Solution by Factorisation

The Zero-Product Rule

Factorisation is the quickest way to solve a quadratic when it factors nicely. It rests on one simple idea:

Zero-Product Rule: If a product of two factors is zero, then at least one of the factors must be zero. That is, if $A \cdot B = 0$, then $A = 0$ or $B = 0$.

So if we can write $ax^2 + bx + c = 0$ as $(x - \alpha)(x - \beta) = 0$ (up to a constant), the roots are simply $x = \alpha$ and $x = \beta$.

Quick example

$x^2 - 5x + 6 = 0 \Rightarrow (x - 2)(x - 3) = 0 \Rightarrow x = 2$ or $x = 3$.

Key Point: Factorise, set each factor to zero, and solve. The two values are the roots.

[Board Important] This method only works when the quadratic factorises over rationals. If it doesn't, use completing the square or the quadratic formula (later sections).

Splitting the Middle Term

To factorise $ax^2 + bx + c$, we split the middle term $bx$ into two terms whose coefficients:

• multiply to $a \times c$, and
• add to $b$.

Worked outline: $2x^2 - 5x + 3 = 0$

1. Here $a \times c = 2 \times 3 = 6$, and we need two numbers multiplying to 6 and adding to $-5$: they are $-2$ and $-3$.
2. Split: $2x^2 - 2x - 3x + 3 = 0$.
3. Group: $2x(x - 1) - 3(x - 1) = 0 \Rightarrow (x - 1)(2x - 3) = 0$.
4. Roots: $x = 1$ and $x = \dfrac{3}{2}$.

Key Point: Find two numbers with product $ac$ and sum $b$. This is the heart of splitting the middle term.

[Board Important] When $a \neq 1$, you must use $a \times c$ (not just $c$) for the product. This is the most common factorisation error.

Equal Roots and Special Forms

Sometimes a quadratic factorises as a perfect square, giving two equal roots.

Perfect-square case

$x^2 - 6x + 9 = 0 \Rightarrow (x - 3)^2 = 0 \Rightarrow x = 3$ (a repeated root).

Difference of squares

$x^2 - 16 = 0 \Rightarrow (x - 4)(x + 4) = 0 \Rightarrow x = 4, -4$.

Common factor first

$2x^2 - 8x = 0 \Rightarrow 2x(x - 4) = 0 \Rightarrow x = 0, 4$. (Take out the common factor before anything else.)

Key Point: Always look for a common factor or a recognisable pattern (perfect square, difference of squares) before splitting the middle term.

[Board Important] $x = 0$ is a perfectly valid root. Don't discard it — when $c = 0$, one root is always 0.

Factorisation with Surds and Fractions

Some board problems involve surd coefficients. The method is the same — find the product and sum.

Surd example

$\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$: here $a \times c = \sqrt2 \times 5\sqrt2 = 10$; split 7 as $2 + 5$. So $\sqrt2 x^2 + 2x + 5x + 5\sqrt2 = \sqrt2 x(x + \sqrt2) + 5(x + \sqrt2) = (x + \sqrt2)(\sqrt2 x + 5) = 0$, giving $x = -\sqrt2$ and $x = -\dfrac{5}{\sqrt2} = -\dfrac{5\sqrt2}{2}$.

Key Point: Even with surds, find two numbers multiplying to $ac$ and adding to $b$. Rationalise the final answers if needed.

[Board Important] After finding roots involving surds, present them in a rationalised, simplified form for full marks.

Solved Examples

Example 1: Simple factorisation

Solve $x^2 - 7x + 12 = 0$.

Solution:

1. Two numbers with product 12 and sum $-7$: $-3$ and $-4$.
2. $(x - 3)(x - 4) = 0$.
3. Roots: $x = 3, 4$.

Final Answer: $x = 3, 4$.

Takeaway: Product $c$, sum $b$ (since $a = 1$).

Example 2: With $a \neq 1$

Solve $2x^2 - 5x + 3 = 0$.

Solution:

1. $a \times c = 6$; numbers multiplying to 6, adding to $-5$: $-2, -3$.
2. $2x^2 - 2x - 3x + 3 = 2x(x-1) - 3(x-1) = (x-1)(2x-3) = 0$.
3. Roots: $x = 1, \dfrac{3}{2}$.

Final Answer: $x = 1, \dfrac{3}{2}$.

Takeaway: Use $a \times c$ for the product when $a \neq 1$.

Example 3: Roots of $6x^2 - x - 2 = 0$

Solve by factorisation.

Solution:

1. $a \times c = -12$; numbers multiplying to $-12$, adding to $-1$: $-4$ and $3$.
2. $6x^2 - 4x + 3x - 2 = 2x(3x - 2) + 1(3x - 2) = (2x + 1)(3x - 2) = 0$.
3. Roots: $x = -\dfrac{1}{2}, \dfrac{2}{3}$.

Final Answer: $x = -\dfrac{1}{2}, \dfrac{2}{3}$.

Takeaway: Negative product means the two numbers have opposite signs.

Example 4: Equal roots (perfect square)

Solve $x^2 - 10x + 25 = 0$.

Solution:

1. $(x - 5)^2 = 0$.
2. $x = 5$ (repeated).

Final Answer: $x = 5, 5$ (equal roots).

Takeaway: A perfect-square trinomial gives two equal roots.

Example 5: Difference of squares

Solve $9x^2 - 25 = 0$.

Solution:

1. $(3x - 5)(3x + 5) = 0$.
2. Roots: $x = \dfrac{5}{3}, -\dfrac{5}{3}$.

Final Answer: $x = \pm\dfrac{5}{3}$.

Takeaway: $a^2 - b^2 = (a - b)(a + b)$.

Example 6: Common factor first

Solve $3x^2 - 12x = 0$.

Solution:

1. Take out $3x$: $3x(x - 4) = 0$.
2. Roots: $x = 0$ and $x = 4$.

Final Answer: $x = 0, 4$.

Takeaway: When $c = 0$, factor out $x$; one root is 0.

Example 7: Surd coefficients

Solve $\sqrt{3}x^2 + 10x + 7\sqrt{3} = 0$.

Solution:

1. $a \times c = \sqrt3 \times 7\sqrt3 = 21$; split 10 as $3 + 7$.
2. $\sqrt3 x^2 + 3x + 7x + 7\sqrt3 = \sqrt3 x(x + \sqrt3) + 7(x + \sqrt3) = (x + \sqrt3)(\sqrt3 x + 7) = 0$.
3. Roots: $x = -\sqrt3$ and $x = -\dfrac{7}{\sqrt3} = -\dfrac{7\sqrt3}{3}$.

Final Answer: $x = -\sqrt3, -\dfrac{7\sqrt3}{3}$.

Takeaway: Surds don't change the method; rationalise the final root.

Example 8: Clearing fractions first

Solve $\dfrac{1}{x} + \dfrac{1}{x - 2} = 3$ (assume $x \neq 0, 2$).

Solution:

1. Combine: $\dfrac{(x-2) + x}{x(x-2)} = 3 \Rightarrow 2x - 2 = 3x(x - 2)$.
2. $2x - 2 = 3x^2 - 6x \Rightarrow 3x^2 - 8x + 2 = 0$.
3. (Does not factor nicely — use the formula later; here we set up correctly.)

Final Answer: $3x^2 - 8x + 2 = 0$ (then solve by formula).

Takeaway: Clear denominators to reach standard form before deciding the method.

Example 9: Word problem by factorisation

The product of two consecutive positive integers is 132. Find them.

Solution:

1. $x(x + 1) = 132 \Rightarrow x^2 + x - 132 = 0$.
2. Numbers with product $-132$, sum 1: $12$ and $-11$. $(x + 12)(x - 11) = 0$.
3. $x = 11$ (reject $-12$ as positive integer needed); integers are 11 and 12.

Final Answer: 11 and 12.

Takeaway: Reject the inadmissible root based on the problem's context.

Example 10: Area word problem

The area of a rectangle is 70 m². Its length is 3 m more than its breadth. Find the dimensions.

Solution:

1. Let breadth $= x$ m. Length $= x + 3$ m. Area: $x(x + 3) = 70 \Rightarrow x^2 + 3x - 70 = 0$.
2. Two numbers with product $-70$ and sum $3$: $10$ and $-7$. So $(x + 10)(x - 7) = 0$.
3. $x = 7$ or $x = -10$. Reject $-10$ (breadth cannot be negative), so $x = 7$.
4. Breadth $= 7$ m, length $= 10$ m.

Final Answer: Breadth 7 m, length 10 m.

Takeaway: Reject the negative root; a length/breadth must be positive.