We know, that by Distributive law \ a\left(b+c\right)=ab+ac\ and\ \left(a+b\right)c=ab+ac . Using this we can derive formula for a lot of algebraic expressions. I have deliberately not given the proof details as it is just simple application of the above law. \\

To find the product of    (x\pm{}a)(x\pm{}b)   

  1.    \left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab   
  2.    \left(x+a\right)\left(x-b\right)=x^2+\left(a-b\right)x-ab   
  3.    \left(x-a\right)\left(x+b\right)=x^2+\left(b-a\right)x-ab   
  4.    \left(x-a\right)\left(x-b\right)=x^2-\left(a+b\right)x+ab   

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Product of Sum and Difference of two terms

  1.    \left(a+b\right)\left(a-b\right)=a^2-b^2   

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Squares of Binomials

  1.    {(a+b)}^2=\ a^2+2ab+b^2   
  2.    {(a-b)}^2=\ a^2-2ab+b^2   
  3.    a^2+b^2=\ {(a+b)}^2-2ab   
  4.    a^2+b^2=\ {(a+b)}^2-2ab   
  5.    a^2+b^2=\ {(a-b)}^2+2ab   
  6.    ab=\ \frac{1}{4}\{\left({a+b)}^2-\ {\left(a-b\right)}^2\right\}   
  7.    a^2+b^2=\ \frac{1}{2}\{{\left(a+b\right)}^2-\ {\left(a-b\right)}^2\}   
  8.    {(a+b)}^2=\ {(a-b)}^2+4ab   
  9.    {\left(a+\frac{1}{a}\right)}^2=\ a^2+\frac{1}{a^2}+2   
  10.    {\left(a-\frac{1}{a}\right)}^2=\ a^2+\frac{1}{a^2}-2   

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Perfect Square Trinomial

We know that    a^2+2ab+b^2=\ {(a+b)}^2 \ \ and \ \ a^2-2ab+b^2=\ {\left(a-b\right)}^2   , therefore every algebraic expression in the above form is called a perfect square.

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Squares of Trinomials

  1.    {(a+b+c)}^2=\left(a^2+b^2+c^2\right)+2(ab+bc+ca)   
  2.    {(a+b-c)}^2=\left(a^2+b^2+c^2\right)+2(ab-bc-ca)   
  3.    {(a-b+c)}^2=\left(a^2+b^2+c^2\right)+2(-ab-bc+ca)   
  4.    {(a-b-c)}^2=\left(a^2+b^2+c^2\right)+2(-ab+bc-ca)   
  5.    \left(a^2+b^2+c^2\right)={(a+b+c)}^2-2(ab+bc+ca)   
  6.    2\left(ab+bc+ca\right)={\left(a+b+c\right)}^2-\left(a^2+b^2+c^2\right)   

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Cubes of Binomials

  1.    {(a+b)}^3=\ a^3+b^3+3ab(a+b)   
  2.    {(a-b)}^3=\ a^3-b^3-3ab(a-b)   
  3.    {a^3+b^3=\ (a+b)}^3-3ab(a+b)   
  4.    {a^3-b^3=\ (a-b)}^3+3ab(a-b)   
  5.    {\left(a+\frac{1}{a}\right)}^3=a^3+\frac{1}{a^3}+3\left(a+\frac{1}{a}\right)   
  6.    {\left(a+\frac{1}{a}\right)}^3=a^3+\frac{1}{a^3}+3\left(a+\frac{1}{a}\right)   
  7.    {\left(a-\frac{1}{a}\right)}^3=a^3-\frac{1}{a^3}-3\left(a-\frac{1}{a}\right)   
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