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Factor the polynomial completely: [tex]x^4 - 81[/tex]

A. [tex](x+9)(x+3)[/tex]
B. [tex]\left(x^2 + 9\right)(x+3)(x-3)[/tex]
C. [tex]\left(x^2 + 9\right)\left(x^2 - 9\right)[/tex]
D. [tex](x-81)(x+1)[/tex]

Sagot :

GinnyAnswer
To factor the polynomial [tex]\( x^4 - 81 \)[/tex] completely, let's explore common factoring techniques, and focus on identifying one that fits the pattern.

First, observe that [tex]\( x^4 - 81 \)[/tex] can be recognized as a difference of squares. Recall the difference of squares formula:
[tex]\[ a^2 - b^2 = (a - b)(a + b) \][/tex]

Here, we can rewrite [tex]\( x^4 - 81 \)[/tex] as:
[tex]\[ x^4 - 81 = (x^2)^2 - 9^2 \][/tex]

Applying the difference of squares formula with [tex]\( a = x^2 \)[/tex] and [tex]\( b = 9 \)[/tex]:
[tex]\[ (x^2)^2 - 9^2 = (x^2 - 9)(x^2 + 9) \][/tex]

Now, we see [tex]\( x^2 - 9 \)[/tex] is also a difference of squares. We can factor it further:
[tex]\[ x^2 - 9 = (x)^2 - (3)^2 = (x - 3)(x + 3) \][/tex]

So putting it all together, we get:
[tex]\[ x^2 - 9 = (x - 3)(x + 3) \][/tex]

Thus, the completely factored form of the original polynomial [tex]\( x^4 - 81 \)[/tex] is:
[tex]\[ x^4 - 81 = (x - 3)(x + 3)(x^2 + 9) \][/tex]

Therefore, the correct answer is:
[tex]\[ b. \left( x^2 + 9 \right)(x + 3)(x - 3) \][/tex]
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