# Proof of $(x+a)(x-b)$ identity in Algebraic Method

The product of binomials $x+a$ and $x-b$ is $(x+a)(x-b)$ and the expansion of the special product can be derived in algebraic method. The expansion of (x+a)(x-b) formula is actually derived by multiplying the algebraic expressions $x+a$ and $x-b$.

### Product form of the Binomials

Multiply the algebraic expressions $x+a$ and $x-b$ for expressing the product of them in mathematical form by multiplying the algebraic expressions.

$(x+a) \times (x-b)$ $\,=\,$ $(x+a)(x-b)$

$\implies$ $(x+a)(x-b)$ $\,=\,$ $(x+a) \times (x-b)$

### Multiply the Algebraic expressions

As per the multiplication of algebraic expressions, multiply each term of the second polynomial by the each term of the first polynomial.

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

$\implies$ $(x+a)(x-b)$ $\,=\,$ $x \times x$ $+$ $x \times (-b)$ $+$ $a \times x$ $+a \times (-b)$

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

In this way, the special product of the multinomials $x+a$ and $x-b$ is expanded as an algebraic expression $x^2-xb+ax-ab$.

### Simplify the Expansion of the Product

Now, the expansion of the special product of the binomials is simplified further to write it in simple form.

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

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

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

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

Therefore, it is successfully proved that the special product of the binomials $x+a$ and $x-b$ is expanded as an algebraic expression $x^2+(a-b)x-ab$ in mathematics. Thus, the expansion of the special product of the binomials $(x+a)(x-b)$ is derived algebraically in algebraic mathematics.

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