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Lesson 6 of 8

Exploring Algebraic Identities · Lesson 6 of 8

Factorisation Without Using Algebra Tiles

“No tiles? No problem—algebra still leaves enough clues to find the factors.”

Learning Objectives

• Factor quadratics by splitting the middle term. • Find two numbers using sum and product conditions. • Handle positive and negative signs correctly. • Factor expressions with leading coefficient other than 1. • Apply factorisation to simple geometric problems.

Once the area model is understood, the same reasoning can be done symbolically. The key is to choose two numbers that split the middle term in exactly the way required for grouping.

Basic patternLaTeX

The Step-by-Step Method

StepAction
1Identify the coefficient of x and the constant term.
2Find two numbers whose sum equals the x-coefficient.
3Check that their product equals the constant term.
4Split the middle term using those numbers.
5Factor by grouping.
Worked Example: x²+7x+12

Problem
Factor x²+7x+12.

  1. 1.We need sum 7 and product 12.
  2. 2.The factor pair 3 and 4 satisfies both.
  3. 3.Split 7x as 3x+4x.
  4. 4.x²+3x+4x+12=x(x+3)+4(x+3).
  5. 5.Therefore the factorisation is (x+3)(x+4).
Worked Example: x²+11x+30

Problem
Factor completely.

  1. 1.We need sum 11 and product 30.
  2. 2.Pairs such as 2 and 15 or 3 and 10 have product 30 but wrong sums.
  3. 3.5 and 6 have sum 11 and product 30.
  4. 4.Therefore x²+11x+30=(x+5)(x+6).

Negative Middle Term

If the constant term is positive but the middle coefficient is negative, both required numbers are negative.

Worked Example: x²−5x+6

Problem
Factor the expression.

  1. 1.We need sum −5 and product +6.
  2. 2.The numbers are −2 and −3.
  3. 3.x²−5x+6=x²−2x−3x+6.
  4. 4.=x(x−2)−3(x−2).
  5. 5.=(x−2)(x−3).

Negative Constant Term

If the constant term is negative, the two required numbers must have opposite signs. The sign of the larger-magnitude number determines the sign of the sum.

Worked Example: r²−r−42

Problem
Factor completely.

  1. 1.We need sum −1 and product −42.
  2. 2.6 and −7 satisfy both conditions.
  3. 3.r²−r−42=r²+6r−7r−42.
  4. 4.=r(r+6)−7(r+6).
  5. 5.=(r+6)(r−7).

Leading Coefficient Greater Than 1

For Ax²+Bx+C, we look for two numbers whose sum is B and whose product is A×C. This allows us to split the middle term before grouping.

Worked Example: 6x²+7x+2

Problem
Factor completely.

  1. 1.A×C=6×2=12.
  2. 2.We need two numbers with sum 7 and product 12: 3 and 4.
  3. 3.6x²+7x+2=6x²+3x+4x+2.
  4. 4.=3x(2x+1)+2(2x+1).
  5. 5.=(3x+2)(2x+1).

Using Factorisation in Geometry

Worked Example: Rectangle Area

Problem
A rectangle has area x²+8x+15. Find possible side expressions.

  1. 1.We need sum 8 and product 15.
  2. 2.3 and 5 satisfy both.
  3. 3.x²+8x+15=(x+3)(x+5).
  4. 4.So possible dimensions are x+3 and x+5.
Worked Example: Rectangular Pool

Problem
A pool has breadth 4 m less than its length and area 96 m². Find its dimensions.

  1. 1.Let the length be x and breadth x−4.
  2. 2.x(x−4)=96, so x²−4x−96=0.
  3. 3.Split −4x as −12x+8x because −12+8=−4 and (−12)(8)=−96.
  4. 4.x²−12x+8x−96=(x−12)(x+8)=0.
  5. 5.So x=12 or x=−8.
  6. 6.A physical length cannot be negative, so length=12 m and breadth=8 m.

Practice Problems

Practice Problems
  1. Factor s²−11s+24.
  2. Factor 10x²−11x−6.
  3. Factor 6x²+7x+2.
  4. Factor r²−r−42.
  5. Factor 16s²+25t²−40st using an identity.
  6. Find possible dimensions of a rectangle with area x²+9x+20.
  7. A rectangle has area 120 m² and breadth 7 m less than length. Form and factor the quadratic equation to find its dimensions.

Key Takeaways

Key Takeaways

• Splitting the middle term depends on both sum and product. • For x²+Bx+C, use numbers with sum B and product C. • For Ax²+Bx+C, use product A×C for the split. • Sign patterns guide the choice of numbers. • Factorisation can solve real area and dimension problems.

Coming Next

Next, we derive cubic identities and use factorisation in more advanced expressions.

Previous · Lesson 5

Factorisation Using Algebra Tiles

Next · Lesson 7

Finding New Identities