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

Contents

- 1 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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