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Sagot :
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]
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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