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

The complete factorization of the term:

`d^4-81` is:

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

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)`

Answer 2

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)`