In this chapter we will learn important formulas which will help to factorize the polynomials.
Note that these formulas are only applicable when the polynomial follow certain pattern.
This means that the formula can’t be used all type of polynomials.
At the end some solved examples are provided which will provide further clarity to the concept.
Factorization using formulas
Given below are formulas which are can be used to factorize the polynomial.
\mathtt{( i) \ a^{2} +2ab+b^{2} =( a+b)^{2}}\\\ \\ \mathtt{( ii) \ a^{2} -2ab+b^{2} =( a-b)^{2}}\\\ \\ \mathtt{( iii) \ ( a+b)( a-b) \ =a^{2} -b^{2}}\\\ \\ \mathtt{( iv) \ ( a-b)\left( a^{2} +ab+b^{2}\right) \ =a^{3} -b^{3}}\\\ \\ \mathtt{( v) \ ( a+b)\left( a^{2} -ab+b^{2}\right) =a^{3} +b^{3}}
You have to remember all the above formulas so that you can recall it while solving the related question.
I’ll suggest you to write these formulas multiple times on paper. This will help you to remember the formulas for long time.
Note:
The formulas are only applicable when polynomial is provided in above format.
Your job is to identify the pattern (if it exists !) and use the formula for factorization.
Given below are solved examples for your reference.
Factorization using Formulas – Solved examples
Example 01
Factorize the below polynomial
\mathtt{\Longrightarrow \ 9x^{2} +6xy\ +\ y^{2}}
Solution
The above polynomial can be written as;
\mathtt{\Longrightarrow \ 9x^{2} +6xy\ +\ y^{2}}\\\ \\ \mathtt{\Longrightarrow \ ( 3x)^{2} +2.( 3x)( y) +( y)^{2}}
Referring the formula;
\mathtt{a^{2} +2ab+b^{2} =( a+b)^{2}}
Here;
a = 3x
b = y
Using the above formula, the polynomial can be simplified as;
\mathtt{\Longrightarrow \ ( 3x)^{2} +2.( 3x)( y) +( y)^{2}}\\\ \\ \mathtt{\Longrightarrow \ ( 3x+y)^{2}}
Hence, \mathtt{( 3x+y)^{2}} is the factorized form of given polynomial.
Example 02
Factorize the below polynomial
\mathtt{\Longrightarrow \ x^{2} +25-10x}
Solution
The above polynomial can be expressed as;
\mathtt{\Longrightarrow \ x^{2} +25-10x}\\\ \\ \mathtt{\Longrightarrow \ ( x)^{2} +( 5)^{2} -2( x)( 5)}\\\ \\ \mathtt{\Longrightarrow \ ( x)^{2} -2( x)( 5) +( 5)^{2}}
Referring the formula;
\mathtt{\ a^{2} -2ab+b^{2} =( a-b)^{2}}
Here;
a = x
b = 5
Using the formula, the polynomial can be factorized as;
\mathtt{\Longrightarrow \ ( x)^{2} -2( x)( 5) +( 5)^{2}}\\\ \\ \mathtt{\Longrightarrow \ ( x\ -\ 5)^{2}}
Hence, the above polynomial is factorized to \mathtt{( x\ -\ 5)^{2}} .
Example 03
Factorize the below polynomial
\mathtt{\Longrightarrow \ x^{4} +2+\frac{1}{x^{4}}}
Solution
\mathtt{\Longrightarrow \ x^{4} +2+\frac{1}{x^{4}}}\\\ \\ \mathtt{\Longrightarrow \ \left( x^{2}\right)^{2} +2\ \left( x^{2}\right)\left(\frac{1}{x^{2}}\right) \ +\frac{1}{\left( x^{2}\right)^{2}}}
Referring the formula;
\mathtt{a^{2} +2ab+b^{2} =( a+b)^{2}}
Here;
a = \mathtt{x^{2}}
b = \mathtt{\frac{1}{x^{2}}}
Now using the formula, we get;
\mathtt{\Longrightarrow \ \left( x^{2} +\frac{1}{x^{2}}\right)^{2}}
Hence, the factorization of given polynomial is complete.
Example 04
Factorize the below polynomial
\mathtt{\Longrightarrow \ 25x^{2} -169}
Solution
\mathtt{\Longrightarrow \ 25x^{2} -169}\\\ \\ \mathtt{\Longrightarrow \ ( 5x)^{2} -( 13)^{2}}
Referring the formula;
\mathtt{( a+b)( a-b) \ =a^{2} -b^{2}}
Here;
a = 5x
b = 13
Using the formula;
\mathtt{\Longrightarrow \ ( 5x)^{2} -( 13)^{2}}\\\ \\ \mathtt{\Longrightarrow \ ( 5x-13) \ ( 5x+13)}
The polynomial has been simplified to (5x-13)(5x+13).
Example 05
Factorize the polynomial.
\mathtt{\Longrightarrow \ 36\ +12x^{2} +x^{4}}
Solution
\mathtt{\Longrightarrow \ 36\ +12x^{2} +x^{4}}\\\ \\ \mathtt{\Longrightarrow \ ( 6)^{2} +2.( 6) .\left( x^{2}\right) +\left( x^{2}\right)^{2}}
Referring the following formula;
\mathtt{a^{2} +2ab+b^{2} =( a+b)^{2}}
Where;
a = 6
b = \mathtt{x^{2}}
Using the above formula;
\mathtt{\Longrightarrow \ ( 6)^{2} +2.( 6) .\left( x^{2}\right) +\left( x^{2}\right)^{2}}\\\ \\ \mathtt{\Longrightarrow \ \left( 6+x^{2}\right)^{2}}
Hence, the polynomial is factorized to \mathtt{\left( 6+x^{2}\right)^{2}}
Example 06
Factorize the below polynomial.
\mathtt{\Longrightarrow \ x^{2} -8xy+16y^{2} -\ 121}
Solution
\mathtt{\Longrightarrow \ \left( x^{2} -8xy+16y^{2}\right) -\ 121}\\\ \\ \mathtt{\Longrightarrow \ \left( x^{2} -8xy+( 4y)^{2}\right) -\ 11^{2}}\\\ \\ \mathtt{\Longrightarrow \ ( x-4y)^{2} -11^{2}}
Now using the formula;
\mathtt{a^{2} -b^{2} \ =( a+b)( a-b) \ }
Where;
a = x – 4y
b = 11
\mathtt{\Longrightarrow \ ( x-4y)^{2} -11^{2}}\\\ \\ \mathtt{\Longrightarrow \ ( x-4y+11) \ ( x-4y-11)}
The polynomial has been factorized to (x – 4y + 11)( x – 4y – 11)
Next chapter : Factorizing \mathtt{a^{2} -b^{2} } identity