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What is the completely factored form of [tex]\( d^4 - 81 \)[/tex]?

Sagot :

To determine the completely factored form of [tex]\(d^4 - 81\)[/tex], we begin by recognizing that [tex]\(81\)[/tex] is a perfect square, as [tex]\(81 = 9^2\)[/tex]. Thus, we can rewrite the expression in a form that highlights this relationship:

[tex]\[ d^4 - 81 = d^4 - 9^2. \][/tex]

We notice that this fits the pattern of a difference of squares, [tex]\(a^2 - b^2\)[/tex], which factors into [tex]\((a - b)(a + b)\)[/tex]. Here, [tex]\(a = d^2\)[/tex] and [tex]\(b = 9\)[/tex]:

[tex]\[ d^4 - 81 = (d^2)^2 - 9^2 = (d^2 - 9)(d^2 + 9). \][/tex]

Next, we can factor the term [tex]\(d^2 - 9\)[/tex] further, as it is also a difference of squares. We apply the same pattern again, where [tex]\(d^2 - 9\)[/tex] can be written as:

[tex]\[ d^2 - 9 = (d - 3)(d + 3). \][/tex]

Thus, substituting back, we have:

[tex]\[ d^4 - 81 = (d^2 - 9)(d^2 + 9) = (d - 3)(d + 3)(d^2 + 9). \][/tex]

The term [tex]\(d^2 + 9\)[/tex] cannot be factored further as a real number expression since it does not fit the pattern for a difference of squares or any other recognizable factoring pattern for real numbers.

Hence, the completely factored form of [tex]\(d^4 - 81\)[/tex] is:

[tex]\[ (d - 3)(d + 3)(d^2 + 9). \][/tex]