## 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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