# 1. Special Products

### What makes these products "special"?

The algebraic products on this page are used all the time later in this chapter, and in a lot of the math you will come across later. They are "special" because they are very common, and they're worth knowing.

If you can recognize these products easily, it makes your life easier later on.

## Special Products involving Squares

The following special products come from multiplying out the brackets. You'll need these often, so it's worth knowing them well.

a(x+y) =ax+ay(Distributive Law)(

x+y)(x−y) =x^{2}−y^{2}(Difference of 2 squares)(

x+y)^{2}=x^{2}+ 2xy+y^{2}(Square of a sum)(

x−y)^{2}=x^{2}− 2xy+y^{2}(Square of a difference)

### Examples using the special products

**Example 1:** Multiply out 2*x*(*a* − 3)

Answer

This one uses the first product above. We just multiply the term outside the bracket (the "2*x*") with the terms inside the brackets (the "*a*" and the "−3").

2

x(a− 3) = 2ax− 6x

**Example 2:** Multiply (7*s* + 2*t*)(7*s* − 2*t*)

Answer

We recognize this one involves the **Difference of 2 squares**:

(7*s* + 2*t*)(7*s* − 2*t*)

= (7

s)^{2}− (2t)^{2}= 49

s^{2}− 4t^{2}

**Example 3:** Multiply (12* *+ 5*ab*)(12 − 5*ab*)

Answer

(12* *+ 5*ab*)(12 − 5*ab*)

= (12)

^{2}− (5ab)^{2}= 144

^{}− 25a^{2}b^{2}

The answer is a difference of 2 squares.

**Example 4:** Expand (5*a* + 2*b*)^{2}

Answer

This one is the **square of a sum of 2 terms**.

(5*a* + 2*b*)^{2}

= (5

a)^{2}+ 2(5a)(2b) + (2b)^{2}= 25

a^{2}+ 20ab+ 4b^{2}

**Example 5: **Expand (*q* − 6)^{2}

Answer

(*q* −
6)^{2}

= (

q)^{2}− 2(q)(6) + (6)^{2}=

q^{2}− 12q+ 36

This example involved the square of a difference of 2 terms.

**Example 6:** Expand (8*x* −* y*)(3*x* + 4*y*)

Answer

This question is not in any of the formats we have above. So we just need to multiply out the brackets, term-by-term.

It's important to recognize when we have a Special Product and when our question is something else.

(8*x* −* y*)(3*x* + 4*y*)

= 8

x(3x+ 4y) −y(3x+ 4y)= 8

x(3x) + 8x(4y) −y(3x) −y(4y)= 24

x^{2}+ 32xy−3xy− 4y^{2}= 24

x^{2}+ 29xy− 4y^{2}

**Example 7:** Expand (*x* + 2 + 3*y*)^{2}

Answer

To expand this, we put it in the form (*a* + *b*)^{2} and expand it using the third rule above, which says:

(

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

I put

a=x+ 2

b= 3y

This gives me:

(*x* + 2 + 3*y*)^{2}

[This is the (

a+b)^{2}step.]

= ([*x* + 2] + 3*y*)^{2}

= [*x* + 2]^{2} + 2[*x* + 2](3*y*) + (3*y*)^{2}

[Here I apply: (

a+b)^{2}=a^{2}+ 2ab+b^{2}]

= [*x*^{2} + 4*x* + 4] + (2*x* + 4)(3*y*) + 9*y*^{2}

[In this row I just expand out the brackets.]

= *x*^{2} + 4*x* + 4 + 6*xy* + 12*y* + 9*y*^{2}

[This is a "tidy up" step.]

I could have chosen the following and obtained the same answer:

a=x

b= 2 + 3y

Try it!

## Special Products involving Cubes

The following products are just the result of multiplying out the brackets.

(

x+y)^{3}=x^{3}+ 3x^{2}y+ 3xy^{2}+y^{3}(Cube of a sum)(

x−y)^{3}=x^{3}− 3x^{2}y+ 3xy^{2 }−y^{3}(Cube of a difference)(

x+y)(x^{2}−xy+y^{2}) =x^{3}+y^{3 }(Sum of 2 cubes)(

x−y)(x^{2}+xy+y^{2}) =x^{3 }−y^{3 }(Difference of 2 cubes)

These are also worth knowing well enough so you recognize the form, and the differences between each of them. (Why? Because it's easier than multiplying out the brackets and it helps us solve more complex algebra problems later.)

**Example 8: ** Expand(2*s* + 3)^{3}

Answer

This involves the Cube of a Sum:

(2*s* + 3)^{3}

= (2*s*)^{3} + 3(2*s*)^{2}(3)
+ 3(2*s*)(3)^{2} + (3)^{3}

= 8*s*^{3} + 36*s*^{2}* + * 54*s *+ 27

### Exercises

Expand:

(1) (*s *+ 2*t*)(*s *− 2*t*)

Answer

Using the Difference of 2 Squares formula

(

x+y)(x−y) =x^{2}−y^{2},

we have:

(*s* + 2*t*)(*s* − 2*t*)

= (

s)^{2}− (2t)^{2}= s

^{2}− 4t^{2}

(2) (*i*_{1} + 3)^{2}

Answer

Using the Square of a Sum formula

(

x+y)^{2}=x^{2}+ 2xy+y^{2},

we have:

(*i*_{1} +
3)^{2}

= (

i_{1})^{2}+ 2(i_{1})(3) + (3)^{2}=

i_{1}^{2}+ 6i_{1}+ 9

(3) (3*x *+ 10*y*)^{2}

Answer

Using the Square of a Sum formula

(

x+y)^{2}=x^{2}+ 2xy+y^{2},

we have:

(3*x *+
10*y*)^{2}

= (3

x)^{2 }+ (2)(3x)(10y) + (10y)^{2}= 9

x^{2}+ 60xy+ 100y^{2}

(4) (3*p *− 4*q*)^{2}

Answer

Using the Square of a Difference formula

(

x−y)^{2}=x^{2}− 2xy+y^{2},

we have:

(3*p *−
4*q*)^{2}

= (3

p)^{2}− (2)(3p)(4q) + (4q)^{2}= 9

p^{2}− 24pq+ 16q^{2}