Multiply the expression $\left( {7a - 9b} \right)$ and$\left( {7a - 9b} \right)$.
(A) $49{a^2} - 126ab + 81{b^2}$
(B) $49{a^2} - 124ab + 81{b^2}$
(C) $49{a^2} - 122ab + 81{b^2}$
(D) $49{a^2} - 120ab + 81{b^2}$

Hint: We have multiplication of two polynomials that are $\left( {7a - 9b} \right)$ and $\left( {7a - 9b} \right)$. Both the polynomials have degree one and have two terms each. Hence, both polynomials are called binomials and linear polynomials in variables a and b. To multiply polynomials, first multiply each term in one polynomial by each term in the other polynomial using distributive law. Then, simplify the resulting polynomial by adding or subtracting the like terms.

Complete step-by-step solution:
So, we have to find the product of terms $\left( {7a - 9b} \right)$ and $\left( {7a - 9b} \right)$.
In the first polynomial, we have two terms and in the second polynomial also we have two terms. Multiply the first term of a polynomial with second polynomial and then the second term with second polynomial, we have,
$ \Rightarrow \left( {7a - 9b} \right)\left( {7a - 9b} \right)$
$ \Rightarrow 7a\left( {7a - 9b} \right) - 9b\left( {7a - 9b} \right)$
Opening the brackets and multiplying, we have,
$ \Rightarrow 49{a^2} - 63ab - 63ab + 81{b^2}$
Adding the like terms, we have,
$ \Rightarrow 49{a^2} - 126ab + 81{b^2}$
Thus we have, $\left( {7a - 9b} \right)\left( {7a - 9b} \right) = 49{a^2} - 126ab + 81{b^2}$.
So, the product of $\left( {7a - 9b} \right)$ and $\left( {7a - 9b} \right)$ is $49{a^2} - 126ab + 81{b^2}$.
Therefore, option (A) is the correct answer.

Note: For avoiding mistakes, write the terms in the decreasing order of their exponent. Thus we obtained a polynomial of degree two. Hence, the obtained polynomial is a quadratic polynomial. When we multiply two polynomials of any degree the obtained polynomial must have degree higher than multiplied individual polynomials. We can check this in the given above problem and it satisfies the condition. Be careful with the sign when you open the brackets. Follow the same procedure for any two polynomials.
We can also solve the problem by using the algebraic identity ${\left( {x - y} \right)^2} = {x^2} - 2xy + {y^2}$, where $x = 7a$ and $y = 9b$.
So, we have, $\left( {7a - 9b} \right)\left( {7a - 9b} \right) = {\left( {7a - 9b} \right)^2}$
$ \Rightarrow \left( {7a - 9b} \right)\left( {7a - 9b} \right) = {\left( {7a} \right)^2} - 2\left( {7a} \right)\left( {9b} \right) + {\left( {9b} \right)^2}$
Computing the squares of the terms, we get,
$ \Rightarrow \left( {7a - 9b} \right)\left( {7a - 9b} \right) = 49{a^2} - 2\left( {7a} \right)\left( {9b} \right) + 81{b^2}$
Simplifying the calculations,
$ \Rightarrow \left( {7a - 9b} \right)\left( {7a - 9b} \right) = 49{a^2} - 126ab + 81{b^2}$
Hence, the answer is the same.