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Difference of Two Squares

Algebra is one of the substantial parts of mathematics, together with geometry, analysis and number theory. In its most general form it is the study of symbol and the rules which manipulates symbols. In algebra we deal with functions, polynomials, equations, and various algebraic properties etc. There are different methods and operations to solve algebraic problems.

One of the most important operations is factoring polynomials. Polynomials can be easily factorized by using algebraic formula. Difference of two square is one of the algebraic formulas which helps to factorized polynomials.
Difference of two square is the product of the sum and difference of the same two terms equals the square of the first term minus the square of the second term.     This method also helpful to find the roots of complex expressions, for example, the root of $x^2$ + 3 = $x^2$ - $i^2$3  = $x^2$ - $(i\sqrt{(3})^2$ = (x + i$\sqrt{3}$)(x + i$\sqrt{3}$). Therefore factors are (x + i$\sqrt{3}$) and (x - i$\sqrt{3}$). We shall go ahead in this section and learn more about difference of two squares algebraic property.

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Definition

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In elementary algebra, difference of squares is defined as: The subtraction of two squares terms is a squared term minus from another squared term i.e. x$^2$ - y$^2$.

This may be factored according to the given below mathematical identity.

x$^2$ - y$^2$ = (x - y)(x + y)

Here x$^2$ - y$^2$ expression is called a difference of two squares. And  (x - y) and (x + y) are the factors of x$^2$ - y$^2$.
Rule for factoring the difference of two squares:
Difference of two squares = Sum of two numbers multiplies their difference.
i.e. x$^2$ - y$^2$ = (x - y)(x + y)

Formula

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The difference of two squares is the product of sum of two numbers and their difference.
Let m and n are two numbers, then difference of their square is written as:

m$^2$ - n$^2$ = (m - n)(m + n)

The proof of this identity is very simple. Let us apply distributive property on the right hand side.

(m - n)(m + n) = $m^2$ + mn - nm + $n^2$   .............(1)

Put mn - nm = 0 (Application of commutative property)

For two variables above result gives a simple proof of AM-GM inequality. This result is commonly used in mathematics. Also it will hold in any commutative ring.
Conversely, if this identity holds in a ring then that ring is commutative. To verify this, apply distributive property to the right hand side of equation (1).

$m^2$ + mn - nm + $n^2$, for this is equal to $m^2$ - $n^2$ we must have mn - nm = 0.
It means ring is commutative.
Difference of two perfect squares is given below:

Difference of two perfect squares = Sum of two numbers $\times$ Difference of two numbers

m$^2$ - n$^2$ = (m - n)(m + n)
Alternate method:
We can also proof this formula from direct calculation.

Solve (m - n)(m + n) using FOIL method

$m^2$ + mn - nm + $n^2$

$m^2$ + mn - mn + $n^2$

$m^2$ - $n^2$ (cancel out common terms)

The difference of two squares formula is valid for real numbers and for complex numbers.

Factor the Difference of Two Squares

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The difference of two squares can be factored. An algebraic term is a perfect square when the exponents of each of the variables are even numbers and the numerical coefficient  is a perfect square.

The binomial $m^2$ - $n^2$  is called a difference of two squares, because $m^2$ is the sqaure of m and $n^2$ is the square of n. We obtain a method for factoring a difference of two squares if we reverse this rule.
To factorize an algebraic expression, follow the below steps:
Step 1: Look for a common factor. (If it is there)
Step 2: Take common factor out and use the difference of two squares formula.
Step 3: Write your answer.
Let us understand this concept with the help of examples:

Example 1: Factorise 4(m - n)$^2$ - 64

Solution: 4(m - n)$^2$ - 64

4 is common, so take it out

4((m - n)$^2$ - 16)

4((m - n)$^2$ - 4$^2$)

Apply difference of two squares formula

4((m - n + 4)(m - n - 4))

Example 2: Factor y$^2$ - 49

Solution: This expression is the combination of two terms, y$^2$ and 49. Both y$^2$ and 49 are perfect squares.

As we are given with difference of two perfect squares, this expression is the difference of two squares.
y$^2$ is the square of y

49  is the square of 7

Using difference of two square formula:

y$^2$ - 49 = y$^2$ - 7$^2$ = (y + 7)(y - 7)

Therefore (y + 7) and (y - 7) are factors of y$^2$ - 49.

Examples

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Below are some solved examples on difference of two squares:

Example 1: Factor y$^4$ - 81

Solution: y$^4$ - 81 = $(y^2)^2$ - $(9)^2$

= $(y^2+9)(y^2-9)$ .....(1)

Again $(y^2-9)$ can be written as $(y^2-3^2)$

$(y^2-3^2)$ = (y + 3)(y - 3)

y$^4$ - 81 = $(y^2+9)(y + 3)(y - 3)$

Example 2: Factorise the following

1) $x^2$ - 25

2) 9$x^2$ - 81

3) $(y-2)^2$ - 9

Solution:
1 $x^2$ - 25 = $x^2$ - $5^2$ = (x + 5)(x - 5)

2) 9$x^2$ - 81 = $(3x)^2$ - 9$^2$ = (3x - 9)(3x + 9)

3) $(y-2)^2$ - 9 = $(y-2)^2$ - $3^2$ = (y - 2 - 3)(y - 2 + 3) = (y - 5)(y + 1)

Example 3: Calculate 57$^2$ - 1 using difference of two square formula.

Solution:  57$^2$ - 1 = 57$^2$ - 1$^2$ = (57 - 1)(57 + 1) = 56 * 58 = 3248

Example 4: Factorize 72 - 8y$^2$

Solution: 72 - 8y$^2$

8(9 - y$^2$) (8 is common)

8(3$^2$ - y$^2$)

8(3 - y)(3 + y) (Difference of two squares)

Example 5: Factor the binomial y$^4$ - 1

Solution: y$^4$ - 1 = $(y^2)^2$ - 1

= ($ y^2$ - 1)($ y^2$ + 1)

= (y - 1)(y + 1)($ y^2$ + 1)

Example 6: Factor 8x$^2$ - 72x$^6$

Solution: To make the problem easier, take out common factors out.

In this case 8x$^2$ is a common factor, that can be factored out.

8x$^2$(1 - 9x$^4$)

In (1 - 9x$^4$), 1 and 9x$^4$ are perfect squares.

9x$^4$ is square of 3x$^2$ and 1 is square of 1 (itself)

Now (1 - 9x$^4$) = (1$^2$ - (3x^2)$^2$) = (1 + 3x$^2$)(1 - 3x$^2$)

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