Junior Cycle Higher Level Mathematics — Factorising
Four methods with area models; then your questions with typed headings and hide/reveal solutions. Applications: solving equations and simplifying algebraic fractions.
1. Highest Common Factor (HCF)
Method. Take out the greatest common numerical factor and the highest common power of each variable.
Area model (1×2). Put the original terms in the two cells. The top headings are the factors (each term divided by the HCF). The left label is the HCF. The sides of the rectangle are the factors in the final bracket.
Example: \(6x^3+9x^2\) → HCF \(=3x^2\).
So \(6x^3+9x^2=3x^2(2x+3)\).
2. By Grouping
Method. Arrange the four terms as \(A,B,C,D\) in a \(2\times2\) array so that the cross products are equal: \(AD = BC\). Then factorise each row/column; a common bracket appears.
Generic area model:
Choose the arrangement so that \(AD = BC\) — this ensures a common bracket appears.
Example: \(5fh - 2h^2 - 6h + 15f\).
Factors: \((5f - 2h)(h + 3)\).
3. Difference of Two Squares (DOTS)
Form. \(a^2-b^2=(a-b)(a+b)\). In the area model, \(+ab\) and \(-ab\) cancel, so there is no middle term.
Example: \(16a^2-9b^2=(4a-3b)(4a+3b)\).
4. Quadratic Expressions \(ax^2+bx+c\)
Method. Find \(m,n\) with \(mn=ac\) and \(m+n=b\). Split \(bx\) into \(mx+nx\) and factor by grouping (2×2 array).
Example: \(6x^2+7x-3\): choose \(m=9,\;n=-2\).
Hence \(6x^2+7x-3=(3x-1)(2x+3)\).
HCF (1×2 model)
Sides (factors): height \(=3x^2\), widths \(=2x,3\). Therefore \(6x^3+9x^2=\boxed{3x^2(2x+3)}\).
Grouping (2×2) with \(AD=BC\). Choose \(A=5fh,\;B=15f,\;C=-2h^2,\;D=-6h\), giving \(AD=BC=-30fh^2\).
Sides (factors): left \(5f, -2h\); top \(h, 3\). Hence \(\boxed{(5f-2h)(h+3)}\).
Quadratic via 2×2 model. Need \(mn=-144,\;m+n=45\Rightarrow (48,-3)\).
Sides (factors): left \(8x,-3\); top \(x,6\). Hence \(\boxed{(8x-3)(x+6)}\).
DOTS.
Sides (factors): left \(5x,7n\); top \(5x,-7n\). Hence \(\boxed{(5x+7n)(5x-7n)}\).
Factor both the numerator and denominator using area models.
Numerator: \(2n^2+n-15\)
We need two numbers whose product is \(2\times(-15)=-30\) and sum is \(+1\). These are \(+6\) and \(-5\).
Sides (factors): Left side \(2n, -5\); top \(n, 3\). Hence numerator \(= (2n - 5)(n + 3)\).
Denominator: \(n^2 - 9\)
This is a difference of two squares: \(n^2 - 3^2\).
Sides (factors): Left side \(n, 3\); top \(n, -3\). Hence denominator \(= (n + 3)(n - 3)\).
Now simplify by cancelling the common factor \((n + 3)\):
\[ \frac{(2n - 5)\,\cancel{(n + 3)}}{(n - 3)\,\cancel{(n + 3)}} = \boxed{\frac{2n - 5}{n - 3}} \] with domain restriction \(n \neq \pm3\).
When simplifying fractions, always state any restrictions where a cancelled factor would be zero.
Factoring Quadratics — Algebra Tile Area Model
Factorisation:
Factorising — Multiple-Choice Quiz
Covers Highest Common Factor (HCF), Grouping, Difference of Two Squares (DOTS), and Quadratics. Two questions of each type; one question shown at a time. Solutions display the relevant area model.
