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)