Why a Quadratic Can Be a Piece of Cake
Ever wondered how a messy curve can become two simple straight lines? That’s the magic of factorisation – it turns a tough quadratic into something you can solve in minutes.
💡 In Simple Words: A quadratic equation is like a hidden treasure box. Factorisation cracks the lock, giving you two easy clues (the factors) that lead straight to the answer.
What is a Quadratic Equation?
A quadratic equation is any equation that can be written as ax² + bx + c = 0, where a, b and c are numbers and a ≠ 0 (otherwise it wouldn’t be quadratic). The highest power of the variable (x) is 2, which is why we call it “quadratic” – think of a square (2×2) shape.
How Does Factorisation Help?
Factorisation means writing the quadratic as a product of two simpler expressions, like (px + q)(rx + s) = 0. If the product of two things is zero, at least one of them must be zero. That’s the zero‑product property – similar to how if a water pipe stops flowing, either the left section or the right section is blocked.
Step‑by‑Step Method to Solve by Factorisation
Follow these four tidy steps whenever the quadratic looks factorable.
- Bring everything to one side. Make sure the equation is in the form ax² + bx + c = 0.
- Look for two numbers that multiply to a·c and add to b. These numbers become the “middle‑term split”.
- Rewrite the middle term using those two numbers and factor by grouping. You’ll end up with two brackets.
- Set each bracket equal to zero and solve for x. Those are your solutions.
Worked Example 1: Simple Numbers
Solve x² - 5x + 6 = 0.
- Step 1: It’s already in the right form.
- Step 2: Find two numbers that multiply to 6 (the constant) and add to -5. Those are -2 and -3.
- Step 3: Rewrite – x² - 2x - 3x + 6 = 0. Group: (x² - 2x) + (-3x + 6) = 0.
- Factor each group: x(x - 2) -3(x - 2) = 0.
- Now factor out the common binomial (x - 2): (x - 3)(x - 2) = 0.
- Step 4: Set each factor to zero → x‑3 = 0 or x‑2 = 0 → x = 3 or x = 2.
Both 3 and 2 satisfy the original equation.
Worked Example 2: Coefficient Not 1
Solve 2x² + 7x + 3 = 0.
- Step 1: Already tidy.
- Step 2: a·c = 2·3 = 6. Need two numbers that multiply to 6 and add to 7 → 1 and 6.
- Step 3: Rewrite – 2x² + 1x + 6x + 3 = 0. Group: (2x² + x) + (6x + 3) = 0.
- Factor each group: x(2x + 1) + 3(2x + 1) = 0.
- Common binomial (2x + 1) gives (x + 3)(2x + 1) = 0.
- Set each factor to zero → x + 3 = 0 → x = -3; 2x + 1 = 0 → x = -½.
Solutions: x = -3, x = -0.5.
Common Mistakes to Avoid
- Skipping the “a·c” step. If you only look for numbers that add to b, you’ll miss many factorable quadratics.
- Wrong sign handling. Remember that a negative product means one number is negative and the other positive.
- Not checking the answer. Plug the roots back into the original equation – a quick sanity check saves marks.
Quick Reference Table
| Step | What to Do | Key Tip |
|---|---|---|
| 1 | Write as ax²+bx+c=0 | Move all terms left |
| 2 | Find numbers p,q with p·q = a·c, p+q = b | Use factor pairs of a·c |
| 3 | Split bx into px + qx and factor by grouping | Group wisely |
| 4 | Apply zero‑product property | Set each bracket = 0 |
📝 Likely Exam Questions
- Solve x² - 4x - 12 = 0 by factorisation.
Answer: (x - 6)(x + 2)=0 → x=6 or x=-2. - Factorise 3x² + 14x + 8 and state the roots.
Answer: (3x+2)(x+4)=0 → x = -2/3, x = -4. - Explain why the zero‑product property works for quadratics.
Answer: If AB=0, at least one of A or B must be 0; applying this to (px+q)(rx+s)=0 gives two simple linear equations. - Given the quadratic 2x² - 9x + 4 = 0, solve by factorisation.
Answer: (2x-1)(x-4)=0 → x=½ or x=4. - Check whether x = 3 is a solution of x² - 7x + 12 = 0.
Answer: Substitute: 9‑21+12=0 → true, so x=3 is a root.