Lesson The difference of squares formula
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<H2>The difference of squares formula</H2> The <B>difference of squares formula</B> is {{{a^2 - b^2 = (a+b)*(a-b)}}}. The formula is valid for any real numbers {{{a}}} and {{{b}}}. In particular, it is valid for all integer numbers. The formula is valid for the complex numbers too. The proof of the formula is very simple. It follows straightforward from the direct calculations: {{{(a + b)*(a-b) = a^2 + b*a - a*b - b^2 = a^2 - b^2}}}. As you see, the distributive and the commutative properties of addition and multiplication operations over the real numbers are used in derivation the formula. The <B><I>difference of squares formula</I></B> is useful in a number of applications. You should memorize it. In some cases, this formula can facilitate calculations and make them mentally without using paper and pencil (or calculator). <H3>Example 1</H3>Calculate {{{99^2-1}}}. <B>Solution</B> {{{99^2-1 = (99+1)*(99-1) = 100*98 = 9800}}}. <H3>Example 2</H3>Calculate {{{999^2-1}}}. <B>Solution</B> {{{999^2-1 = (999+1)*(999-1) = 1000*998 = 998000}}}. <H3>Example 3</H3>Calculate {{{49^2-1}}}. <B>Solution</B> {{{49^2-1 = (49+1)*(49-1) = 50*48 = 2400}}}. <H3>Do yourself</H3>Apply the <B><I>difference of squares formula</I></B> to check that {{{51^2-1 = 2600}}}, {{{101^2-1 = 10200}}}. The <B><I>difference of squares formula</I></B> is applicable not only to numbers. It is applicable for binomials too. For example, {{{x^2 - 1 = (x + 1)*(x-1)}}}. You can check validity of the last formula directly by performing all relevant calculations: opening the brackets, multiplying the terms and combining the like terms. You will get the same result. It is not surprising, because addition and multiplication operations over polynomials have the same distributive and commutative properties as over the real numbers. Thus, the <B><I>difference of squares formula</I></B> is simply the useful shortcut formula. It may help you when you need to factor polynomials. <H3>Example 4</H3>Factor the binomial {{{4x^2-9}}}. <B>Solution</B> {{{4x^2-9 = ((2x)^2-3)*((2x)^2+3) = (2x+3)*(2x-1)}}}. <H3>Example 5</H3>Factor the binomial {{{49x^2-81}}}. <B>Solution</B> {{{49x^2-81 = ((7x)^2-9)*((7x)^2+9) = (7x+9)*(7x-9)}}}. <B>Note</B>. Pay attention how the brackets are used to group monomials according the pattern of the <B><I>difference of squares formula</I></B>. <H3>Example 6</H3>Factor the binomial {{{x^4-1}}}. <B>Solution</B> {{{x^4-1 = (x^2-1)*(x^2+1) = (x-1)*(x+1)*(x^2+1)}}}. <H3>Example 7</H3>Factor the binomial {{{a^4-b^4}}}. <B>Solution</B> {{{a^4-b^4 = (a^2-b^2)*(a^2+b^2) = (a-b)*(a+b)*(a^2+b^2)}}}. You got the decomposition of the bi-quadratic binomial {{{a^4-b^4}}} into the product of the two linear binomials {{{a-b}}} and {{{a+b}}} and one quadratic binomial {{{a^2+b^2}}}. <BLOCKQUOTE><B>Note 1</B>. You can get another decomposition of the binomial {{{a^4-b^4}}} by rolling up the factors {{{a+b}}} and {{{(a^2+b^2)}}} of the previous factorization into the product {{{(a+b)*(a^2+b^2) = a^3 + a^2*b + a*b^2 + b^3}}}. Then you get {{{a^4-b^4 = (a^2-b^2)*(a^2+b^2) = (a-b)*(a+b)*(a^2+b^2) = (a-b)*(a^3 + a^2*b + a*b^2 + b^3)}}}.</BLOCKQUOTE> <BLOCKQUOTE><B>Note 2</B>. The quadratic binomial {{{a^2+b^2}}} can not be factored into the product of linear binomials on {{{a}}} and {{{b}}} over the real numbers. For those acquainted with the complex numbers, this quadratic binomial can be factored over the complex numbers: {{{a^2+b^2 = (a+bi)*(a-bi)}}}, where {{{i=sqrt(-1)}}} is the imaginary complex numbers unit. Indeed, {{{(a+bi)*(a-bi) = a^2 - i^2*b^2 = a^2 + b^2}}}. But any factorization of the binomial {{{a^2+b^2}}} into the product of linear binomials on {{{a}}} and {{{b}}} is not possible over the real numbers.</BLOCKQUOTE><BLOCKQUOTE><B>Note 3</B>. In contrast to the quadratic binomial {{{a^2+b^2}}}, the bi-quadratic binomial {{{a^4+b^4}}} can be factored into the product of two polynomials of lesser degree. Some trick is used: add and distract the monomial {{{2a^2b^2}}} to/from the binomial {{{a^4+b^4}}}, then apply the shortcut <B><I>square of the sum formula</I></B> and the <B><I>difference of squares formula</I></B>. You get {{{a^4+b^4 = (a^4+2a^2*b^2+b^4) - 2a^2*b^2 = (a^2+b^2)^2 - (sqrt(2)*ab)^2 = (a^2+b^2+sqrt(2)*ab)*(a^2+b^2-sqrt(2)*ab)}}}.</BLOCKQUOTE> <H3>Example 8</H3>Factor the expression {{{a^2+2ab+b^2-c^2}}}. <B>Solution</B> First, apply the <B><I>square of the sum formula</I></B> and then apply the <B><I>difference of squares formula</I></B>. You get {{{a^2+2ab+b^2-c^2 = (a+b)^2-c^2 = (a+b+c)*(a+b-c)}}}. <H3>Summary</H3>The <B><I>difference of squares formula</I></B> {{{a^2 - b^2 = (a+b)*(a-b)}}} is useful shortcut multiplication formula. At the same time, you can use it to factor binomials when applicable. For similar lessons see <A HREF=http://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/The_square_of_a_sum.lesson>The square of the sum formula</A> and <A HREF=http://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/The_square_of_a_difference.lesson>The square of the difference formula</A> under the current topic in this site. The <B><I>difference of squares formula</I></B> is widely used in rationalizing fractions by making their denominator free of square roots. For details and examples see the lesson <A HREF=http://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/HOW-TO-make-the-denominator-of-a-fraction-free-of-square-roots.lesson>HOW TO rationalize a fraction by making its denominator free of square roots</A> under the current topic in this site. It is used also in simplifying rational expressions. For details and examples see the lesson <A HREF=http://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/Simplifying-rational-expressions-with-the-use-the-shortcut-multiplication-formulas.lesson>Simplifying rational expressions with the use the shortcut multiplication formulas</A> under the current topic in this site. For the <B>list of all shortcut quadratic multiplication formulas</B> see the lesson <A HREF=http://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/Review-of-shortcut-quadratic-multiplication-formulas.lesson>OVERVIEW of shortcut quadratic multiplication formulas</A> under the current topic in this site. For factoring the binomials of the third degree {{{a^3-b^3}}}, {{{x^3-a^3}}}, {{{a^3+b^3}}}, {{{x^3+a^3}}} see the lessons <A HREF=http://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/The-difference-of-cubes-formula.lesson>The difference of cubes formula</A>, <A HREF=http://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/The-sum-of-cubes-formula.lesson>The sum of cubes formula</A> under the current topic in this site. For factoring the binomials of high degrees {{{x^n-a^n}}}, {{{x^n+a^n}}} see the lessons <A HREF=http://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/Divisibility-of-the-Binomial-%28x%5En-a%5En%29-by-%28x-a%29.lesson>Factoring the binomials {{{x^n-a^n}}}</A>, <A HREF=http://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/Factoring-the-binomials-x%5En+a%5En-for-odd-degrees.lesson>Factoring the binomials {{{x^n+a^n}}} for odd degrees</A>, <A HREF=http://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/Review-of-Factoring-the-binomials-x%5En-a%5En-and-x%5En+a%5En.lesson>OVERVIEW of Factoring the binomials {{{x^n-a^n}}} and {{{x^n+a^n}}}</A> under the current topic in this site. Use this file/link <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A> to navigate over all topics and lessons of the online textbook ALGEBRA-I.
