Why factorising quadratics matters

Ever wondered why some math problems feel like puzzles you can crack with a simple trick? Factorising turns a tough equation into a friendly product you can split apart.

💡 In Simple Words: A quadratic equation is an expression where the highest power of x is 2, like ax²+bx+c=0. By factorising we rewrite it as a product of two simpler expressions, then use the zero‑product rule (if AB=0, then A=0 or B=0) to find x.

What is a quadratic equation?

A quadratic equation looks like ax² + bx + c = 0 where a, b, c are numbers and a ≠ 0. Think of it as a hill‑shaped curve on a graph; the points where it crosses the x‑axis are the solutions.

How to solve by factorisation – the 5‑step recipe

  • Step 1: Bring everything to one side so the equation equals 0.
  • Step 2: Look for two numbers that multiply to a·c (product of a and c) and add to b.
  • Step 3: Split the middle term bx using those two numbers and factor by grouping.
  • Step 4: Apply the zero‑product rule – set each bracket equal to 0.
  • Step 5: Solve the resulting simple linear equations.
graph TD A[Write equation = 0] --> B[Factorise the quadratic] --> C[Apply zero product property] --> D[Set each factor = 0] --> E[Solve each linear equation] --> F[Collect solutions]

Worked example 1

Solve x² – 5x + 6 = 0.

Step 1: It’s already =0.

Step 2: We need two numbers that multiply to 6 (a·c) and add to –5. Those are –2 and –3.

Step 3: Rewrite –5x as –2x –3x:
x² – 2x – 3x + 6 = 0

Step 4: Group and factor:
(x² – 2x) + (‑3x + 6) = 0
x(x – 2) –3(x – 2) = 0

Now we have a common factor (x – 2):
(x – 3)(x – 2) = 0

Step 5: Set each bracket to 0:
x – 3 = 0 → x = 3
x – 2 = 0 → x = 2

So the solutions are x = 2 and x = 3.

Worked example 2 (with a leading coefficient ≠ 1)

Solve 2x² + 7x + 3 = 0.

Step 1: Already =0.

Step 2: a·c = 2×3 = 6. We need two numbers that multiply to 6 and add to 7 – they are 1 and 6.

Step 3: Split 7x:
2x² + 1x + 6x + 3 = 0

Step 4: Group:
(2x² + x) + (6x + 3) = 0
x(2x + 1) + 3(2x + 1) = 0

Factor out the common binomial (2x + 1):
(x + 3)(2x + 1) = 0

Step 5: Solve:
x + 3 = 0 → x = –3
2x + 1 = 0 → x = –½

Solutions: x = –3 and x = –½.

Quick reference table

StepWhat to doWhy it works
1Move all terms to one side → 0Standard form for factorisation
2Find two numbers with product a·c and sum bThese numbers split the middle term
3Rewrite and groupCreates a common factor
4Factor out the common binomialTurns the quadratic into a product
5Apply zero‑product ruleEach factor set to zero gives a solution

Common pitfalls to avoid

  • Forgetting to make the constant term zero before factorising.
  • Mixing up the signs when the product a·c is negative.
  • Skipping the grouping step – it’s the bridge to factorisation.

📝 Likely Exam Questions

  1. Solve x² – 4x – 5 = 0.
    Answer: (x – 5)(x + 1)=0 → x=5 or x=‑1.
  2. Factorise and solve 3x² – 11x + 6 = 0.
    Answer: (3x – 2)(x – 3)=0 → x=2/3 or x=3.
  3. Given that (x – 2) is a factor of 2x² – kx + 8, find k.
    Answer: Substitute x=2 → 2·4 – 2k + 8 =0 → 8 – 2k + 8 =0 → k=8.
  4. Write the quadratic equation whose roots are –1 and 4.
    Answer: (x +1)(x –4)=0 → x² –3x –4 =0.
  5. Solve 4x² – 12x + 9 = 0 by factorisation.
    Answer: (2x –3)²=0 → x = 3/2 (double root).
#ICSE#Class 10#Math#Quadratic Equations#Factorisation