More factorising methods

You now have every main tool: com­mon fac­tors, group­ing, the iden­ti­ties and split­ting the mid­dle term. Harder expres­sions usu­ally need two or three of these tools in a row, plus a small trick to get started. This les­son col­lects the most use­ful tricks: split­ting a term to make group­ing work, adding and sub­tract­ing the same term, the spe­cial case $a + b + c = 0$, and fac­toris­ing again and again until noth­ing is left to split.

Split­ting a term so that group­ing works#

Exam­ple 1#

Fac­torise $x^{3} - 7x + 6$.

There are three terms and no com­mon fac­tor. Split $-7x$ as $-x - 6x$. The rea­son for this choice: $x^{3} - x$ has the fac­tor $x - 1$, and so does $-6x + 6$.

$$\begin{aligned} x^{3} - 7x + 6 &= x^{3} - x - 6x + 6 \\ &= x(x^{2} - 1) - 6(x - 1) \\ &= x(x - 1)(x + 1) - 6(x - 1) \\ &= (x - 1)[x(x + 1) - 6] \\ &= (x - 1)(x^{2} + x - 6) \end{aligned}$$

The qua­dratic still fac­torises. For $x^{2} + x - 6$ we need prod­uct $-6$ and sum $1$: the num­bers $3$ and $-2$.

$$x^{2} + x - 6 = x^{2} + 3x - 2x - 6 = x(x + 3) - 2(x + 3) = (x + 3)(x - 2)$$

So $x^{3} - 7x + 6 = (x - 1)(x + 3)(x - 2)$.

Adding and sub­tract­ing the same term#

Some­times an expres­sion is "almost" a per­fect square. Add the term that would com­plete the square, and sub­tract it again so that noth­ing has changed. The result is often a dif­fer­ence of two squares.

Exam­ple 2#

Fac­torise $x^{4} + x^{2} + 1$.

A per­fect square would be $(x^{2} + 1)^{2} = x^{4} + 2x^{2} + 1$, with $2x^{2}$ in the mid­dle. We have only $x^{2}$, so add $x^{2}$ and take it away again.

$$\begin{aligned} x^{4} + x^{2} + 1 &= (x^{2})^{2} + 2x^{2} + 1 - x^{2} \\ &= (x^{2} + 1)^{2} - x^{2} \\ &= (x^{2} + 1 + x)(x^{2} + 1 - x) \\ &= (x^{2} + x + 1)(x^{2} - x + 1) \end{aligned}$$

Exam­ple 3#

Fac­torise $x^{4} + 4$.

There is no mid­dle term at all. To com­plete the square $(x^{2} + 2)^{2}$ we need $2(x^{2})(2) = 4x^{2}$. Add it and sub­tract it:

$$\begin{aligned} x^{4} + 4 &= (x^{2})^{2} + (2)^{2} + 2(x^{2})(2) - 4x^{2} \\ &= (x^{2} + 2)^{2} - (2x)^{2} \\ &= (x^{2} + 2 + 2x)(x^{2} + 2 - 2x) \\ &= (x^{2} + 2x + 2)(x^{2} - 2x + 2) \end{aligned}$$

The trick works here because the term we sub­tracted, $4x^{2}$, is itself a per­fect square, $(2x)^{2}$.

When three things add up to zero#

There is an iden­tity for the sum of three cubes:

$$\text{if } a + b + c = 0, \text{ then } a^{3} + b^{3} + c^{3} = 3abc$$

It saves a great deal of work when­ever three cubed brack­ets add up to zero.

Exam­ple 4#

Fac­torise $(x - y)^{3} + (y - z)^{3} + (z - x)^{3}$.

Let $a = x - y$, $b = y - z$ and $c = z - x$. Then

$$(x - y) + (y - z) + (z - x) = 0$$

because every let­ter appears once with a plus sign and once with a minus sign. So by the iden­tity,

$$(x - y)^{3} + (y - z)^{3} + (z - x)^{3} = 3(x - y)(y - z)(z - x)$$

Expand­ing all three cubes would take a page; the iden­tity takes one line.

Keep going until noth­ing splits#

Exam­ple 5#

Fac­torise $xy^{9} - yx^{9}$.

Take out the com­mon fac­tor $xy$ first. What is left is a dif­fer­ence of two squares, and it keeps pro­duc­ing more of them.

$$\begin{aligned} xy^{9} - yx^{9} &= xy(y^{8} - x^{8}) \\ &= xy\left((y^{4})^{2} - (x^{4})^{2}\right) \\ &= xy(y^{4} + x^{4})(y^{4} - x^{4}) \\ &= xy(y^{4} + x^{4})(y^{2} + x^{2})(y^{2} - x^{2}) \\ &= xy(y^{4} + x^{4})(y^{2} + x^{2})(y + x)(y - x) \end{aligned}$$

Each dif­fer­ence of squares split again; each sum of squares stayed as it was. Stop only when no fac­tor can be bro­ken fur­ther.

Exam­ple 6: spot­ting a com­mon bracket#

Fac­torise $(x + 2)(x^{2} + 25) - 10x^{2} - 20x$.

The last two terms have the com­mon fac­tor $10x$, and what is left is $(x + 2)$, the same bracket as at the start.

$$\begin{aligned} (x + 2)(x^{2} + 25) - 10x^{2} - 20x &= (x + 2)(x^{2} + 25) - 10x(x + 2) \\ &= (x + 2)(x^{2} + 25 - 10x) \\ &= (x + 2)(x^{2} - 10x + 25) \\ &= (x + 2)(x - 5)^{2} \end{aligned}$$

The last step uses the per­fect square $a^{2} - 2ab + b^{2}$ with $a = x$ and $b = 5$.

A check­list for any expres­sion#

1. Take out any com­mon fac­tor.
2. Count the terms. Two terms: try a dif­fer­ence of squares, or a sum or dif­fer­ence of cubes. Three terms: try a per­fect square, then split­ting the mid­dle term. Four or more: try group­ing, or a per­fect cube.
3. If a bracket repeats, replace it by a sin­gle let­ter.
4. If noth­ing fits, try split­ting a term, or adding and sub­tract­ing a term.
5. Look at every fac­tor again and ask whether it splits fur­ther.
6. Check by expand­ing.

Prac­tice#

Fac­torise.

1. $x^{3} - 13x + 12$
2. $x^{4} + x^{2}y^{2} + y^{4}$
3. $4x^{4} + 1$
4. $(a - 2b)^{3} + (2b - 3c)^{3} + (3c - a)^{3}$
5. $x^{8} - 1$
6. $a^{2}x^{2} + (ax^{2} + 1)x + a$
7. $a(a + b)^{3} - 3a^{2}b(a + b)$
8. $x^{3} - 6x^{2} + 11x - 6$ (hint: write it as $x^{3} - x^{2} - 5x^{2} + 5x + 6x - 6$)

Answers#