FACTORING POLYNOMIALS USING ALGEBRAIC IDENTITIES

On this page "factoring polynomials using algebraic identities",we are going see clear explanation of factoring trinomials with two different variables. 

Factoring polynomials using algebraic identities (a+b)²and (a-b)²

Example 1:

Factorize 9x² - 24xy + 16y² 

Solution:

Step 1:

To factor these kind of expressions first we try to write the first and last term as squares. 

Step 2:

Since it is possible,we have to split the middle term as shown the below picture.

We write the middle term 24xy as the multiple of 2 x first term x last term 

Step 3:

Comparing these terms with the algebraic identity.

(3x - 4y) (3x -4y) are the factors 

Example 2:

Factorize 4x² + 12xy + 9y²

Solution:

Step 1:

To factor these kind of expressions first we try to write the first and last term as squares. 

              = 4x² + 12xy + 9y²

             = 2² x² + 12 xy + 3²y²

             = (2x)² + 12 xy + (3y)²

Step 2:

Since it is possible,we can to split the middle term as the multiple of 2 times product of first and last term

             = (2x)² + 2 (2x) (3y) + (3y)²

Step 3:

Comparing these terms with the algebraic identity.

a² + 2ab + b² = (a + b)²

              = (2x + 3y)²

(2x + 3y)(2x + 3y) are the factors 

Example 3:

Factorize 16a² - 8a + 1

Solution:

Step 1:

To factor these kind of expressions first we try to write the first and last term as squares. 

              = 16a² - 8a + 1

             = 4² a² - 8a + 1²

             = (4a)² - 8a + (1)²

Step 2:

Since it is possible,we can to split the middle term as the multiple of 2 times product of first and last term

             = (4a)² - 2 (4a) (1) + (1)²

Step 3:

Comparing these terms with the algebraic identity.

a² - 2ab + b² = (a - b)²

              = (4a - 1)²

Factoring polynomials using algebraic identities a²- b²

Example 4:

Factorize 16a² - 9b²

Solution:

              = 16a² - 9b²

             = 4² a² - 3²b²

             = (4a)² - (3b)²

The above algebraic expression exactly matches the identity a² - b²

Formula for a² - b² is (a + b) (a - b). In the above expression we have "4a" instead of "a" and "3b" instead of "b".

             = (4a + 3b) (4a - 3b)

Example 5:

Factorize (a + b)² - (a - b)²

Solution:

Let x = a + b  and y = a - b

              = x² - y²

The above algebraic expression exactly matches the identity a² - b²

Formula for a² - b² is (a + b) (a - b). In the above expression we have "x" instead of "a" and "y" instead of "b".

             = (x + y) (x - y)

             = [(a + b) + (a - b)] [(a + b) - (a -b)]

             = (a + b + a - b) (a + b - a + b)

             = 2a (2b)

             = 4 a b

Factoring polynomials using algebraic identities a³+b³and a³-b³

Example 6:

Factorize 8x³ - 125y³

Solution:

Step 1:

Let us try to write the numbers 8 and 125 in term of cube.

               8 = 2³ and 125 = 5³

              = 2³ x³ - 5³y³

              = (2x)³ - (5y)³

Step 2:

The above algebraic expression exactly matches the identity a³ - b³.

Formula for a³ - b³ is (a - b) (a² + ab + b²)

             = (2x - 5y) [ (2x)² + (2x) (5y) + (5y)²]

             = (2x - 5y) [ 4x² + 10xy + 25y²]

Example 7:

Factorize 27x³ + 64y³

Solution:

Step 1:

Let us try to write the numbers 27 and 64 in term of cube.

               27 = 3³ and 64 = 4³

              = 3³ x³ + 4³y³

              = (3x)³ + (4y)³

Step 2:

The above algebraic expression exactly matches the identity a³ + b³.

Formula for a³ + b³ is (a + b) (a² - ab + b²)

             = (3x + 4y) [ (3x)² - (3x) (4y) + (4y)²]

             = (3x + 4y) [ 9x² - 12xy + 16y²]

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