Factorization

Factoring is expanding in reverse: it bundles a spread-out polynomial back into a product of factors, like turning a mixture back into its ingredients. That is how x²+6x+9 becomes (x+3)². You will meet the perfect-square and difference-of-squares identities and learn to spot two numbers whose product and sum give the constant and middle term. Slide a, b, and the constant to watch the pieces build one square.

What is Factorization? — Reverse of Expansion

🔄 Expand ↔ Factor

①Expand: (a+b)² → a²+2ab+b² (product → sum)

②Factor: a²+2ab+b² → (a+b)² (sum → product)

③Factoring is reading multiplication formulas backward!

Factor Using Identities

Perfect Square

a²+2ab+b² = (a+b)²

Check the middle term equals 2ab

Difference of Squares

a²−b² = (a+b)(a−b)

Difference of squares → sum × difference

🔍 How to Spot the Form

①Three terms with both ends perfect squares → suspect perfect square

②Two terms, both squared → difference of squares

③Confirm middle term = 2 × √(first) × √(third)

x²+(a+b)x+ab Form

Constant (pq)6

Find Factor Pair

x²+(p+q)x+pq = (x+p)(x+q)

Product = constant, sum = coefficient of x

🧩 Strategy

①List every divisor pair of the constant

②Pick the pair whose sum matches the x coefficient

③Check negative pairs too!

Common Factor & Substitution

Pull Out Common Factor

ma+mb = m(a+b)

Always extract the common factor first

📌 Order of Factoring

①Step 1: Pull out the common factor

②Step 2: Apply identities to the rest

③Step 3: Repeat until no more factoring is possible

④e.g., 2x²+4x+2 = 2(x²+2x+1) = 2(x+1)²

Work It Out

Example 1

Factor x²+6x+9.

Check both ends are perfect squares: x²=(x)², 9=3².

x²+6x+9 = x²+2·x·3+3²

The middle term 6x = 2·x·3, so it is a perfect square.

= (x+3)²

▸ (x+3)²

If the middle term equals 2×(product of the roots of the ends), it is a perfect square.

Example 2

Factor x²−7x+12.

Find two numbers with product 12 and sum −7: −3 and −4.

(−3)×(−4)=12, (−3)+(−4)=−7

Group into the form (x+p)(x+q).

= (x−3)(x−4)

▸ (x−3)(x−4)

If the product is positive and the sum is negative, both numbers are negative.

Exam Wrap-up

Core Identities

a²±2ab+b² = (a±b)², a²−b² = (a+b)(a−b)

Factoring = reversed multiplication formulas

G9 school-exam type

Which is the correct factorization of x²−25?

①(x+5)²

②(x−5)²

③(x+5)(x−5)

④(x+25)(x−1)

⑤5(x−5)

▸ ③ (x+5)(x−5)

Note x²−25 = x²−5² is a difference of squares.

x²−25 = x²−5²

Apply a²−b²=(a+b)(a−b).

= (x+5)(x−5)

🎯 Exam Points

①Always pull out the common factor first

②Perfect square: middle term = 2 × (√first × √third)

③Difference of squares: subtraction form → apply directly

④Factor pair: pick divisors of constant whose sum matches

⑤Habit: re-expand to verify your factoring!

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Polynomial Multiplication

Quadratic Equations

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