Math Doubts

Proof of $(x-a)(x-b)$ identity in Geometric Method

The (x-a)(x-b) formula can be derived in geometrical approach by splitting a square as three different rectangles. Let’s start deriving the expansion of special product of binomials $x-a$ and $x-b$ in algebraic form.

Express Area of Square

area of a square for (x-a)(x-b) formula

Take, a square and it is taken that the length of each side is $x$. Now, evaluate the area of a square mathematically.

$Area$ $\,=\,$ $x \times x$

$\implies$ $Area$ $\,=\,$ $x^2$

Therefore, the area of this square is equal to $x^2$ mathematically and remember this.

Express Areas of Rectangles

areas of rectangles for (x-a)(x-b) identity
  1. Now, divide the square horizontally as two different rectangles. If the width of one rectangle is $a$, then the width of the remaining rectangle is equal to $x-a$.
  2. Similarly, divide the upper rectangle vertically as two different rectangles. If length of one rectangle is $b$, then the length of remaining rectangle is $x-b$.

Thus, a square whose area is $x^2$, is split as three different rectangles. Now, write the area of each rectangle in mathematical form.

areas of rectangles for (x-a)(x-b) formula
  1. The length and width of first rectangle are $x-b$ and $x-a$ respectively, then its area is $(x-b)(x-a)$ in product form.
  2. The length and width of second rectangle are $b$ and $x-a$ respectively, then its area in product form is $b(x-a)$.
  3. The length and width of third rectangle are $a$ and $x$ respectively. Therefore, the area of this rectangle is equal to $ax$.

This information is used to derive the expansion for the special product of polynomials $(x-a)(x-b)$ geometrically.

Proving the Algebraic identity

The square whose area is $x^2$, is divided as three different rectangles geometrically. Therefore, the area of a square is equal to sum of the areas of the rectangles.

$x^2$ $\,=\,$ $(x-b)(x-a)$ $+$ $b(x-a)$ $+$ $ax$

$\implies$ $x^2$ $\,=\,$ $(x-a)(x-b)$ $+$ $b(x-a)$ $+$ $ax$

$\implies$ $x^2$ $\,=\,$ $(x-a)(x-b)$ $+$ $bx-ba$ $+$ $ax$

$\implies$ $x^2$ $\,=\,$ $(x-a)(x-b)$ $+$ $ax+bx$ $-$ $ba$

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

$\implies$ $x^2$ $\,=\,$ $(x-a)(x-b)$ $+$ $(a+b)x$ $-$ $ab$

$\implies$ $x^2$ $-$ $(a+b)x$ $+$ $ab$ $\,=\,$ $(x-a)(x-b)$

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

Therefore, it is proved that the product of special binomials $x-a$ and $x-b$ is equal to the algebraic expression $x^2-(a+b)x+ab$.

This algebraic identity can also be derived directly in alternative method in geometry. The rectangle whose area is $(x-b)(x-a)$ can be obtained by subtracting the rectangles whose areas are $b(x-a)$ and $ax$ from the square whose area is $x^2$.

$(x-a)(x-b)$ $\,=\,$ $x^2$ $-$ $\Big(b(x-a)+ax\Big)$

$\implies$ $(x-a)(x-b)$ $\,=\,$ $x^2$ $-$ $\Big(bx-ba+ax\Big)$

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

$\implies$ $(x-a)(x-b)$ $\,=\,$ $x^2$ $-$ $\Big((a+b)x-ab\Big)$

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

Math Doubts

A best free mathematics education website for students, teachers and researchers.

Maths Topics

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

Maths Problems

Learn how to solve the maths problems in different methods with understandable steps.

Learn solutions

Subscribe us

You can get the latest updates from us by following to our official page of Math Doubts in one of your favourite social media sites.

Copyright © 2012 - 2021 Math Doubts, All Rights Reserved