Factorising (VCE SSCE Mathematical Methods): Revision Notes

Worked Example: Factorising with Common Factors

Factorise $9x^2 + 81x$

Solution:

The common factor $9x$ has been extracted from both terms. We can check this answer by expanding: $9x(x + 9) = 9x^2 + 81x$ ✓

Worked Example: Common Factors with Multiple Variables

Factorise $2a^2 - 8ax^2$

Solution:

The common factor $2a$ has been taken outside the brackets.

Worked Example: Identifying All Common Factors

Factorise $7x^2y - 35xy^2$

Solution:

Here, the common factor is $7xy$. Notice that both numerical coefficients (7 and 35) share a factor of 7, and both terms contain at least one $x$ and one $y$.

Worked Example: Factorising by Grouping

Factorise $x^3 + 4x^2 - 3x - 12$

Solution:

Explanation:

First, we group the four terms into two pairs

Next, we factorise each pair separately

Finally, we notice that $(x + 4)$ is common to both terms, so we extract it

Worked Example: Difference of Two Squares with Common Factor

Factorise $3x^2 - 75$

Solution:

Explanation:

First, we extract the common factor of 3

Then we recognise that $x^2 - 25$ is a difference of two squares: $x^2 - 5^2$

We apply the identity with $a = x$ and $b = 5$

Worked Example: Recognising Perfect Squares

Factorise $9x^2 - 36$

Solution:

We first take out the common factor 9, then apply the difference of squares identity to $x^2 - 4 = x^2 - 2^2$.

Worked Example: Complex Difference of Squares

Factorise $(x - y)^2 - 16y^2$

Solution:

Explanation:

We recognise this as a difference of two squares with $a = (x - y)$ and $b = 4y$

Applying the identity: $a^2 - b^2 = (a + b)(a - b)$

Simplifying the brackets gives our final answer

Worked Example: Monic Quadratic

Factorise $x^2 - 2x - 8$

Solution:

Explanation:

We want $x^2 - 2x - 8 = (x + a)(x + b)$

When expanded, this gives $x^2 + (a + b)x + ab$

Therefore, we need: $ab = -8$ and $a + b = -2$

The values that work are $a = -4$ and $b = 2$

Check: $-4 \times 2 = -8$ ✓ and $-4 + 2 = -2$ ✓

Worked Example: Cross-Multiplication Method

Factorise $6x^2 - 13x - 15$

Solution:

Method:

We list possible factors of $6x^2$: could be $6x$ and $x$, or $3x$ and $2x$

We list factors of $-15$: could include $+5$ and $-3$, or $-5$ and $+3$, etc.

We try combinations until the cross-products add to give $-13x$:

• $6x \times (-3) = -18x$
• $x \times 5 = 5x$
• Sum: $-18x + 5x = -13x$ ✓

Worked Example: The ac Method

Factorise $6x^2 - 13x - 15$ using the grouping method

Solution:

First, find $ac = 6 \times (-15) = -90$ and $b = -13$

We need two numbers that multiply to $-90$ and add to $-13$

These numbers are $-18$ and $5$: $(-18) \times 5 = -90$ and $-18 + 5 = -13$

Now we split the middle term:

Worked Example: Another Non-Monic Quadratic

Factorise $8x^2 + 2x - 15$

Solution:

$ac = 8 \times (-15) = -120$ and $b = 2$

We need two numbers that multiply to $-120$ and add to $2$

These are $12$ and $-10$

Worked Example: Extracting Common Factors First

Factorise $2x^2 + 6x - 20$

Solution:

Explanation:

First, we extract the common factor of 2

This leaves us with a simpler monic quadratic to factorise

We find two numbers that multiply to $-10$ and add to $3$: these are $5$ and $-2$

Worked Example: Substitution Technique

Factorise $(x + 1)^2 - 2(x + 1) - 3$

Solution:

Let $a = x + 1$ to make the expression easier to work with

Substituting back $a = x + 1$:

This substitution technique is useful when we have a repeated expression in brackets.