Question

What is the completely factored form of d4 − 81?
(d + 3)(d − 3)(d + 3)(d − 3)
(d2 + 9)(d + 3)(d − 3)
(d2 + 9)(d − 3)(d − 3)
(d2 + 9)(d2 − 9)

Answer 1

For this case we must factor the following expression:

$d ^ 4-81$

Rewriting the expression:

$(d ^ 2) ^ 2-9 ^ 2$

We factor using the formula of the square difference:

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

Where:

$a = d ^ 2\\b = 9$

So:

$(d ^ 2 + 9) (d ^ 2-9)$

From the second term we have:

$d ^ 2-3 ^ 2 = (d-3) (d + 3)$

Finally, the factored expression is:

$(d ^ 2 + 9) (d-3) (d + 3)$

Answer:

$(d ^ 2 + 9) (d-3) (d + 3)$

Answer 2

Answer:

The complete factorization of the term:

              $d^4-81$ is:

        $(d-3)(d+3)(d^2+9)$

Step-by-step explanation:

To factor a term means to express is as a product of distinct factors i.e. multiples.

We are asked to factor the algebraic expression which is given by:

$d^4-81$

We could write this expression as:

$(d^2)^2-(3^2)^2=(d^2)^2-(9)^2$

We know that:

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

i.e.

$d^4-81=(d^2-9)(d^2+9)\\\\i.e.\\\\d^4-81=(d^2-3^2)(d^2+9)\\\\i.e.\\\\d^4-81=(d-3)(d+3)(d^2+9)$