Group Symbols and Binomials

Different Types of Parentheses
Different Types of Parentheses

Removing Group Symbols

 

Sometimes, algebraic expressions are written in group symbols such as:

2(a+3)

...in which you are asked to expand and simplify as seen below:

= 2 \times a + 2 \times 3
 = 2a + 6

In these cases, the distributive law is used to remove group symbols:

a(b+c)=ab+ac

Examples:

 

-(2x-5)\\ =(-1)(2x-5)\\ =(-1)\times 2x+\left( -1 \right) \times -5\\ =-2x+5

 

{ 5a }^{ 2 }(4+3ab-c)\\ ={ 5a }^{ 2 }\times 4+{ 5a }^{ 2 }\times 3ab-{ 5a }^{ 2 }\times c\\ =20{ a }^{ 2 }+15{ a }^{ 3 }b-5{ a }^{ 2 }c

 

2(b-5)-(b+1)\\ =2b-10-b-1\\ =b-11

Binomial Products

 

A binomial expression consists of two elements, for example x+3.

A set of two binomial expressions multiplied together is called a binomial product.

Binomial products are multiplied in FOIL order, that is:

  1. First elements of both expressions.
  2. Outer elements of both expressions.
  3. Inner elements of both expressions.
  4. Last elements of both expressions.

Hence, the products are multiplied as follows:

(a+b)(x+y)=ax+ay+bx+by

Examples:

 

(a+4)(b-2)=ab-2a+4b-8

 

(2{ a }^{ 2 }-4)(-5+6b)\\ =2{ a }^{ 2 }\times -5\quad +\quad 2{ a }^{ 2 }\times 6b\quad +\quad -4\times -5\quad +\quad -4\times 6b\\ =-10{ a }^{ 2 }+12{ a }^{ 2 }b+20-24b

 

Sometimes we see a trinomial that consists of three elements, and when multiplied it works the same way.

(a+b)(x+y+z)=ax+ay+az+bx+by+bz

For example:

(x+4)(2x-3y-1)\\ =\quad 2{ x }^{ 2 }-3xy-x+8x-12y-4\\ =\quad 2{ x }^{ 2 }-3xy+7x-12y-4

Perfect Squares

 

Several characteristics of binomial products are shown below:

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

This is because:

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

The same can be done when:

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

...and the proof as follows:

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

(Remember: two negatives multiplied will yield a positive result)

There is also a the difference of 2 squares as seen below:

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

Reflection: What is the difference between { (a-b) }^{ 2 } and { a }^{ 2 }-{ b }^{ 2 } ?

  

 Examples:

Expand and simplify the following:

 

{ (3x+4) }^{ 2 }\\ =\quad { (3x) }^{ 2 }+2(3x)(4)+{ 4 }^{ 2 }\\ =\quad 9{ x }^{ 2 }+24x+16

 

(3y-4)(3y+4)\\ =\quad { (3y) }^{ 2 }-{ 4 }^{ 2 }\\ =\quad { 9y }^{ 2 }-16

Factorising

 

When factorising a binomial product, take out the highest common numbers (or pronumerals) that perfectly divide both elements without leaving a remainder. This also applies to trinomials, which will be discussed further in another section.

To factorise an expression, we use the distributive law.

ax+bx=x(a+b)

(Reflection: Do you remember what HCF - Highest Common Factor - means?)

Examples:

 

Factorise:

 

1. 3x+12

Show Answer

The highest common factor is 3.

3x+12 = 3(x+4)

 

2. { p }^{ 3 }-3{ p }^{ 2 }

Show Answer

The common factor is {p}^{2}.

{ p }^{ 3 }-3{ p }^{ 2 } = { p }^{ 2 }(p-3)

 

3. 16{ x }^{ 5 }{ y }^{ 3 }-8{ x }^{ 2 }{ y }^{ 2 }+2{ x }^{ 4 }y

Show Answer

If there are several common factors, take out all the factors. In this case, the common factor is 2{x}^{2}{y}.

16{ x }^{ 5 }{ y }^{ 3 }-8{ x }^{ 2 }{ y }^{ 2 }+2{ x }^{ 4 }y\\ =\quad 2{ x }^{ 2 }y(8{ x }^{ 3 }{ y }^{ 2 }-4y+{ x }^{ 2 })

 

 

Sometimes you will see expressions that are grouped in pairs, and therefore they could be factorised in pairs:

4. 2(a-7)+a(a-7)

Show Answer

2(a-7)+a(a-7)\\ =\quad (2+a)(a-2) 

 

5. 2x-4+6y-3xy

Show Answer

2x-4+6y-3xy\\ =\quad 2(x-2)+3y(2-x)\\=\quad 2(x-2)-3y(-1)(x-2) \qquad <- notice\quad  the\quad  sign\quad  changes\\ =\quad 2(x-2)+3y(x-2)\\ =\quad (2+3y)(x-2) 

 

 

A special case of binomial products is:

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

 

6. Factorise { d }^{ 2 }-36

Show Answer

{ d }^{ 2 }-36\\=\quad { d }^{ 2 }-{ 6 }^{ 2 }\\ =\quad (d+6)(d-6) 

 

7. Factorise { (a+3) }^{ 2 }-{ (b-1) }^{ 2 }

Show Answer

{ (a+3) }^{ 2 }-{ (b-1) }^{ 2 }\\ =\quad [(a+3)+(b-1)][(a+3)-(b-1)]\\ =\quad (a+3+b-1)(a+3-b+1)\\ =\quad (a+b+2)(a-b+4) 

 

 

There is also an interesting way to factorise the sums and differences of 2 cubes.

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

 

8. Factorise 8{ x }^{ 3 }+27{ y }^{ 3 }

Show Answer

8{ x }^{ 3 }+27{ y }^{ 3 } \\ =\quad { (2x) }^{ 3 }+{ (3y) }^{ 3 }\\ =\quad (2x+3y)[{ (2x) }^{ 2 }-(2x)(3y)+{ (3y) }^{ 2 }]\\ =\quad (2x+3y)(4{ x }^{ 2 }-6xy+9{ y }^{ 2 })