In this article explained about factoring polynomials by using the basic algebraic expressions like factoring a sum of cubes, factoring a difference of cubes, the sum and difference of two squares
Contents
 1 Factoring Polynomials by using Basic Algebraic Expressions
Factoring Polynomials by using Basic Algebraic Expressions
The basic algebraic equations are true for all values of the variables. So we can also use of these algebraic identities to factorise the algebraic expressions
Algebraic Identities
Here is a list of Identities
Identity – 1 (Difference and addition of squares or Product of binomials)
( a + b )^{2} = a^{2} + 2ab + b^{2}
( a – b )^{2} = a^{2} – 2ab + b^{2}
a^{2} – b^{2} = ( a + b) (a – b)
(x +a ) ( x + b) = x^{2} + (a +b) x + ab
Factoring polynomials practice on Identity – 1
Ex – 1 : Evaluate 105 x 107 without multiplying directly
Solution : 105 x 107 = (100 + 5) ( 100 +7)
By using (x +a ) ( x + b) = x^{2} + (a +b) x + ab
= 100^{2} + (5+7)(100) + ( 5 x 7)
= 10000 + 1200 + 35 = 11235
Ex – 2: Evaluate 1004 x 996 without multiplying directly
Solution : 1004 x 996 = (1000 + 4) (1000 – 4)
By using a^{2} – b^{2} = ( a + b) (a – b)
= 1000^{2} – 4^{2} = 1000000 16 = 999984
Ex – 3: Expand (p/3 –q)^{2}
Solution : Use ( a – b )^{2} = a^{2} – 2ab + b^{2}
A = p/3 & b = q
= (p/3 –q)^{2 }= (p/3)^{2} – 2(p/3) (q) + q^{2}
Ex – 4 : Find the value x^{2} + y^{2} if xy = 10 and x +y = 7
Solution : We know that ( x + y )^{2} = x^{2} + 2xy + y^{2}
x^{2} + y^{2 }= ( x + y )^{2} – 2 xy
= 7^{2} – 2 (10) = 49 – 20 = 29
Ex – 5 : By using Algebraic Identities factorise the following

 49 x^{2} + 70 xy + 25 y^{2}
 4 x^{2} – 4 y^{2} + 4x + 1
 5 – 20 a^{2}
 a^{2 }+ 5a + 3
Solution :
1 . 49 x^{2} + 70 xy + 25 y^{2}
By using ( a + b )^{2} = a^{2} + 2ab + b^{2}
Here a = 7x & b = 5y
49 x^{2} + 70 xy + 25 y^{2} = (7x + 5y) (7x + 5y)
2. 4 x^{2} – 4 y^{2} + 4x + 1
= 4 x^{2 } + 4x + 1 – 4 y^{2}
Using ( a + b )^{2} = a^{2} + 2ab + b^{2} for 4 x^{2 } + 4x + 1 ( i.e a = 2x & b = 1)
= (2x +1)^{2} – 4 y^{2}
Now using a^{2} – b^{2} = ( a + b) (a – b) ( here – a = 2x +1 & b = 2y )
= (2x +1 + 2y) ( 2x +1 – 2y)
So 4 x^{2 } + 4x + 1 – 4 y^{2 }= (2x + 2y +1 ) ( 2x – 2y +1 )
3. 5 – 20 a^{2}
5 – 20 a^{2}
= 5 ( 1 – 4 a^{2} )
Using a^{2} – b^{2} = ( a + b) (a – b) ( Here a = 1 & b = 2a )
= 5 ( 1 + 2a) (1 – 2a)
4.
5. a^{2 }+ 5a + 3
= a^{2 }+ 5a + 3
= a^{2 }+ 2 (5/2) a + 3
= a^{2 }+ 2 a (5/2) + (5/2)^{2} – (5/2)^{2} +3
By using a^{2} – b^{2} = ( a + b) (a – b)
Identity – 2 (Difference and addition of Cubic Polynomial )
a^{3} + b^{3} = (a + b) ( a^{2 }–ab + b^{2 }) = (a + b) [ (a + b)^{2} – 3ab ]
a^{3} – b^{3} = (a – b) ( a^{2 }+ab + b^{2 }) = (a – b) [ (a – b)^{2} – 3ab ]
(a + b )^{3 }= a^{3}+ 3a^{2}b + 3ab^{2 }+ b^{3 }= a^{3} + b^{3 } + 3ab ( a +b)
(a – b)^{3 }= a^{3} – 3a^{2}b + 3ab^{2 }– b^{3 }= a^{3} – b^{3 }– 3ab ( a – b)
Factoring polynomials practice on Identity – 2
Ex – 6 : Evaluate (998)^{3} without multiplying directly
Solution : (998)^{3} = (1000 – 2 )^{3}
By using (a – b)^{3 }= a^{3} – b^{3 }– 3ab ( a – b)
Here a = 1000 & b = 2
(998)^{3} = (1000 – 2 )^{3}
= 1000000000 – 8 – (3 x 2 x 1000 x 998)
= 1000000000 – 8 – 5988000 = 994011992
Ex – 7: Factorising Algebraic Expressions 8 a^{3} + 27b^{3} + 36 a^{2}b + 54 ab^{2}
Solution : The above expression can be written as
= 8 a^{3} + 27b^{3} + 36 a^{2}b + 54 ab^{2}
= (2 a)^{3} + (3b)^{3} + 18ab ( 2a + 3b)
= ( 2a + 3b)^{3}
= ( 2a + 3b) ( 2a + 3b) ( 2a + 3b)
Ex – 8 : Factorising polynomial – 125 x^{4} + 216 xy^{3}
Solution: 125 x^{4} + 216 xy^{3}
The above expression can be written as
= x ( 125 x^{3} + 216 y^{3 })
= x [ (5 x)^{3} + (6 y)^{3 }]
= x (5x + 6y) [ (5x)^{2} – (5x)(6y) + (6y)^{2} ]
= x (5x + 6y) (25x^{2} – 30xy + 36y^{2} )
Ex – 9 : By using Algebraic Identities factorise the following
 x^{6} – y^{6}
 x ^{6} – 7 x^{3} – 8
 (2x + 3y)^{3} – (2x – 3y)^{3}
Solution
1 . x^{6} – y^{6}
The above expression can be written as (x^{3}) ^{2 }– (y^{3})^{2}
By using a^{2} – b^{2} = ( a + b) (a – b)
Here a = x^{3} & b = y^{3}
= (x^{3} – y^{3} ) (x^{3} + y^{3} )
Using a^{3} – b^{3} = (a – b) ( a^{2 }+ab + b^{2 }) & a^{3} + b^{3} = (a + b) ( a^{2 }–ab + b^{2 })
x^{6} – y^{6} = (x – y) ( x^{2 }+ xy + y^{2 }) (x + y) ( x^{2 }– xy + y^{2 })
2 . x ^{6} – 7 x^{3} – 8
Let x^{3} = a then above equation can be written as
x ^{6} – 7 x^{3} – 8
= a ^{2} – 7 a 8
= a ^{2} – 8 a + a – 8
= a( a – 8) + 1 ( a – 8)
= ( a – 8) ( a + 1)
= ( x^{3} – 8) ( x^{3} + 1)
= ( x^{3} – 2^{3} ) ( x^{3} + 1^{3})
Using a^{3} – b^{3} = (a – b) ( a^{2 }+ab + b^{2 }) & a^{3} + b^{3} = (a + b) ( a^{2 }–ab + b^{2 })
=( x – 2) ( x^{2 }+2x + ^{4 }) ( x + 1) ( x^{2 }– x + 1 )
3 . (2x + 3y)^{3} – (2x – 3y)^{3}
Let (2x + 3y) = a & (2x – 3y) = b then
a^{3} – b^{3} = (a – b) ( a^{2 }+ab + b^{2 })
Now substituent the values of a & b
= (2x + 3y – 2x + 3y) [ (2x + 3y)^{2} + (2x + 3y)(2x – 3y) + (2x – 3y)^{2 }]
= 6y ( 12 x^{2} + 9 y^{2} )
= 18 y ( 4 x^{2} + 3 y^{2})
Identity – 3 (Difference and addition of quartic or bi quadratic)
( a + b)^{4} = a^{4} + b^{4 }+ 4 a^{3} b + 6 a^{2} b^{2} + 4 ab^{3}
( a – b)^{4} = a^{4} + b^{4 }– 4 a^{3} b + 6 a^{2} b^{2} – 4 ab^{3}
a^{4} – b^{4} = ( a + b) (a + b) (a^{2} +b^{2})
a^{5} – b^{5} = ( a – b) (a^{2} +b^{2 }+ ab + a^{2}b^{2} + ab^{3} )
Identity – 4 (Difference and addition of trinomials)
a^{3} + b^{3} + c^{3} = (a + b + c ) ( a^{2 }+ b^{2 }+ c^{2} – ab – bc – ca) + 3 abc
Square of a Trinomial
(a + b + c)^{2 }= a^{2 }+ b^{2 }+ c^{2 }+ 2ab + 2ac + 2bc
(a – b + c)^{2 }= a^{2 }+ b^{2 }+ c^{2 }– 2ab + 2ac – 2bc
(a + b – c)^{2} = a^{2} + b^{2} + c^{2} + 2ab – 2bc – 2ca
(a – b – c)^{2} = a^{2} + b^{2} + c^{2} – 2ab + 2bc – 2ca
Factoring polynomials practice on Identity – 3 & 4
Ex – 10 : Factorising polynomial 4 a^{2} + b^{2} + c^{2} – 4ab – 2bc + 4ac^{ }
Solution:
4 a^{2} + b^{2} + c^{2} – 4ab – 2bc + 4ac
= (2a) ^{2} + b^{2} + C^{2} – (2a)b – 2bc + 2(2a)c
Using (a – b + c)^{2 }= a^{2 }+ b^{2 }+ c^{2 }– 2ab + 2ac – 2bc
= ( 2a – b + c )^{2}
= ( 2a – b + c ) ( 2a – b + c )
Ex – 11 : Factorising polynomial 27 a^{3} + b^{3} + c^{3} 9abc
Solution: 27 a^{3} + b^{3} + c^{3} 9abc
The polynomial can be written as
= (3a)^{3} + b^{3} + c^{3} – 3 (3a)bc
By using identity a^{3} + b^{3} + c^{3} + 3 abc = (a + b + c ) ( a^{2 }+ b^{2 }+ c^{2} – ab – bc – ca)
= (3a + b + c) ( 9a^{2 }+ b^{2 }+ c^{2} – 3ab – bc – 3ca)
Ex – 12 : If x + y +z = 6 , xyz = 8 & x^{2 }+ y^{2 }+ z^{2 } = 12 then find the value of x^{3} + y^{3} + z^{3}
Solution: We know the identity
(a + b + c)^{2}^{ }= a^{2 }+ b^{2 }+ c^{2 }+ 2ab + 2ac + 2bc
(6) ^{2} = 12 + 2 ( xy + yz + zx )
( xy + yz + zx ) = 24 /2 = 12
a^{3} + b^{3} + c^{3} = (a + b + c ) ( a^{2 }+ b^{2 }+ c^{2} – ab – bc – ca) + 3 abc
So x^{3} + y^{3} + z^{3 }
= 6 (12 – (12)) + 3 (8) = 0 + 24 = 24
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