Questions: Factor the binomial completely. x^8-81 x^4 Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. x^8-81 x^4= (Factor completely.) B. The polynomial is prime.

Factor the binomial completely.
x^8-81 x^4

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A. x^8-81 x^4= (Factor completely.)
B. The polynomial is prime.

Transcript text: Factor the binomial completely. \[ x^{8}-81 x^{4} \] Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. $x^{8}-81 x^{4}=$ $\square$ (Factor completely.) B. The polynomial is prime.

Solution

Solution Steps

To factor the binomial $ x^8 - 81x^4 $ completely, we can use the following steps:

Recognize that $ x^8 - 81x^4 $ can be factored by treating it as a difference of squares.
Rewrite $ x^8 $ as $ (x^4)^2 $ and $ 81x^4 $ as $ (9x^2)^2 $.
Apply the difference of squares formula: $ a^2 - b^2 = (a - b)(a + b) $.
Continue factoring any resulting expressions that can be further factored.

Step 1: Recognize the Difference of Squares

The given expression is $ x^8 - 81x^4 $. We can recognize this as a difference of squares: \[ x^8 - 81x^4 = (x^4)^2 - (9x^2)^2 \]

Step 2: Apply the Difference of Squares Formula

Using the difference of squares formula $ a^2 - b^2 = (a - b)(a + b) $, we can factor the expression: \[ (x^4)^2 - (9x^2)^2 = (x^4 - 9x^2)(x^4 + 9x^2) \]

Step 3: Factor Further

We can further factor $ x^4 - 9x^2 $ as another difference of squares: \[ x^4 - 9x^2 = x^2(x^2 - 9) = x^2(x - 3)(x + 3) \] The term $ x^4 + 9x^2 $ cannot be factored further over the real numbers.