Practice Maths

Solving Quadratic Equations by Factorising

Key Ideas

Key Terms

quadratic equation
Has the form ax² + bx + c = 0. It can have 0, 1, or 2 solutions.
Null Factor Law
If A × B = 0, then A = 0 or B = 0 (or both). This is the key to solving by factorising.
Step 1
Rearrange so one side is 0.   Step 2: Factorise the non-zero side.   Step 3: Apply the Null Factor Law.
roots
Of the equation or the x-intercepts of the parabola y = ax² + bx + c.
StepActionExample: x² = 5x − 6
1Rearrange to = 0x² − 5x + 6 = 0
2Factorise(x − 2)(x − 3) = 0
3Null Factor Lawx − 2 = 0  or  x − 3 = 0
4Solve eachx = 2  or  x = 3
Hot Tip — Always = 0 First! The Null Factor Law only works when one side is exactly zero. If you have (x−2)(x+3) = 6, you cannot say x−2=6 or x+3=6. You must first expand, rearrange to get = 0, then re-factorise.

Worked Example — Non-monic equation

Solve 2x² + x − 6 = 0.

Product-sum: ac = −12, need sum = 1. Use 4 and −3.

2x² + 4x − 3x − 6 = 0 ⇒ 2x(x + 2) − 3(x + 2) = 0 ⇒ (2x − 3)(x + 2) = 0

x = 3/2  or  x = −2

Worked Example — Equation not in standard form

Solve x² = 3x + 10.

Rearrange: x² − 3x − 10 = 0 ⇒ (x − 5)(x + 2) = 0

x = 5  or  x = −2

From Expression to Equation

In the previous lesson you learned to factorise quadratic expressions. Now we use that skill to solve quadratic equations — statements like x² + 5x + 6 = 0 where we need to find the value (or values) of x that make it true.

A quadratic equation can have two solutions, one solution (a repeated root), or no real solutions. We’ll see how to determine this later with the discriminant, but for now we focus on equations that factorise nicely.

The Null Factor Law

The Null Factor Law is the core idea: if the product of two factors is zero, then at least one of the factors must be zero. In symbols: if A × B = 0, then A = 0 or B = 0.

This might seem obvious, but it’s powerful. If we can write a quadratic in the form (x − p)(x − q) = 0, then either x − p = 0 (giving x = p) or x − q = 0 (giving x = q). Those are our two solutions.

The crucial point is that this only works when the right-hand side is exactly zero. If you have (x + 3)(x − 2) = 12, you cannot split it into x + 3 = 12 or x − 2 = 12. You must expand, rearrange to equal zero, then re-factorise.

The Solving Process

Follow this four-step method every time:

Step 1: Rearrange. Move everything to one side so the equation equals zero. For example, x² + 3x = 18 becomes x² + 3x − 18 = 0.

Step 2: Factorise. Factorise the left-hand side completely. x² + 3x − 18 = (x + 6)(x − 3).

Step 3: Apply the Null Factor Law. Set each factor equal to zero: x + 6 = 0 or x − 3 = 0.

Step 4: Solve each linear equation. x = −6 or x = 3.

Always check your answers by substituting back into the original equation, not just the rearranged form.

Special Case: Equations with No Middle Term

When the equation has no bx term (e.g. x² − 9 = 0), you can factorise using the difference of two squares: (x−3)(x+3) = 0, giving x = 3 or x = −3. Alternatively, isolate x² and take square roots: x² = 9 ⇒ x = ±3. Don’t forget the ± sign!

Setting Up from Word Problems

Many real-world problems lead to quadratic equations. The key steps are: define a variable, write an equation from the context, rearrange to standard form, factorise, solve, and then check that your solution makes sense in the context. (A length cannot be negative, for example.)

Common exam mistake: Dividing by x. If you have x² = 5x and divide both sides by x, you get x = 5 and lose the solution x = 0. Instead, rearrange: x² − 5x = 0, factorise: x(x−5) = 0, solve: x = 0 or x = 5. Both solutions are valid.

Mastery Practice

  1. Each quadratic equation is already in standard form. Solve by factorising. Fluency

     EquationSolutions
    (a)x² + 5x + 6 = 0 
    (b)x² − 7x + 12 = 0 
    (c)x² − 4x − 21 = 0 
    (d)x² − 16 = 0 
    (e)x² − 6x + 9 = 0 
    (f)x² + 3x = 0 

  2. Rearrange each equation to standard form (= 0), then solve by factorising. Fluency

     EquationSolutions
    (a)x² = 4x + 12 
    (b)x² + 10 = 7x 
    (c)2x² = 8x 
    (d)3x² + 5x = 2 

  3. Solve each non-monic quadratic equation. Fluency

     EquationSolutions
    (a)2x² + 7x + 3 = 0 
    (b)3x² − 5x − 2 = 0 
    (c)4x² − 9 = 0 
    (d)6x² − x − 2 = 0 

  4. Take out the HCF first, then solve. Fluency

    1. 3x² − 12x = 0
    2. 5x² − 45 = 0
    3. 2x² + 6x + 4 = 0
    4. 4x² − 16x + 16 = 0

  5. Find and fix the student’s error. Understanding

    Student’s working. A student was asked to solve x² − 5x = 0. They wrote: “Divide both sides by x: x − 5 = 0, so x = 5.”
    1. Identify the error the student made and explain why it is wrong.
    2. Solve the equation correctly using factorising.
    3. What solution did the student lose by dividing by x?

  6. Positive dimensions. Understanding

    Geometry Problem. A right-angled triangle has legs of length x and (x − 3), and hypotenuse of length (x + 3) cm.
    1. Apply Pythagoras’ theorem to write a quadratic equation in x.
    2. Solve the equation to find x.
    3. State the lengths of all three sides.

  7. Consecutive integers. Understanding

    Number Problem. The product of two consecutive positive integers is 56.
    1. Let the smaller integer be n. Write an equation for the product.
    2. Rearrange to standard form and solve by factorising.
    3. State the two consecutive integers.

  8. Rectangular area from a quadratic. Understanding

    Architecture. A rectangular room has a length that is 4 m more than its width. The area of the room is 96 m².
    1. Let the width be w metres. Write a quadratic equation for the area.
    2. Solve to find the width, discarding any solution that is not physically valid.
    3. State the dimensions of the room and verify your answer.

  9. Solve each equation, rearranging and expanding as needed. Problem Solving

    Multi-step equations. These are not in standard form and require expansion before solving.
    1. (x + 3)(x − 1) = 5
    2. (2x − 1)(x + 4) = (x + 2)(x + 3)
    3. x(x + 5) = (x + 1)(x + 2) + 2

  10. Projectile motion. Problem Solving

    Physics application. A ball is launched upward from a cliff and its height above ground (in metres) after t seconds is given by h = −5t² + 20t + 60.
    1. Find when the ball hits the ground (h = 0) by solving −5t² + 20t + 60 = 0.
    2. Which solution must be rejected and why?
    3. Find the height of the ball at t = 1 and t = 3 seconds.
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