 15 \\
\end{align}\)
Note that this is similar to collecting like terms, \(xa - 3a = a(x - 3) \). Instead of \(a\) being the common term, \((2x + 5)\) is common.