The subtraction of sixteen from the square of $x$ is the given algebraic expression in this problem and it should be factorized to express it as a product of two or more factors. So, let’s learn how to factorize $x$ square minus sixteen in this math problem.

There are two terms in the algebraic expression $x$ square minus sixteen. So, it is a binomial.

$x^2-16$

Let’s analyze the nature of the given algebraic expression to choose a suitable factoring method.

- The first term is $x$ square, and it is already in perfect square form.
- The second term is $16$.
- There is a minus sign between the terms.

If the second term is converted in square form, then the polynomial $x$ square minus sixteen becomes an algebraic expression in the form of the difference of two perfect squares.

The second term sixteen should be expressed in perfect square form to factorize the polynomial $x$ square minus sixteen. So, let’s use the factoring the numbers technique to write the number $16$ as a product of two factors but the factors should be same to express the number sixteen in square form.

$\implies$ $x^2-16$ $=\,\,$ $x^2-4 \times 4$

Now, use exponentiation to express the product of the factors four and four in exponential notation.

$=\,\,$ $x^2-4^2$

The polynomial $x$ square minus sixteen is converted as $x$ square minus four square.

- The two terms in the algebraic expression $x$ square minus four square are in perfect square form.
- There is a minus sign between the two terms $x$ square and four square.

The two properties of the expression $x$ square minus four square exactly matches with the difference or two perfect square form. So, it can be factorized by using the difference of two squares formula and let’s learn how to factorise $x$ square minus $16$ by using the difference of squares method.

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

Let’s take $a = x$ and $b = 4$. Now, substitute the values of $a$ and $b$ in the $a$ square minus $b$ square factoring formula to factorise the $x$ square minus four square.

$\implies$ $x^2-4^2$ $\,=\,$ $(x+4)(x-4)$

$\,\,\,\therefore\,\,\,\,\,\,$ $x^2-16$ $\,=\,$ $(x+4)(x-4)$

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