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    You are at:Home»Pure Math»Algebra»Factorising Algebraic Expressions | Factoring Binomial & Trinomial Formule
    Factoring Polynomials by using Algebraic Expressions | How to Factor Algebraic Equations | all math tricks
    Algebra

    Factorising Algebraic Expressions | Factoring Binomial & Trinomial Formule

    sivaalluriBy sivaalluriDecember 16, 2018Updated:February 12, 2025No Comments10 Mins Read

    This article explained 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

    Table of Contents

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    • Factoring Polynomials by Using Basic Algebraic Expressions
      • Identity – 1 (Difference and addition of squares or Product of binomials)
        • Factoring polynomials practice on Identity – 1
      • Identity – 2 (Difference and addition of Cubic Polynomial )
        • Factoring polynomials practice on Identity – 2
      • Identity – 3 (Difference and addition of  quartic or bi quadratic)
      • Identity – 4 (Difference and addition of trinomials)
        • Factoring polynomials practice on Identity – 3 & 4

    Factoring Polynomials by Using Basic Algebraic Expressions

    The basic algebraic equations are true for all values of the variables. So we can also use 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  = a2 + 2ab + b2

    ( a  – b )2  = a2 – 2ab + b2

    a2 – b2 = ( a + b) (a – b)

    (x +a ) ( x + b) = x2  + (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) = x2  + (a +b) x  + ab

    = 1002  + (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 a2 – b2 = ( a + b) (a – b)

    = 10002  – 42 =  1000000 -16 = 999984

    Ex – 3:  Expand  (p/3 –q)2

    Solution : Use ( a  – b )2  = a2 – 2ab + b2

    A = p/3 & b = q

    = (p/3 –q)2 = (p/3)2  – 2(p/3) (q) + q2

     = \frac{p^2}{9} \ + \ \frac{2pq}{3} \ + q^2

    Ex – 4 : Find the value x2 + y2  if  xy = 10 and x +y = 7

    Solution : We know that ( x  + y )2  = x2 + 2xy + y2

    x2  + y2  = ( x  + y )2    –  2 xy

    =  72  – 2 (10) = 49 – 20 = 29

    Ex – 5 : By using  Algebraic Identities factorise the following

      1. 49 x2 + 70 xy + 25 y2
      2. 4 x2 – 4 y2 + 4x + 1
      3. 5 – 20 a2
      4.  \frac{a^4}{4} + \frac{4}{a^4} + 1
      5. a2 + 5a + 3

    Solution :

    1 . 49 x2 + 70 xy + 25 y2

    By using ( a  + b )2  = a2 + 2ab + b2

    Here a = 7x & b = 5y

    49 x2 + 70 xy + 25 y2 = (7x + 5y)  (7x + 5y)

    2. 4 x2 – 4 y2 + 4x + 1

    = 4 x2  + 4x + 1 – 4 y2

    Using ( a  + b )2  = a2 + 2ab + b2 for 4 x2  + 4x + 1  ( i.e a = 2x & b = 1)

    = (2x +1)2 – 4 y2

    Now using a2 – b2 = ( a + b) (a – b)         ( here –  a = 2x +1 &  b = 2y )

    = (2x +1 + 2y) ( 2x +1 – 2y)

    So   4 x2  + 4x + 1 – 4 y2   = (2x + 2y +1 ) ( 2x – 2y +1 )

    3.  5 – 20 a2

    5 – 20 a2

    = 5 ( 1 – 4 a2 )

    Using  a2 – b2 = ( a + b) (a – b)     ( Here    a  = 1  & b = 2a  )

    =  5 ( 1 + 2a) (1 – 2a)

    4.   \frac{a^4}{4} + \frac{4}{a^4} + 1

     = \frac{a^4}{4} + \frac{4}{a^4} + 2 - 1

     = \left[ \frac{a^2}{2} \right]^2 + \left[ \frac{2}{a^2} \right]^2 + 2 \left[ \frac{a^2}{2} \right] \left[ \frac{2}{a^2} \right] - 1

     = \left[ \frac{a^2}{2} + \frac{2}{a^2} \right]^2 - (1)^2

     = \left[ \frac{a^2}{2} + \frac{2}{a^2} + 1 \right] \left[ \frac{a^2}{2} + \frac{2}{a^2} - 1 \right]

    5.  a2 + 5a + 3

    = a2 + 5a + 3

    = a2 +  2 (5/2) a + 3

    = a2 +  2 a  (5/2)  + (5/2)2  – (5/2)2 +3

     = \left[ a + \frac{5}{2} \right]^2 - \frac{25}{4} + 3

     = \left[ a + \frac{5}{2} \right]^2 - \frac{13}{4}

     = \left[ a + \frac{5}{2} \right]^2 - \left[ \frac{\sqrt{13}}{2} \right]^2

    By using a2 – b2 = ( a + b) (a – b)

     = \left[ a + \frac{5}{2} - \frac{\sqrt{13}}{2} \right] \left[ a + \frac{5}{2} + \frac{\sqrt{13}}{2} \right]

    Identity – 2 (Difference and addition of Cubic Polynomial )

    a3 + b3  =         (a + b) ( a2 –ab + b2 )  = (a + b) [ (a + b)2 – 3ab ]

    a3 –  b3  =         (a – b) ( a2 +ab + b2 ) = (a – b) [ (a – b)2 – 3ab ]

     (a + b )3   =      a3+ 3a2b + 3ab2 + b3  = a3 +  b3 + 3ab ( a +b)

    (a – b)3   =         a3 – 3a2b + 3ab2 – b3  = a3 –  b3 – 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   =    a3 –  b3 – 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 a3 + 27b3 + 36 a2b + 54 ab2

    Solution : The above expression can be written as

    = 8 a3 + 27b3 + 36 a2b + 54 ab2

    = (2 a)3 + (3b)3 + 18ab ( 2a + 3b)

    = ( 2a + 3b)3

    = ( 2a + 3b) ( 2a + 3b) ( 2a + 3b)

    Ex – 8 : Factorising polynomial –  125 x4 + 216 xy3

    Solution:  125 x4 + 216 xy3

    The above expression can be written as

    = x ( 125 x3 + 216 y3 )

    = x  [ (5 x)3 + (6 y)3 ]

    = x (5x + 6y) [ (5x)2 – (5x)(6y) + (6y)2 ]

    = x (5x + 6y)  (25x2 – 30xy + 36y2 )

    Ex – 9 : By using  Algebraic Identities factorise the following

    1. x6 – y6
    2. x 6 – 7 x3 – 8
    3. (2x + 3y)3  – (2x – 3y)3

    Solution

    1 . x6 – y6

    The above expression can be written as (x3) 2 – (y3)2

    By using a2 – b2 = ( a + b) (a – b)

    Here a = x3 & b = y3

    = (x3 – y3 ) (x3 + y3 )

    Using a3 –  b3   =   (a – b) ( a2 +ab + b2 )  &  a3 + b3  =  (a + b) ( a2 –ab + b2 )

    x6 – y6 = (x – y) ( x2 + xy + y2 ) (x + y) ( x2 – xy  + y2 )

    2 .  x 6 – 7 x3 – 8

    Let  x3 = a then the above equation can be written as

    x 6 – 7 x3 – 8

    = a 2 – 7 a- 8

    = a 2 – 8 a + a – 8

    = a( a  – 8) + 1 ( a – 8)

    = ( a  – 8) ( a + 1)

    = ( x3 – 8) ( x3 + 1)

    = ( x3 – 23 ) ( x3 + 13)

    Using a3 –  b3   =   (a – b) ( a2 +ab + b2 )  &  a3 + b3  =  (a + b) ( a2 –ab + b2 )

    =( x – 2) ( x2 +2x + 4 ) ( x + 1) ( x2  – x + 1 )

    3 . (2x + 3y)3  – (2x – 3y)3

    Let  (2x + 3y) = a &  (2x –  3y) = b then

    a3 –  b3  =   (a – b) ( a2 +ab + b2 )

    Now substituent the values of a & b

    = (2x + 3y  – 2x +  3y) [ (2x + 3y)2 + (2x + 3y)(2x – 3y) + (2x – 3y)2 ]

    = 6y ( 12 x2 + 9 y2 )

    = 18 y ( 4 x2 + 3 y2)

    Identity – 3 (Difference and addition of  quartic or bi quadratic)

    ( a + b)4  = a4 + b4 + 4 a3 b + 6 a2 b2 + 4 ab3

    ( a – b)4  = a4 + b4 – 4 a3 b + 6 a2 b2 – 4 ab3

    a4 – b4 = ( a + b) (a + b) (a2 +b2)

    a5 – b5 = ( a – b) (a2 +b2 + ab + a2b2 + ab3 )

    Identity – 4 (Difference and addition of trinomials)

    a3 + b3  +  c3 =  (a + b + c ) ( a2  + b2  + c2 – ab – bc – ca) + 3 abc

    Square of a Trinomial

    (a + b + c)2  = a2 + b2 + c2 + 2ab + 2ac + 2bc

    (a –  b + c)2  = a2 + b2 + c2 – 2ab + 2ac – 2bc

    (a + b – c)2 = a2 + b2 + c2 + 2ab – 2bc – 2ca

    (a – b – c)2 = a2 + b2 + c2 – 2ab + 2bc – 2ca

    Factoring polynomials practice on Identity – 3 & 4

    Ex – 10 : Factorising polynomial  4 a2 + b2 + c2 – 4ab – 2bc + 4ac

    Solution:

    4 a2 + b2 + c2 – 4ab – 2bc + 4ac

    = (2a) 2 + b2 + C2 – (2a)b – 2bc + 2(2a)c

    Using  (a –  b + c)2  = a2 + b2 + c2 – 2ab + 2ac – 2bc

    = ( 2a – b + c )2

    = ( 2a – b + c ) ( 2a – b + c )

    Ex – 11 : Factorising polynomial  27 a3 + b3 + c3 -9abc

    Solution: 27 a3 + b3 + c3 -9abc

    The polynomial can be written as

    = (3a)3 + b3 + c3 – 3 (3a)bc

    By using identity a3 + b3  +  c3 + 3 abc =  (a + b + c ) ( a2  + b2  + c2 – ab – bc – ca)

    = (3a + b + c) ( 9a2  + b2  + c2 – 3ab – bc – 3ca)

    Ex – 12 : If x + y +z = 6 ,  xyz = 8 &  x2  + y2  + z2  = 12 then find the value of  x3 + y3  +  z3

    Solution: We know the identity

    (a + b + c)2  = a2 + b2 + c2 + 2ab + 2ac + 2bc

      (6) 2 = 12 + 2 ( xy + yz + zx ) 

    ( xy + yz + zx )  = 24 /2 = 12

    a3 + b3  +  c3 =  (a + b + c ) ( a2  + b2  + c2 – ab – bc – ca) + 3 abc

    So   x3 + y3  +  z3

    = 6 (12 – (12)) +  3 (8)  = 0 + 24 = 24

    Dear friends Thanks for reading. I hope you liked this article “Factorising Algebraic Expressions”. Give feedback, comments and please don’t forget to share them.

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