Quadratic Factorizer with Cross Method

Enter any quadratic expression ax² + bx + c and instantly see the factored form (px + r)(qx + s) with a visual cross method diagram and step-by-step working!

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Enter the coefficients a, b, and c to factorize your quadratic expression.

Quadratic Factorizer

Enter coefficients to factor ax² + bx + c into (px + r)(qx + s)

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x²

What is Quadratic Factorization?

Factorization is the process of breaking down an expression into simpler factors that multiply together to give the original expression.

For quadratic expressions of the form $ax^2 + bx + c$ , we want to write them as:

$(px + r)(qx + s)$

where $p \times q = a$ , $r \times s = c$ , and $ps + qr = b$ .

💡 Why Factorize?

Factorization is essential for:

Solving quadratic equations: If $ax^2 + bx + c = 0$ , then the roots are where each factor equals zero
Simplifying algebraic fractions: Helps cancel common factors
Sketching parabolas: Easily find x-intercepts from factored form

The Cross Method Explained

The cross method (also called criss-cross method) is a systematic way to factorize quadratics, especially when $a \neq 1$ .

How the Cross Method Works

For $ax^2 + bx + c$ :

Left side of cross: Find factor pairs of $a$
Right side of cross: Find factor pairs of $c$
Cross multiply: Multiply diagonally and add
Match: Find the combination where cross products sum to $b$

Visual Representation

    px ——————— r
       ╲   ╱
        ╲ ╱
         ╳      Cross multiply: (p × s) + (q × r) = b
        ╱ ╲
       ╱   ╲
    qx ——————— s

    Factors of a    Factors of c

Method 1: Simple Quadratics (a = 1)

When $a = 1$ , factorization is simpler. For $x^2 + bx + c$ :

Find two numbers that:

Multiply to give c (the constant term)
Add to give b (the coefficient of x)

Example 1: x² + 5x + 6

Problem:

Factorize x² + 5x + 6

Step 1: Find two numbers that multiply to 6 and add to 5

Numbers	Product	Sum
1, 6	6 ✓	7 ✗
2, 3	6 ✓	5 ✓

Step 2: Write the factors

$x^2 + 5x + 6 = (x + 2)(x + 3)$

Check: $(x + 2)(x + 3) = x^2 + 3x + 2x + 6 = x^2 + 5x + 6$ ✓

Example 2: x² − 7x + 12

Problem:

Factorize x² − 7x + 12

Step 1: Find two numbers that multiply to 12 and add to −7

Since the product is positive and sum is negative, both numbers must be negative.

Numbers	Product	Sum
−1, −12	12 ✓	−13 ✗
−2, −6	12 ✓	−8 ✗
−3, −4	12 ✓	−7 ✓

Step 2: Write the factors

$x^2 - 7x + 12 = (x - 3)(x - 4)$

Check: $(x - 3)(x - 4) = x^2 - 4x - 3x + 12 = x^2 - 7x + 12$ ✓

Method 2: Non-Monic Quadratics (a ≠ 1)

When $a \neq 1$ , use the cross method or factor by grouping.

The AC Method with Cross

For $ax^2 + bx + c$ :

Calculate the product $a \times c$
Find two numbers $m$ and $n$ where $m \times n = ac$ and $m + n = b$
Use the cross diagram to find the correct factor pairs

Example 3: 2x² + 7x + 3

Problem:

Factorize 2x² + 7x + 3

Step 1: Identify coefficients: $a = 2$ , $b = 7$ , $c = 3$

Step 2: Calculate $a \times c = 2 \times 3 = 6$

Step 3: Find two numbers that multiply to 6 and add to 7

$m = 1$ , $n = 6$ → $1 \times 6 = 6$ ✓, $1 + 6 = 7$ ✓

Step 4: Set up the cross method

Factors of $a = 2$ : Try 2 and 1
Factors of $c = 3$ : Try 3 and 1

    2x ——————— 1
       ╲   ╱
        ╲ ╱
         ╳      (2 × 3) + (1 × 1) = 6 + 1 = 7 ✓
        ╱ ╲
       ╱   ╲
     x ——————— 3

Step 5: Write the factors

$2x^2 + 7x + 3 = (2x + 1)(x + 3)$

Check: $(2x + 1)(x + 3) = 2x^2 + 6x + x + 3 = 2x^2 + 7x + 3$ ✓

Example 4: 6x² + 11x − 10

Problem:

Factorize 6x² + 11x − 10

Step 1: Identify coefficients: $a = 6$ , $b = 11$ , $c = -10$

Step 2: Calculate $a \times c = 6 \times (-10) = -60$

Step 3: Find two numbers that multiply to −60 and add to 11

$m = 15$ , $n = -4$ → $15 \times (-4) = -60$ ✓, $15 + (-4) = 11$ ✓

Step 4: Set up the cross method

Factors of 6: Try 3 and 2, or 6 and 1
Factors of −10: Try 5 and −2, or −5 and 2, etc.

Testing with 3, 2 for $a$ and 5, −2 for $c$ :

    3x ——————— (−2)
       ╲   ╱
        ╲ ╱
         ╳      (3 × 5) + (2 × −2) = 15 − 4 = 11 ✓
        ╱ ╲
       ╱   ╲
    2x ——————— 5

Step 5: Write the factors

$6x^2 + 11x - 10 = (3x - 2)(2x + 5)$

Check: $(3x - 2)(2x + 5) = 6x^2 + 15x - 4x - 10 = 6x^2 + 11x - 10$ ✓

Method 3: Factor by Grouping

This is an alternative to the cross method for non-monic quadratics.

Example 5: 3x² − 10x + 8

Problem:

Factorize 3x² − 10x + 8 using grouping

Step 1: Find $a \times c = 3 \times 8 = 24$

Step 2: Find two numbers that multiply to 24 and add to −10

$m = -4$ , $n = -6$ → $(-4) \times (-6) = 24$ ✓, $(-4) + (-6) = -10$ ✓

Step 3: Split the middle term $3x^2 - 10x + 8 = 3x^2 - 4x - 6x + 8$

Step 4: Group and factor $= x(3x - 4) - 2(3x - 4)$

Step 5: Factor out the common bracket $= (3x - 4)(x - 2)$

Check: $(3x - 4)(x - 2) = 3x^2 - 6x - 4x + 8 = 3x^2 - 10x + 8$ ✓

Special Cases

Perfect Square Trinomials

$a^2 + 2ab + b^2 = (a + b)^2$ $a^2 - 2ab + b^2 = (a - b)^2$

Example: x² + 6x + 9

This is $(x)^2 + 2(x)(3) + (3)^2 = (x + 3)^2$

Recognition tip: Check if $c$ is a perfect square and if $b = 2\sqrt{c}$

$c = 9 = 3^2$ ✓
$b = 6 = 2 \times 3$ ✓

Therefore: $x^2 + 6x + 9 = (x + 3)^2$

Difference of Squares

$a^2 - b^2 = (a + b)(a - b)$

Example: 4x² − 25

This is $(2x)^2 - (5)^2 = (2x + 5)(2x - 5)$

Recognition tip: No middle term ( $b = 0$ ) and both terms are perfect squares.

When Factorization is Not Possible

Not all quadratics can be factorized over integers.

⚠️ Check the Discriminant

For $ax^2 + bx + c$ , calculate the discriminant: $\Delta = b^2 - 4ac$

If $\Delta < 0$ : Cannot be factorized over real numbers
If $\Delta$ is not a perfect square: Cannot be factorized over integers (use quadratic formula instead)
If $\Delta = 0$ : Perfect square trinomial
If $\Delta > 0$ and is a perfect square: Can be factorized over integers

Example: x² + 2x + 5

Problem:

Can x² + 2x + 5 be factorized?

Check discriminant: $\Delta = 2^2 - 4(1)(5) = 4 - 20 = -16$

Since $\Delta < 0$ , this quadratic cannot be factorized over real numbers.

(You would need to use the quadratic formula with complex numbers.)

Common Mistakes to Avoid

❌ Mistake 1: Wrong Signs

Pay careful attention to signs!

For $x^2 - 5x + 6$ :

Wrong: $(x + 2)(x + 3)$ gives $x^2 + 5x + 6$ ✗
Right: $(x - 2)(x - 3)$ gives $x^2 - 5x + 6$ ✓

❌ Mistake 2: Not Checking the Answer

Always expand your factors to verify!

If you get $(2x + 3)(x + 4)$ for $2x^2 + 11x + 12$ :

Check: $(2x + 3)(x + 4) = 2x^2 + 8x + 3x + 12 = 2x^2 + 11x + 12$ ✓

❌ Mistake 3: Forgetting Factor Pairs

For $a \times c$ , consider ALL factor pairs including negatives.

For $x^2 + x - 12$ where $c = -12$ :

Factor pairs: $(1, -12), (-1, 12), (2, -6), (-2, 6), (3, -4), (-3, 4)$
The pair that adds to $b = 1$ is $(4, -3)$

Quick Reference

Quadratic Type	Factorization Pattern
$x^2 + bx + c$	$(x + m)(x + n)$ where $mn = c$ and $m + n = b$
$ax^2 + bx + c$	$(px + r)(qx + s)$ where $pq = a$ , $rs = c$ , $ps + qr = b$
$a^2 + 2ab + b^2$	$(a + b)^2$
$a^2 - 2ab + b^2$	$(a - b)^2$
$a^2 - b^2$	$(a + b)(a - b)$

Practice Strategy for O-Levels

Start with simple quadratics ( $a = 1$ ) until the process becomes automatic
Learn to recognize special cases (perfect squares, difference of squares)
Master the cross method for non-monic quadratics
Always check by expanding your answer
Time yourself - aim to factorize simple quadratics in under 30 seconds

More O-Level Resources

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