An algebraic expression whose highest power of the variable is 2.
Standard form: \(ax^{2}+bx+c\) where \(a\neq 0\).
A quadratic equation is formed by setting a quadratic expression to zero: \(ax^2 + bx + c = 0\). The values of \(x\) that satisfy it are called roots; they are where the parabola crosses the x-axis.
\(x^2 - 5x + 6 = 0\) → roots \(x = 2, 3\).
Factorisation + Zero-Product Property lets us read roots instantly.
If \(AB = 0\), at least one factor must be zero.
Rewrite \(ax^{2}+bx+c\) as \((A)(B)=0\) by factorising.
Solve \(A = 0\) and \(B = 0\) to get both roots quickly.
Look for factor pairs of \(a \times c\) to spot brackets fast.
Middle-term splitting helps factorise quadratics when \(a = 1\).
Choose two numbers whose product equals \(c\) and sum equals \(b\).
Rewrite \(bx\) as \(mx + nx\) using those numbers.
Group terms in pairs and factor their GCFs.
Write the common bracket, set each factor to zero, solve for \(x\).
Example—\(x^{2}+7x+12\): numbers are 3 and 4 because \(3\times4=12\) and \(3+4=7\).
Example 1 – Easy coefficients
Solve \(x^{2}+5x+6=0\) by factorisation.
Product \(=6\), sum \(=5\). Choose \(2\) and \(3\).
\(2 \times 3 = 6,\; 2 + 3 = 5\)
Rewrite the quadratic using \(2\) and \(3\).
\(x^{2}+2x+3x+6=0\)
Choosing the right pair of numbers makes factorisation quick and error-free.
Group terms, extract the common bracket, then apply the zero-product rule.
\(x(x+2)+3(x+2)=0\\(x+2)(x+3)=0\\x=-2 \text{ or } -3\)
Roots: \(x=-2\) or \(x=-3\)
Both values satisfy \(x^{2}+5x+6=0\).
Which factorised form solves \(x^{2}+7x+12=0\)?
Find two numbers that add to 7 and multiply to 12.
Great! \(4+3=7\) and \(4\\times3=12\).
Recheck the sum and product of your chosen pair.
Arrange the four steps needed to factorise \(x^{2}+8x+15\) by middle-term splitting.
Step 1
Step 2
Step 3
Step 4
Remember: 5 × 3 = 15 and 5 + 3 = 8.
Coefficient a ≠ 1 | Factorisation Method
Solve \(2x^{2}+7x+3=0\) by factorisation.
\(a \times c = 2 \times 3 = 6\).
\(ac = 6\)
Numbers with product 6 and sum 7 are 6 and 1. Rewrite as \(2x^{2}+6x+x+3=0\).
\(2x^{2}+6x+x+3=0\)
For \(a>1\), always start with \(ac\). The splitting method then works just like when \(a=1\).
\(2x(x+3)+1(x+3)=0 \Rightarrow (x+3)(2x+1)=0\).
\((x+3)(2x+1)=0\)
Roots: \(x=-3\) or \(x=-\\dfrac{1}{2}\)
Factorisation complete; learning outcome achieved.
Write the quadratic as \(ax^{2}+bx+c=0\).
Factorise into two brackets \( (A)(B)=0 \) and set each to zero.
Split the middle term: numbers multiply to \(c\) and add to \(b\).
If \(a\neq1\), use the product \(a\times c\) instead.
Expand to verify the factors match the original equation.
Practise on varied quadratics; test both \(a=1\) and \(a\neq1\), then check roots by substitution.
Thank You!
We hope you found this lesson informative and engaging.