In this chapter we will the formula to expand the trinomial with power 2.

After completing the chapter, you will be able to expand any given trinomial with power 2 with shortcut techniques.

The solved problems are also given at the end for your practice.

Before moving on with the formulas, let us revise the basics first.

## What are trinomials ?

The **algebraic expressions containing three terms** are called** trinomial**.

Given below are examples of trinomial with power 2.

\mathtt{\Longrightarrow \ ( x+y+z)^{2}}\\\ \\ \mathtt{\Longrightarrow \ ( 2x-2+z)^{2}}\\\ \\ \mathtt{\Longrightarrow \ \left( a+\frac{1}{5} -b\right)^{2}}

I hope you understood the above concept. Let us now look at the formula which we will use to expand trinomial with power 2.

## Formula for trinomial with power 2.

Let us consider the trinomial \mathtt{\ ( a+b+c)^{2}}

The above trinomial can be expressed as;

\mathtt{\Longrightarrow \ ( a+( b+c))^{2}}

Referring the formula;

\mathtt{( x+y)^{2} =x^{2} +2xy+y^{2}}

Using the formula, we get;

\mathtt{\Longrightarrow \ ( a+( b+c))^{2}}\\\ \\ \mathtt{\Longrightarrow \ a^{2} +2a( b+c) +( b+c)^{2}}\\\ \\ \mathtt{\Longrightarrow \ a^{2} +2ab+2ac\ +\ b^{2} +c^{2} +2bc}\\\ \\ \mathtt{\Longrightarrow \ a^{2} +b^{2} +c^{2} +2ab+2bc+2ca}

**Hence, we get the following formula;**

\mathtt{( a+b+c)^{2} =\ a^{2} +b^{2} +c^{2} +2ab+2bc+2ca}

This is **very important formula**. Please memorize this as it will help you solve problems fast and easy way.

### Other related formulas

Using the above mentioned formula, one can also derive different variations by changing variables from left to right.

On such formula is;

\mathtt{a^{2} +b^{2} +c^{2} =\ ( a+b+c)^{2} -2ab-2bc-2ca}

This formula is not important,** so you don’t need to memorize it**. I have put this expression just to give an idea about different variations of the same formula.

I hope you understood the above concepts. Let us now look at the examples for practice.

## \mathtt{( a+b+c)^{2}} formula – Solved problems

**Example 01**

Expand the expression \mathtt{( x+2y+5)^{2}}

**Solution**

Referring the formula;

\mathtt{( a+b+c)^{2} =\ a^{2} +b^{2} +c^{2} +2ab+2bc+2ca}

Here;

a = x

b = 2y

c = 5

Using the formula, we get;

\mathtt{\Longrightarrow \ x^{2} +( 2y)^{2} +5^{2} +2( x)( 2y) +2( 2y)( 5) +2( 5)( x)}\\\ \\ \mathtt{\Longrightarrow \ x^{2} +4y^{2} +25+4xy\ +20y+10x}

Hence, the above expression is the solution.

**Example 02**

Expand the expression \mathtt{( x-y+3)^{2}}

**Solution**

Referring the formula;

\mathtt{( a+b+c)^{2} =\ a^{2} +b^{2} +c^{2} +2ab+2bc+2ca}

Here;

a = x

b = -y

c = 3

Putting the values in formula, we get;

\mathtt{\Longrightarrow \ x^{2} +( -y)^{2} +3^{2} +2( x)( -y) +2( -y)( 3) +2( 3)( x)}\\\ \\ \mathtt{\Longrightarrow \ x^{2} +y^{2} +9-2xy-6y+6x}

Hence, the above expression is the solution.

**Example 03**

Expand the expression \mathtt{( 5-y-z)^{2}}

**Solution**

Again it is the question involving trinomial with power 2.

Referring the formula;

\mathtt{a^{2} +b^{2} +c^{2} =\ ( a+b+c)^{2} -2ab-2bc-2ca}

Here;

a = 5

b = -y

c = -z

Putting the values;

\mathtt{\Longrightarrow \ 5^{2} +( -y)^{2} +( -z)^{2} +2( 5)( -y) +2( -y)( -z) +2( -z)( 5)}\\\ \\ \mathtt{\Longrightarrow \ 25+y^{2} +z^{2} -10y+2yz-10z}

Hence, the above expression is the solution.

**Example 04**

Expand the expression \mathtt{\left( x+\frac{1}{2} +\frac{y}{3}\right)^{2}}

**Solution**

Here;

a = x

b = 1/2

c = y/3

Putting the values in square of trinomial formula, we get;

\mathtt{\Longrightarrow \ x^{2} +\left(\frac{1}{2}\right)^{2} +\left(\frac{y}{3}\right)^{2} +2( x)\left(\frac{1}{2}\right) +2\left(\frac{1}{2}\right)\left(\frac{y}{3}\right) +2\left(\frac{y}{3}\right)( x)}\\\ \\ \mathtt{\Longrightarrow \ x^{2} +\frac{1}{4} +\frac{y^{2}}{9} +x+\frac{y}{3} +\frac{2xy}{3}}\\\ \\ \mathtt{\Longrightarrow \ x^{2} +\frac{y^{2}}{9} +\frac{2xy}{3} +x+\frac{y}{3} \ +\frac{1}{4}}

Hence, the above expression is the solution.

**Example 05**

If x + y + z = 10 and xy + yz + zx = 15.

Find value of \mathtt{x^{2} +y^{2} +z^{2}}

**Solution**

Using the formula;

\mathtt{( x+y+z)^{2} =\ x^{2} +y^{2} +z^{2} +2( xy+yz+zx)}

Putting the values;

\mathtt{10^{2} =x^{2} +y^{2} +z^{2} +2\ ( 15)}\\\ \\ \mathtt{100\ =x^{2} +y^{2} +z^{2} \ +\ 30}\\\ \\ \mathtt{\ \ x^{2} +y^{2} +z^{2} \ =\ 100\ -\ 30}\\\ \\ \mathtt{x^{2} +y^{2} +z^{2} \ =\ 70}

Hence, **70 is the required solution**.

**Example 06**

Expand the expression \mathtt{\left( x^{4} +x^{2} +x\right)^{2}}

**Solution**

Referring the formula;

\mathtt{( a+b+c)^{2} =\ a^{2} +b^{2} +c^{2} +2ab+2bc+2ca}

Where;

a = \mathtt{x^{4}}

b = \mathtt{x^{2}}

c = x

Putting the values;

\mathtt{\Longrightarrow \ \left( x^{4}\right)^{2} +\left( x^{2}\right)^{2} +( x)^{2} +2\left( x^{4}\right)\left( x^{2}\right) +2\left( x^{2}\right)( x) +2( x)\left( x^{4}\right)}\\\ \\ \mathtt{\Longrightarrow \ x^{8} +x^{4} +x^{2} +2x^{6} +2x^{3} +2x^{5}}\\\ \\ \mathtt{Rearranging\ the\ terms}\\\ \\ \mathtt{\Longrightarrow \ x^{8} +2x^{6} +2x^{5} +x^{4} +2x^{3} +x^{2}}

Hence, the above expression is solution.

**Next chapter** : **Formula for (x + y)(x – y)**