Discover a world of knowledge and community-driven answers at IDNLearn.com today. Discover in-depth answers from knowledgeable professionals, providing you with the information you need.

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]
Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. Find precise solutions at IDNLearn.com. Thank you for trusting us with your queries, and we hope to see you again.