# 2. Common Factor and Difference of Squares

**Factoring** means writing an expression as the product of
its simplest factors.

### Example 1: Factoring a number

14 = 7 × 2

[7 and 2 are the **simplest factors** of 14. We can't break it down any more than this.]

### Example 2: Factoring an algebraic expression

3

x+ 15 = 3(x+ 5)

This means that the **factors** of 3*x* + 15 are

3, and

(

x+ 5)

To be able to factor successfully, we need to recognise the formulas from Section 1. So it's a good idea to learn those formulas well!

## Factoring Difference of Two Squares

To factor the difference of 2 squares, we just apply the formula given in Section 1 - Special Products in reverse. That is:

x^{2}−y^{2}= (x+y)(x−y)

### Example 3: Factoring difference of 2 squares

Factor 36*s*^{2} − 121*t*^{2}

Answer

We first recognize it is a difference of 2 squares, then we use the formula given above.

36*s*^{2} − 121*t*^{2}

= (6

s)^{2 }− (11t)^{2}= (6

s− 11t)(6s+ 11t)

### Exercises

Factor the following:

**(1)** 18*p*^{3} − 3*p*^{2}

Answer

We note that 3*p*^{2} divides evenly into the 2 terms in the question. So the factorization is given by:

18*p*^{3} − 3*p*^{2} =
3*p*^{2}(6*p* − 1)

**(2) **5*a* + 10*ax* − 5*ay* +
20*az*

Answer

5*a* + 10*ax* − 5*ay* + 20*az* =
5*a*(1 + 2*x* − *y* + 4*z*)

**(3)** 36*a*^{2}*b*^{
2} − 169*c*^{2}

Answer

This is a difference of 2 squares.

36*a*^{2}*b*^{2} −
169*c*^{2}

= (6

ab)^{2}− (13c)^{2}= (6

ab+ 13c)(6ab− 13c)

**(4)** (*a* −
*b*)^{2} − 1

Answer

Once again, we recognize this as a difference of 2 squares.

(*a* − *b*)^{2} − 1 =
(*a* − *b*)^{2} −
1^{2}

Put *X *=* a *−* b* and *Y* = 1

So (*a* − *b*)^{2} − 1

=

X^{2}−Y^{2}= (

X+Y)(X−Y)= (

a−b+ 1)(a−b− 1)

**(5) ***y*^{4} − 81

Answer

*y*^{4} − 81

= (

y^{2})^{2}− (9)^{2}= (

y^{2}+ 9)(y^{2}− 9)= (

y^{2}+ 9)(y+ 3)(y− 3)

**(6) ***r*^{2}* *− *s*^{2 }+ 2*st ** *− *t*^{2}

Answer

We recognize that this involves 2 differences of two squares. We group it as follows:

*r*^{2}* *− *s*^{2 }+ 2*st ** *− *t*^{2}

= *r*^{2}* *− (*s*^{2 }− 2*st *+ *t*^{2})

We recognize that *s*^{2 }− 2*st *+ *t*^{2} is a square, and equals (*s* − *t*)^{2}. So we can factor our expression as follows:

*r*^{2}* *− *s*^{2 }+ 2*st ** *− *t*^{2}

= *r*^{2}* *− (*s*^{2 }− 2*st *+ *t*^{2})

= *r*^{2}* *− (*s* − *t*)^{2}

[This is also a difference of 2 squares.]

= [*r* − (*s* − *t*)][*r* + (*s* − *t*)]

= (*r* − *s* + *t*)(*r* + *s* − *t*)