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

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.

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

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

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

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

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