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

The expansion of (x-a)(x-b) formula can be derived in algebraic approach by multiplying the two difference basis different binomials $x-a$ and $x-b$. According to multiplication of algebraic expressions, the special product of them can be obtained in mathematics.

Product form of the Binomials

Multiply the algebraic expressions $x-a$ and $x-b$ to express their product in mathematical the multiplication of algebraic expressions.

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

Multiply the Algebraic expressions

According to the multiplication of algebraic expressions, multiply each term in the second polynomial by the each term in the first polynomial.

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

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

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

Thus, 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

The expansion of the special product of the binomials can be further simplified for expressing 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 proved that the special product of the binomials $x-a$ and $x-b$ can be expanded as an algebraic expression $x^2-(a+b)x+ab$ in mathematics. In this way, the expansion of the special product of binomials $(x-a)(x-b)$ can be derived in algebra.

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