Solving Quadratics by Factorising

Solving Quadratic Equations by Factorising

A quadratic equation has the form \( ax^2 + bx + c = 0 \). Factorising — rewriting the left side as a product of two linear factors — is the most direct algebraic method when the quadratic factorises neatly over the integers. The key tool is the zero-product property: if \( AB = 0 \) then either \( A = 0 \) or \( B = 0 \) (or both).

Method

  1. Rearrange so the equation is in the form \( ax^2 + bx + c = 0 \) (everything on one side, zero on the other).
  2. Factorise the left side into two linear brackets.
  3. Apply the zero-product property: set each factor equal to zero and solve.

Factorising only works neatly when the discriminant \( b^2-4ac \) is a perfect square. If not, use the quadratic formula (see B1664).

Special Case: Difference of Two Squares

\[ x^2 - k^2 = (x-k)(x+k) = 0 \implies x = k \text{ or } x = -k \]

This applies whenever the equation has no middle term and the constant is a perfect square (or can be made one).

Worked Examples

Solve \( x^2 - 5x + 6 = 0 \).
\[ (x-2)(x-3)=0 \implies x=2 \text{ or } x=3 \]
Solve \( x^2 + x - 12 = 0 \).
\[ (x+4)(x-3)=0 \implies x=-4 \text{ or } x=3 \]
Solve \( x^2 - 16 = 0 \) (difference of two squares).
\[ (x-4)(x+4)=0 \implies x=4 \text{ or } x=-4 \]
Solve \( 2x^2 + 5x - 3 = 0 \).

Find two numbers that multiply to \( 2 imes (-3) = -6 \) and add to \( 5 \): these are \( 6 \) and \( -1 \).

\[ 2x^2 + 6x - x - 3 = 2x(x+3) - 1(x+3) = (2x-1)(x+3)=0 \] \[ x=\frac{1}{2} \text{ or } x=-3 \]
Solve \( 3x^2 - 10x + 8 = 0 \).
\[ (3x-4)(x-2)=0 \implies x=\frac{4}{3} \text{ or } x=2 \]
Solve \( x^2 = 3x + 10 \).

Rearrange to zero: \( x^2 - 3x - 10 = 0 \).

\[ (x-5)(x+2)=0 \implies x=5 \text{ or } x=-2 \]
Solve \( 2x^2 = x + 3 \).

Rearrange: \( 2x^2 - x - 3 = 0 \).

\[ (2x-3)(x+1)=0 \implies x=\frac{3}{2} \text{ or } x=-1 \]

 Key Takeaways

  • Always rearrange to \( ax^2+bx+c=0 \) (zero on one side) before factorising.
  • Zero-product property: \(AB=0 Rightarrow A=0\) or \(B=0\). This only works when one side is zero.
  • Difference of two squares: \(x^2-k^2=(x-k)(x+k)\) — use when there is no \(x\) term.
  • For \(ax^2+bx+c\) with \(a eq 1\): find two numbers multiplying to \(ac\) and summing to \(b\), then split and group.
  • Always check both solutions by substituting back into the original equation.
  • A quadratic equation has at most two solutions. It can have zero, one (repeated) or two distinct solutions.