# Factoring in Algebra

## Factors

Numbers have factors:

And expressions (like **x ^{2}+4x+3**) also have factors:

## Factoring

Factoring (called "**Factorising**" in the UK) is the process of **finding the factors**:

Factoring: Finding what to multiply together to get an expression.

It is like "splitting" an expression into a multiplication of simpler expressions.

Example: factor 2y+6

Both 2y and 6 have a common factor of 2:

- 2y is 2 × y
- 6 is 2 × 3

So you can factor the whole expression into:

2y+6 = 2(y+3)

So, 2y+6 has been "factored into" 2 and y+3

Factoring is also the opposite of Expanding:

## Common Factor

In the previous example we saw that 2y and 6 had a common factor of **2**

But to do the job properly make sure you have the **highest common factor**, including any variables

Example: factor 3y^{2}+12y

Firstly, 3 and 12 have a common factor of 3.

So you could have:

3y^{2}+12y = 3(y^{2}+4y)

But we can do better!

3y^{2} and 12y also share the variable y.

Together that makes 3y:

- 3y
^{2}is 3y × y - 12y is 3y × 4

So you can factor the whole expression into:

3y^{2}+12y = 3y(y+4)

Check: **3y(y+4) = 3y × y + 3y × 4 = 3y ^{2}+12y**

## More Complicated Factoring

### Factoring Can Be Hard !

The examples have been simple so far, but factoring **can** be very tricky.

Because you have to figure **what got multiplied** to produce the expression you are given!

It can be like trying to find out what ingredients went into a cake to make it so delicious. It is sometimes not obvious at all! |

### Experience Helps

But the more experience you get, the easier it becomes.

### Example: Factor **4x**^{2} - 9

^{2}- 9

Hmmm... I can't see any common factors.

But if you know your Special Binomial Products you might see it as the **"difference of squares"**:

Because **4x ^{2}** is

**(2x)**, and

^{2}**9**is

**(3)**,

^{2}so we have:

4x^{2} - 9 = (2x)^{2} - (3)^{2}

And that can be produced by the difference of squares formula:

(a+b)(a-b) = a^{2} - b^{2}

Where **a** is 2x, and **b** is 3.

So let us try doing that:

(2x+3)(2x-3) = (2x)^{2} - (3)^{2} = 4x^{2} - 9

Yes!

So the factors of **4x ^{2} - 9** are

**(2x+3)**and

**(2x-3)**:

Answer: 4x^{2} - 9 = (2x+3)(2x-3)

How can you learn to do that? By getting lots of practice, and knowing "Identities"!

### Remember these Identities

Here is a list of common "Identities" (including the **"difference of squares"** used above).

It is worth remembering these, as they can make factoring easier.

a^{2} - b^{2} |
= | (a+b)(a-b) |

a^{2} + 2ab + b^{2} |
= | (a+b)(a+b) |

a^{2} - 2ab + b^{2} |
= | (a-b)(a-b) |

a^{3} + b^{3} |
= | (a+b)(a^{2}-ab+b^{2}) |

a^{3} - b^{3} |
= | (a-b)(a^{2}+ab+b^{2}) |

a^{3}+3a^{2}b+3ab^{2}+b^{3} |
= | (a+b)^{3} |

a^{3}-3a^{2}b+3ab^{2}-b^{3} |
= | (a-b)^{3} |

There are many more like those, but those are the simplest ones.

## Advice

The factored form is usually best.

When trying to factor, follow these steps:

- "Factor out" any common terms
- See if it fits any of the identities, plus any more you may know
- Keep going till you can't factor any more

You can also use computers! There are Computer Algebra Systems (called "CAS") such as *Axiom, Derive, Macsyma, Maple, Mathematica, MuPAD, Reduce* and many more that are good at factoring.

## More Examples

I said that experience helps, so here are more examples to help you on the way:

### Example: w^{4} - 16

An exponent of 4? Maybe we could try an exponent of 2:

w^{4} - 16 = (w^{2})^{2 }- 4^{2}

Yes, it is the difference of squares

w^{4} - 16 = (w^{2}^{ }+ 4)(w^{2}^{ }- 4)

And "(w^{2}^{ }- 4)" is another difference of squares

w^{4} - 16 = (w^{2}^{ }+ 4)(w^{ }+ 2)(w^{ }- 2)

That is as far as I can go (unless I use imaginary numbers)

### Example: 3u^{4} - 24uv^{3}

Remove common factor "3u":

3u^{4} - 24uv^{3} = 3u(u^{3} - 8v^{3})

Then a difference of cubes:

3u^{4} - 24uv^{3} = 3u(u^{3} - (2v)^{3})

= 3u(u-2v)(u^{2}+2uv+4v^{2})

That is as far as I can go.

### Example: z^{3} - z^{2} - 9z + 9

Try factoring the first two and second two separately:

z^{2}(z-1) - 9(z-1)

Wow, (z-1) is on both, so let us use that:

(z^{2}-9)(z-1)

And z^{2}-9 is a difference of squares

(z-3)(z+3)(z-1)

That is as far as I can go.