Difference Of Squares
The difference of two squares can also be illustrated geometrically as the difference of two square areas in a plane
. In the diagram, the shaded part represents the difference between the areas of the two squares, i.e. a^2-b^2. The area of the shaded part can be found by adding the areas of the two rectangles; a(a-b)+ b(a-b) , which can be factorized to (a+b)(a-b). Therefore a^2 - b^2 = (a+b)(a-b)
Another geometric proof proceeds as follows: We start with the figure shown in the first diagram below, a large square with a smaller square removed from it. The side of the entire square is a, and the side of the small removed square is b. The area of the shaded region is a^2 - b^2. A cut is made, splitting the region into two rectangular pieces, as shown in the second diagram. The larger piece, at the top, has width a and height a-b. The smaller piece, at the bottom, has width a-b and height b. Now the smaller piece can be detached, rotated, and placed to the right of the larger piece. In this new arrangement, shown in the last diagram below, the two pieces together form a rectangle, whose width is a+b and whose height is a-b. This rectangle's area is (a+b)(a-b) . Since this rectangle came from rearranging the original figure, it must have the same area as the original figure. Therefore, a^2 - b^2 = (a+b)(a-b).Any odd number can be expressed as difference of two squares.
a^2 - b^2 = (a -b ) ( a+b)
(a-b)(a+b) we expand this using FOIL method
Steps: (a - b) (a + b) = a* ( a+b ) - b *( a+b)
= (a* a + a*b) - (b*a - b*b)
= a^2 + ab - ba - b^2
= a^2 + ab - ab - b^2
= a^2 - b^2
Therefore, a^2 - b^2 = ( a + b) ( a - b)
We take the difference or subtraction of two perfect squares .
Steps :
We try to find any common factors between the terms , if there are any then take it out as common.
We try to find whether the remaining terms are perfect squares or not.
If the terms are perfect square then we apply the formula and simplify.
Check that the terms are in lowest form , if any of the factors are found to be still as perfect square then repeat Step 3.
Examples-difference of Squares
1) x^2 - 25
Solution:
No common factors are present.
x^2 is perfect square of x and 25 is perfect square of 5.
We plug in a =x and b = 5 in the standard formula
= x^2 - 5^2 = (x -5) (x +5), we get difference of squares.
2) 9x^2 - 64y^2
Solution :
9x^2 = 3^2 x^2 a perfect square
64y^2 = 8^2 y^2 a perfect square
We plug in a = 3x and b= 8y in the formula
= (3x)^2 - (8y)^2 = (3x - 8y) ( 3x +8y), we get difference of squares.
3) 25x^2 - 10x + 1 - 36y^2
Solution:
Here the first three terms are the terms of square of a binomial . We find the binomial first.
25x^2 is 5^2 x^2 , 1 = 1^2 middle term is negative , so that is a binomial with the second term negative
=(5x)^2 - 2(5x) (1) + 1^2 - (6y)^2
=[ 5x -1 ]^2 - (6y)^2 we plug in a = (5x-1) and b = 6y
= [(5x -1) - 6y] [(5x -1 ) +6y]
=(5x-6y -1) (5x +6y -1) ,difference of squares.
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4) 32x^4 - 2
Solution:
2 is the common factor , we remove 2 from both the terms
32x^4 - 2 =2( 16 x 4 -1) 16x^4 = 4^2(x^2)^2 a perfect square and 1 = 1^2 a perfect square
= 2 [ (4x^2)^2 - 1} we plug in a = 4x^2 and b = 1
= 2[ (4x^2 -1 ) (4x^2 +1) ]
= 2 [ ( 4x^2 +1) ( (2x)^2 - 1^2 )] we can see that 4x^2 = 2^2 x^2 a perfect square , 1 = 1^2 .We apply the formula once again
= 2 [ (4x^2 +1) (2x -1)(2x+1) ]
5) (x +1)^2 - (y-1)^2
Solution: Let x +1 = p, y-1 = q
(x +1)^2 - (y-1)^2 = p^2 - q^2 = (p-q) (p+q) difference of perfect squares of p and q
(x +1)^2 - (y-1)^2 = [ x+1- ( y-1) ] [x +1 + (y-1)] we plug in the values of p and q
= [x +1-y+1][x+1+y-1} simplify
= (x-y +2) (x+y)
Some Complex Examples-difference of Squares
1) x^4 + 4
Solution:The terms are not difference of two perfect squares , we complete the squares. It becomes square of a binomial .
x^4 + 4 = x^4 + 4x^2 +4 - 4x^2 ( adding and subtracting 4x^2 )
= (x^2)^2 + 2(x^2) (2) + 2^2 - 4x^2 Square of a binomial
= [ x^2 +2]^2 -(2x)^2
suppose a = x^2 +2 , b = 2x then (a^2 - b^2) = (a-b)(a+b), we plug in the values in the formula
= (x^2 + 2 - 2x)(x^2 +2 + 2x) Rearrange
= ( x^2 -2x +2) ( x^2 +2x +2), we get difference of squares.
2) x^8 -y^8
Solution: x^8 -y^8 = (x^4)^2 - (y^4)^2 x^8 is (x^4)^2 , x^4 is (x^2)^2 and x^2 is x square
=(x^4 + y^4) (x^4 - y^4) y^8 is (y^4)^2 , y^4 is (y^2)^2 and y^2 is y square
=(x^4 + y^4)[ (x^2)^2 - (y^2)^2]
=(x^4 + y^4) [ (x^2 +y^2) (x^2 - y^2)]
=(x^4 +y^4) (x^2 +y^2) (x+y)(x-y)
We apply the formula 3 times for the difference terms , the addition terms remain the same
3) ( x^2 + x + 1) ( x^2 + x - 1) = { ( x^2 + 1) + x} { ( x^2 + 1) - x} This is reverse of difference of squares formula
= ( x^2 + 1) 2 - (x)^2 ( a = x^2 + 1, b = x), we plug in (a-b)(a+b) = a^2 - b^2
We use the square of binomial formula to expand the first term
= x^4 + 2 x^2 + 1 - x^2 [ ( x^2 + 1) 2 = x^4 + 2* x^2* 1 + 1^2 = x^4 + 2x^2 + 1}
= x^4 + x^2 + 1
4) (2x + 3y)( 2x - 3y)( 4x^2 + 9y^2)
Solution: (2x + 3y)( 2x - 3y)( 4x^2 + 9y^2)
= { ( (2x + 3y)( 2x - 3y)} ( 4x^2 + 9y^2) This is reverse of difference of squares formula
= { (2x)^2 - (3y)^2 } ( 4x^2 + 3y^2) first two factors a = 2x , b = 3y , we plug in the formula
= ( 4x^2 - 9y^2 ) ( 4x^2 + 9y^2) we have a = 4x^2 and b = 9y^2 we apply the formula again
= ( 4x^2)^2 - ( 9y^2)^2 Simplify
= 16 x^4 - 81 y^4
Finding product of big numbers using difference of square formula
5) 68 x 72
Solution: 68 x 72 = (70 - 2) ( 70 + 2) a = 70 , b = 2
= (70)^2 - (2)^2
= 4900- 4
= 4896
7) 128^2 - 77^2
Solution: 128^2 - 77^2 = (128 + 77) ( 128 -77) a = 128 , b = 77
= 205 x 51
= 10455.
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