Factorising Algebraic Expressions | Factoring Binomial & Trinomial Formule

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

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  = 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  + y= ( 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 x + 4x + 1 – 4 y2

Using ( a  + b )2  = a2 + 2ab + b2 for 4 x + 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 x + 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 - [ \sqrt{13} / 2 ] ^2

By using a2 – b2 = ( a + b) (a – b)

= \left [ a + \frac{5}{2} - \sqrt{13} / 2 \right ] \ \left [ a + \frac{5}{2} + \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 + b= a3 +  b3 + 3ab ( a +b)

(a – b)3   =         a3 – 3a2b + 3ab2 – b= 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 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) ( x– 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 ) ( a+ b+ c2 – ab – bc – ca) + 3 abc

Square of a Trinomial

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

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

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

Ex – 12 : If x + y +z = 6 ,  xyz = 8 &  x+ y+ z = 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 ) ( a+ b+ c2 – ab – bc – ca) + 3 abc

So   x3 + y3  +  z3

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

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My self Sivaramakrishna Alluri. Thank you for watching my blog friend

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