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Course: Mathematics — Second Preparatory
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Curriculum

Mathematics — Second Preparatory

Text lesson

Factoring by Grouping

Mathematics — Second Preparatory

Factoring by Grouping

Worked examples

Worked examples

Comparison aspectMain ideaKeywords
Example 1Factor xy + 2x + 3y + 6Factor
Solution
  • Group the first two terms and the last two terms: x(y + 2) + 3(y + 2). The common bracket is (y + 2)
  • so the factoring is (x + 3)(y + 2). Check: (x + 3)(y + 2) = xy + 2x + 3y + 6
Groupfirsttermslast
Example 2Factor x³ + 2x² + 3x + 6Factor
Solution
  • The first two terms give x²(x + 2)
  • and the last two terms give 3(x + 2). The common bracket is (x + 2)
  • so the factoring is (x + 2)(x² + 3)
firsttermsgivelast
Example 3Factor 2ab − 6a + 3b − 9Factor
Solution
  • The first group gives 2a(b − 3)
  • and the second group gives 3(b − 3). The common bracket is (b − 3)
  • so the factoring is (b − 3)(2a + 3). Check: 2ab + 3b − 6a − 9
firstgroupgivessecond
Example 4Factor xy − 3x − 2y + 6Factor
Comparison aspectMain ideaKeywords
Solution
  • The first group gives x(y − 3)
  • and the second group gives −2(y − 3). Note the negative sign when taking out −2. The common bracket is (y − 3)
  • so the factoring is (x − 2)(y − 3). Check: xy − 3x − 2y + 6
firstgroupgivessecond
Example 5Factor x³ − x² + x − 1Factor
Solution
  • Group the first two terms to get x²(x − 1)
  • and group the last two terms to get 1(x − 1). The common bracket is (x − 1)
  • so the factoring is (x − 1)(x² + 1). Check: x³ + x − x² − 1 = x³ − x² + x − 1
Groupfirsttermslast
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