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taurus [48]
3 years ago
13

What is the completely factored form of d^4 - 81

Mathematics
2 answers:
White raven [17]3 years ago
5 0

Answer:

(d - 3)(d + 3)(d² + 9)

Step-by-step explanation:

d^{4} - 81 ← can be factored as a difference of squares

a² - b² = (a - b)(a + b), thus

d^{4} - 81

= (d²)² - 9²

= (d² - 9)(d² + 9)

Note that d² - 9 is also a difference of square, so

= (d - 3)(d + 3)(d² + 9) ← in factored form

Flauer [41]3 years ago
4 0

Answer:

Step-by-step explanation:

d^4-81

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

Using formula

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

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victus00 [196]

Answer:

88

Step-by-step explanation:

Given:

(h⁴ + h² – 2) ÷ (h + 3).

We could obtain the remainder using the remainder theorem :

That is the remainder obtained when (h⁴ + h² – 2) is divided by (h + 3).

Using the reminder theorem,

Equate h+3 to 0 and obtain the value of h at h+3 = 0

h + 3 = 0 ; h = - 3

Substituting h = - 3 into (h⁴ + h² – 2) to obtain the remainder

h⁴ + h² – 2 = (-3)⁴ + (-3)² - 2 = 81 + 9 - 2 = 88

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Answer:

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Step-by-step explanation:

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