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 + 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

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

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

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

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+ 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

x⁶ – y⁶
x ⁶ – 7 x³ – 8
(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) ( 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

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