How do you multiply ${(3x - 1)^2}$ ?

Hint:As we know that to multiply means to increase in number especially greatly or in multiples.
Multiplicand refers to the number multiplied and multiplier refers to the number that multiplies the first number.
Here in this question we have to find the product of two polynomials, we just multiply
each term of the first polynomial by each term of the second polynomial and then simplify or if there is any algebraic identity possible we can apply that.

Complete step by step solution:
Here we have ${(3x - 1)^2}$; we know the difference formula ${(a - b)^2} = {a^2} - 2ab + {b^2}$
In this case $a = 3x,b = - 1$, by applying the formula and substituting the values, we get
$ = {(3x)^2} + 2(3x)( - 1) + {( - 1)^2}$$ \Rightarrow 9{x^2} - 6x + 1$, Also remember that ${( - 1)^2} = - 1 \times - 1 = + 1$
Hence the required answer of ${(3x - 1)^2}$ is $9{x^2} - 6x + 1$

Note: We should know that the algebraic identity used in the above solution is called the square of the difference of the terms formula, which is also a binomial expression or also called the special binomial product rule. It is expanded as the subtraction of two times the product of two terms from the sum of the squares of the given terms. We know that it can also be expanded as the term i.e.
$(3x - 1)(3x - 1)$ and the first $x$ is multiplied with second polynomial $(3x - 1)$ and then $( - 1)$ is multiplied with the second polynomial again, and then simplified. It will give the same result. Also we should be careful with the positive and negative sign as in the above solution twice of $( - 1)$ gives the positive value not the negative value.