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Evgen [1.6K]
3 years ago
11

What is the completely factored form of d4 − 81?

Mathematics
2 answers:
lilavasa [31]3 years ago
4 0

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)

saw5 [17]3 years ago
3 0
<h2>Answer:</h2>

The complete factorization of the term:

              d^4-81 is:

        (d-3)(d+3)(d^2+9)

<h2>Step-by-step explanation:</h2>

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)

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