Group Symbols and Binomials 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. $3x+12$

The highest common factor is 3. $3x+12 = 3(x+4)$

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

The common factor is ${p}^{2}$. ${ p }^{ 3 }-3{ p }^{ 2 } = { p }^{ 2 }(p-3)$

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

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. $2(a-7)+a(a-7)$ $2(a-7)+a(a-7)\\ =\quad (2+a)(a-2)$

5. $2x-4+6y-3xy$ $2x-4+6y-3xy\\ =\quad 2(x-2)+3y(2-x)\\=\quad 2(x-2)-3y(-1)(x-2) \qquad <- notice\quad the\quad sign\quad changes\\ =\quad 2(x-2)+3y(x-2)\\ =\quad (2+3y)(x-2)$

A special case of binomial products is: $(a+b)(a-b)={a}^{2}-{b}^{2}$

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

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

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