MathematicsChapter 4: Factorization and Algebraic Manipulation
Factoring By Grouping
When four terms do not share one monomial, try grouping.
Learning objectives
- Explain Factoring By Grouping in simple mathematical language.
- Apply two and two correctly.
- Apply watch the signs correctly.
- Use the symbols, conditions, and formulae accurately.
- Solve a short example and check the answer.
- Recognise and correct common mistakes.
Definition
When four terms do not share one monomial, try grouping. Pair them so each pair has its own common factor, and hope the two brackets that remain are the same. The pattern (a + b)(c + d) expands to ac + ad + bc + bd, so grouping is that expansion in reverse. If the pairs give matching brackets, you have factored.
Where / symbols
- grouping
- When four terms do not share one monomial, try grouping.
- (a + b)(c + d)
- The pattern (a + b)(c + d) expands to ac + ad + bc + bd, so grouping is that expansion in reverse.
- Grouping
- When four terms do not share one monomial, try grouping.
- identical
- The leftover binomials must be identical (or exact opposites, which you fix with a minus).
- regroup
- You may regroup as (ac + bc) + (ad + bd) if the first pairing fails.
Overview
Split four terms into two pairs, then factor each pair:
Two and Two
From ac + ad + bc + bd, group (ac + ad) + (bc + bd) = a(c + d) + b(c + d) = (a + b)(c + d). The shared binomial (c + d) becomes a common factor of the two groups. You may regroup as (ac + bc) + (ad + bd) if the first pairing fails. Same product, different pairing.
Examples x^3 + x^2 + x + 1 = (x^2 + 1)(x + 1) · ax + ay + bx + by = (a + b)(x + y) · 2x + 2y + zx + zy = (2 + z)(x + y)
Watch the Signs
If a pair is minus, factor the minus with the monomial so the leftover brackets match. Example: ax - ay - bx + by = a(x - y) - b(x - y) = (a - b)(x - y). If you factor +b from the last two terms by mistake, the brackets will not match and grouping looks 'broken' when only the sign was wrong.
Examples ax - ay - bx + by = (a - b)(x - y) · x^3 - x^2 + x - 1 = (x^2 + 1)(x - 1) · 6xy - 3x + 4y - 2 = (3x + 2)(2y - 1)
WHAT — Core idea
When four terms do not share one monomial, try grouping. Pair them so each pair has its own common factor, and hope the two brackets that remain are the same. The pattern (a + b)(c + d) expands to ac + ad + bc + bd, so grouping is that expansion in reverse. If the pairs give matching brackets, you have factored.
WHY — Why it matters
Grouping is not limited to four terms, but four is the school pattern. If no pairing produces equal brackets, grouping is the wrong tool; try another identity. Always expand your answer to check.
HOW — Method and examples
1. TWO AND TWO: From ac + ad + bc + bd, group (ac + ad) + (bc + bd) = a(c + d) + b(c + d) = (a + b)(c + d). 2. WATCH THE SIGNS: If a pair is minus, factor the minus with the monomial so the leftover brackets match.
HOW — Answer in an exam
The leftover binomials must be identical (or exact opposites, which you fix with a minus). If they differ, regroup or take a common minus. Do not invent a common factor that only two of four terms share and then stop.
Explanation
Grouping is not limited to four terms, but four is the school pattern. If no pairing produces equal brackets, grouping is the wrong tool; try another identity. Always expand your answer to check.
In simple words
The leftover binomials must be identical (or exact opposites, which you fix with a minus). If they differ, regroup or take a common minus. Do not invent a common factor that only two of four terms share and then stop.
Key points
- grouping
- (a + b)(c + d)
Exam preparation
Exam Focus
Quick revision
Revision Zone
Review the key facts, formulas and checks before you move on.Factoring By Grouping
Sticky Notes · Quick Revision
WHAT
- When four terms do not share one monomial, try grouping.
Key Points
- When four terms do not share one monomial, try grouping.
- From ac + ad + bc + bd, group (ac + ad) + (bc + bd) = a(c + d) + b(c + d) = (a + b)(c + d).
- If a pair is minus, factor the minus with the monomial so the leftover brackets match.
- Grouping is not limited to four terms, but four is the school pattern.
HOW
- 1. TWO AND TWO: From ac + ad + bc + bd, group (ac + ad) + (bc + bd) = a(c + d) + b(c + d) = (a + b)(c + d).
- 2. WATCH THE SIGNS: If a pair is minus, factor the minus with the monomial so the leftover brackets match.
Examples
- x3 + x2 + x + 1 = (x2 + 1)(x + 1)
- ax + ay + bx + by = (a + b)(x + y)
- 2x + 2y + zx + zy = (2 + z)(x + y)
- ax - ay - bx + by = (a - b)(x - y)
- x3 - x2 + x - 1 = (x2 + 1)(x - 1)
Common Mistake
- The leftover binomials must be identical (or exact opposites, which you fix with a minus).
- If they differ, regroup or take a common minus.
Exam Tip
- The leftover binomials must be identical (or exact opposites, which you fix with a minus). If they differ, regroup or take a common minus. Do not invent a common factor that only two of four terms share and then stop.
60-second revision
- When four terms do not share one monomial, try grouping.
- From ac + ad + bc + bd, group (ac + ad) + (bc + bd) = a(c + d) + b(c + d) = (a + b)(c + d).
- If a pair is minus, factor the minus with the monomial so the leftover brackets match.
- Grouping is not limited to four terms, but four is the school pattern.
- The leftover binomials must be identical (or exact opposites, which you fix with a minus).
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