FACTORISING USING SPECIAL IDENTITY

Special Identities

Special identities have three different kind with a fixed format that has to be applied all the time when solving using this method. The formats would be: <br>

PERFECT SQUARES

à   (a+b) ² = a²+2ab+b^{2  <br>}

DIFFERENCE OF 2 SQUARES

à   (a-b) ² = a²-2ab+b^{2 <br>} 

à   (a+b)(a-b) = a²-b^{2  <br>} 

For special identity, only certain expressions that meet the criteria would be able to be solved using these methods.<Br>

The criteria are: <br>

à The first term has to be a perfect square<br>

àThe last term has to be a constant and also a perfect square <br>

Only then will you be able to apply this method on. <br> 

An example would be: <br>

<br>

x² +12x+36 <br>

Since that x² is a perfect square and 36 is a perfect square and a constant, we can thus apply our special identity to this expression. <br>

x² +12x+36 is the same as a²+2ab+b². Thus, we know that the factorized form would be (a+b)^{2 <br>} 

a = x        b = 6 <br>

x² +12x+36 in its factorized form would then be (x+6)^2<br> 

The next example would be using the second perfect squares.

y²-6yz+49

Since y is a perfect square and 49 is also a perfect square, we can apply the second special identity.

y²-6yz+49 is the same as a²-2ab+b²

Thus, a = y           b = 7

The factorised form would then be (y-7)

The third one would be:

(11y+10)(11y-10)

The expression above is the same as the difference of squares so thus we can apply the special identity.

a = 11y     b = 10

The factorised form of (11y+10)(11y-10) would be 121y²-100.

Some of the more complex questions that could be solved using

Special identity would be:

 Evaluate the following without the use of calculator: 659²-341²

= (660-1)²-(340+1)²

= (660²-2(660)(1)+1²)-(340²+2(340)(1)+1²)

= (435600-1320+1)-(115600+680+1)

= 434281-116281

= 318000