${(x+a)}{(x+b)}$ $\,=\,$ $x^2+(a+b)x+ab$

The $(x+a)(x+b)$ algebraic identity can be derived geometrically by the concept of areas of rectangle and square.

- Take a rectangle. Divide it as two parts horizontally and the lengths of them are $x$ and $a$.
- Now, divide the same rectangle vertically but the length of one part should be $x$ and take the length of remaining part is $b$.
- The length and width of whole rectangle are $x+a$ and $x+b$ respectively. Therefore, the area of the rectangle is ${(x+a)} \times {(x+b)}$ geometrically.

The geometrical approach splits a rectangle as a square and three different small rectangles. Now, calculate area of every geometric shape mathematically.

- The length of side of square is $x$. So, the area of square is $x^2$
- Length and width of first rectangle are $b$ and $x$ respectively. So, area of the rectangle is $bx$.
- Length and width of second rectangle are $x$ and $a$ respectively. So, area of this rectangle is $xa$.
- Length and width of third rectangle are $b$ and $a$ respectively. So, area of this rectangle is $ba$.

It is derived that the area of a whole rectangle is ${(x+a)}{(x+b)}$. Actually, the same rectangle is divided as a square and three small different rectangles. So, the area of rectangle should be equal to the sum of the areas of one square and three different rectangles.

${(x+a)}{(x+b)}$ $\,=\,$ $x^2+bx+xa+ba$

$\implies$ ${(x+a)}{(x+b)}$ $\,=\,$ $x^2+bx+ax+ab$

$\implies$ ${(x+a)}{(x+b)}$ $\,=\,$ $x^2+ax+bx+ab$

$\,\,\, \therefore \,\,\,\,\,\,$ ${(x+a)}{(x+b)}$ $\,=\,$ $x^2+(a+b)x+ab$

Latest Math Topics

Latest Math Problems

Email subscription

Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers.
Know more

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.