In this chapter we will learn different methods to factorize polynomials with solved examples.
What is Factorization ?
The breaking up of number or polynomial into its individual constituents is called factorization.
The individual component generated during factorization are called factors of that particular number.
For example;
Consider the factorization of number 15.
15 = 3 x 5
It tells that number 15 is formed by multiplication of number 3 and 5.
Hence, 3 and 5 are the factors of number 15.
There can be multiple factorization for single number.
For example, consider the factorization of number 12.
(i) 12 = 1 x 12
(ii) 12 = 2 x 6
(iii) 12 = 4 x 3
(iv) 12 = 2 x 2 x 3
All the above factorizations are correct and valid.
For simplicity purpose, try to factorize number to into its prime number components.
Here, the option (iv) is simple to understand as all the components have been broken into prime numbers 2, 2 and 3.
Factorization of Polynomials
Here we try to factorize the polynomials into individual components.
Consider the below polynomial;
⟹ ab + ac
Note that component ” a ” is common in both the addition terms.
The above polynomial can be factorized as;
⟹ a . (b + c)
Hence, after factorization we get two individual components ” a ” and ” b + c”
How to factorize polynomials ?
In this chapter we will learn four methods to factorize polynomial.
(a) HCF method
(b) Group method
(c) Quadratic equation method
(d) Formula method
Note that you cannot use any one technique to factor all kinds of polynomial.
Each technique works for particular type of problem only. Hence, you need to practice all the above methods to be able to solve all kind of problems.
HCF method to factorize polynomial
In this method, we will identify the common elements in each component of polynomial and then separate it using brackets.
Let us understand the concept with example;
Example 01
Factorize \mathtt{6x^{2} +2xy}
Solution
Follow the below steps;
Break the polynomial into small components.
The above polynomial can be expressed as;
\mathtt{\Longrightarrow \ 6x^{2} +2xy}\\\ \\ \mathtt{\Longrightarrow \ 2.3\ x.x\ +\ 2.x.y}
Now separate the common elements present in both components.
\mathtt{\Longrightarrow \ 2.3\ x.x\ +\ 2.x.y}\\\ \\ \mathtt{\Longrightarrow \ 2x\ ( 3x+y)}
Hence, the above expression is the factorized form of given polynomial.
You can check the solution by multiplying the the terms back. After multiplication you will get back the same polynomial.
\mathtt{\Longrightarrow \ 2x\ ( 3x+y)}\\\ \\ \mathtt{\Longrightarrow \ 2x.3x\ +\ 2x.y}\\\ \\ \mathtt{\Longrightarrow \ 6x^{2} +2xy}
Example 02
Factorize \mathtt{18x^{3} y^{2} +81x^{2} y^{2}}
Solution
To solve question faster first find HCF of numbers 18 & 81.
HCF (18, 81) = 9
Among numbers, 9 is the common digit present in both components.
Now the polynomial is expressed as;
\mathtt{\Longrightarrow \ 18x^{3} y^{2} +81x^{2} y^{2}}\\\ \\ \mathtt{\Longrightarrow \ 9.2.x^{3} y^{2} +9.9.x^{2} y^{2}}
Note that variables \mathtt{x^{2} y^{2}} are common among both variables.
Taking \mathtt{\ 9x^{2} y^{2}} common in the given polynomial, we get;
\mathtt{\Longrightarrow \ 9x^{2} y^{2} \ ( 2x\ +9) \ }
Hence, the above expression is the solution.
Example 03
Factorize \mathtt{20x^{3} +5x^{6} -50x^{10}}
Solution
(a) Common Constant
Let’s find the HCF of numbers
HCF (20, 5, 50 ) = 5
Hence, number 5 is common among all the components.
(b) Common Variable
Note that variable \mathtt{x^{3}} is common in all component.
(c) Separate the common components.
\mathtt{\Longrightarrow \ 20x^{3} +5x^{6} -50x^{10}}\\\ \\ \mathtt{\Longrightarrow \ 5x^{3}\left( 4+x^{3} -10x^{7}\right)}
Hence, the above expression is the solution.
Example 04
Factorize \mathtt{\ 2x^{2} y+8x^{5} y^{2} -13y^{3}}
Solution
(a) Common constant
HCF (2, 8, 13) = 1
There is no common number present in three components.
(b) Common variable
Only variable y is common among the three components.
(c) Separate the common components
\mathtt{\Longrightarrow \ 2x^{2} y+8x^{5} y^{2} -13y^{3}}\\\ \\ \mathtt{\Longrightarrow \ y\ \left( 2x^{2} +8x^{5} y-13y^{2}\right)}
Hence, the above expression is the factorized form.
Example 05
Factorize \mathtt{\ 4x^{5} +12x^{3} +4x}
Solution
(a) Find common constant.
HCF (4, 12, 4) = 4
Number 4 is common among given component.
(b) Find common variable
Variable ” x ” is common among the components.
(c) Separate common terms
\mathtt{\Longrightarrow \ 4x^{5} +12x^{3} +4x}\\\ \\ \mathtt{\Longrightarrow \ 4x\ \left( x^{4} +3x^{2} +1\right)}
Hence, the above expression is the factorized form.
Group method of factorization
In this method we separate the components in different group and then factorize individually.
After individual factorization, we again get the common component for further factorization.
Let us understand the concept with example.
Example 01
Factorize \mathtt{\ 6x^{3} +2x^{2} +21x+7}
Solution
Group the first and last two terms and factorize.
\mathtt{\Longrightarrow \ 6x^{3} +2x^{2} +21x+7}\\\ \\ \mathtt{\Longrightarrow \left( 6x^{3} +2x^{2}\right) +( 21x+7)}\\\ \\ \mathtt{\Longrightarrow \ 2x^{2}( 3x+1) \ +7( 3x+1)}
Note that 3x+1 is again the common term.
Factorizing further, we get;
\mathtt{\Longrightarrow \ 2x^{2}( 3x+1) \ +7( 3x+1)}\\\ \\ \mathtt{\Longrightarrow \ \ ( 3x+1)\left( 2x^{2} +7\right)}
Hence, the above expression is the factorized form.
Example 02
Factorize \mathtt{x^{6} -3x+5x^{7} -15x^{2}}
Solution
Group first and last two terms and then factorize.
\mathtt{\Longrightarrow \left( \ x^{6} -3x\right) +\left( 5x^{7} -15x^{2}\right)}\\\ \\ \mathtt{\Longrightarrow \ x\left( x^{5} -3\right) +5x^{2}\left( x^{5} -3\right)}
Note \mathtt{x^{5} -3} is the common term in expression.
Factorizing further, we get;
\mathtt{\Longrightarrow \ x^{6} -3x+5x^{7} -15x^{2}}\\\ \\ \mathtt{\Longrightarrow \left( x^{6} -3x\right) +\left( 5x^{7} -15x^{2}\right)}\\\ \\ \mathtt{\Longrightarrow \ x\left( x^{5} -3\right) \ +5x^{2}\left( x^{5} -3\right)}\\\ \\ \mathtt{\Longrightarrow \ \ \left( x^{5} -3\right)\left( x+5x^{2}\right)}
Note that in second component, the variable x is common in both terms.
Factorizing further, we get;
\mathtt{\Longrightarrow \ \ \left( x^{5} -3\right)\left( x+5x^{2}\right)}\\\ \\ \mathtt{\Longrightarrow \ \left( x^{5} -3\right) .x.( 1+5x)}
Hence, the above expression is the solution.
Example 03
Factorize \mathtt{2y^{4} -10y^{2} -6y^{2} +30}
Solution
Form groups of first and last two entities and then factorize.
\mathtt{\Longrightarrow \ 2y^{4} -10y^{2} -6y^{2} +30}\\\ \\ \mathtt{\Longrightarrow 2y^{2}\left( y^{2} -5\right) -6\left( y^{2} -5\right)}\\\ \\ \mathtt{\Longrightarrow \ \left( y^{2} -5\right) \ \left( 2y^{2} -6\right)}
Note that in second component, number 2 is common.
Factorizing further, we get;
\mathtt{\Longrightarrow \ \left( y^{2} -5\right) \ \left( 2y^{2} -6\right)}\\\ \\ \mathtt{\Longrightarrow \ \left( y^{2} -5\right) .2\ .\left( y^{2} -3\right)}\\\ \\ \mathtt{\Longrightarrow \ 2\ \left( y^{2} -5\right)\left( y^{2} -3\right)}
Hence, the above expression is the solution.
Factorizing quadratic equation
The expression in the form of \mathtt{ax^{2} +bx\ +\ c} is called quadratic equation.
Here we will learn basic method to factorize quadratic equation. The advanced method of solving quadratic equation is covered in higher classes.
Let’s understand the method with example.
Example 01
Factorize \mathtt{x^{2} +2x\ -63}
Solution
(i) Multiply coefficients of \mathtt{x^{2}} with last number having no variable.
Hence, multiplying first and last terms.
⟹ 1 x 63
⟹ 63
(ii) Now find factors of 63 such that after addition/subtraction of the factors we can get value equal to coefficient of x ( here number 2)
Some possible factors of 63 are;
(a) 63 ⟹ 1 x 63
(b) 63 ⟹ 3 x 21
(c) 63 ⟹ 7 x 9
Note that the third factor 7 x 9 is suitable because when we subtract 9 – 7 = 2, which is equal to the middle coefficient.
(iii) Rewriting the quadratic equation.
Express the coefficient of x in form of selected factors ( 9 – 7 ) and then factorize.
\mathtt{\Longrightarrow \ x^{2} +2x\ -63}\\\ \\ \mathtt{\Longrightarrow \ x^{2} +\ ( 9-7) x-63}\\\ \\ \mathtt{\Longrightarrow \ x^{2} +9x-7x-63}
Forming group of first and last two terms.
\mathtt{\Longrightarrow \ x^{2} +9x-7x-63}\\\ \\ \mathtt{\Longrightarrow \left( x^{2} +9x\right) -( 7x+63)}\\\ \\ \mathtt{\Longrightarrow \ x\ ( x+9) -7( x+9)}\\\ \\ \mathtt{\Longrightarrow \ ( x-7)( x+9)}
Hence, (x – 7)(x + 9) is the solution.
Example 02
Factorize \mathtt{\ x^{2} -7x\ -30}
Solution
The above expression is in form of quadratic equation.
(i) Multiply first and last terms
⟹ 1 x 30
⟹ 30
(ii) Find the value of factors such that after addition/division of factors we get value 7.
Possible factors of 30 are;
(a) 30 = 1 x 30
(b) 30 = 2 x 15
(c) 30 = 5 x 6
(d) 30 = 10 x 3
The fourth set of factor is suitable because after subtracting we get number 7.
i.e. 10 – 3 = 7
(iii) Rewriting the quadratic equation.
Express the coefficient 7 as (10 – 3) in the middle.
\mathtt{\Longrightarrow \ x^{2} -7x\ -30}\\\ \\ \mathtt{\Longrightarrow \ x^{2} -( 10-3) x\ -30}\\\ \\ \mathtt{\Longrightarrow \ x^{2} -10x+3x-30}
Group the first and last two terms.
\mathtt{\Longrightarrow \ x^{2} -10x+3x-30}\\\ \\ \mathtt{\Longrightarrow \left( x^{2} -10x\right) +( 3x-30)}\\\ \\ \mathtt{\Longrightarrow \ x( x-10) \ +3( x-10)}\\\ \\ \mathtt{\Longrightarrow \ ( x-10) \ ( x+3)}
Hence, the above expression is the solution.
Formulas for factorizing polynomials
Given below are general formulas used to factorize polynomial
\mathtt{a^{2} -b^{2} =\ ( a-b)( a+b)}\\\ \\ \mathtt{a^{2} +2ab+b^{2} =\ ( a+b)^{2}}\\\ \\ \mathtt{a^{2} -2ab+b^{2} =( a-b)^{2}}\\\ \\ \mathtt{a^{3} -b^{3} =( a-b)\left( a^{2} +ab+b^{2}\right)}\\\ \\ \mathtt{a^{3} +b^{3} =( a+b)\left( a^{2} -ab+b^{2}\right)}
You have to remember all the above formulas as the related questions can be directly asked in the exams.
Given below are some examples related to factorizing polynomials using formulas.
Example 01
Factorize \mathtt{x^{2} -4}
Solution
The expression can be expressed as;
\mathtt{\Longrightarrow \ x^{2} -4}\\\ \\ \mathtt{\Longrightarrow \ x^{2} -2^{2}}
Using formula;
\mathtt{a^{2} -b^{2} =\ ( a-b)( a+b)}
\mathtt{\Longrightarrow \ x^{2} -2^{2}}\\\ \\ \mathtt{\Longrightarrow \ ( x-2) \ ( x+2)}
Hence, the above expression is the factorized form.
Example 02
Factorize \mathtt{8x^{3} -27}
Solution
The above expression can be expressed as;
\mathtt{\Longrightarrow \ 8x^{3} -27}\\\ \\ \mathtt{\Longrightarrow \ ( 2x)^{3} -( 3)^{3}}
The expression is in the form of;
\mathtt{a^{3} -b^{3} =( a-b)\left( a^{2} +ab+b^{2}\right)}
Using the formula, we get;
\mathtt{\Longrightarrow \ ( 2x)^{3} -( 3)^{3}}\\\ \\ \mathtt{\Longrightarrow \ ( 2x-3)\left(( 2x)^{2} +( 2x)( 3) +3^{2}\right)}\\\ \\ \mathtt{\Longrightarrow \ ( 2x-3)\left( 4x^{2} +6x+9\right)}
Hence, the above expression is the solution