Question

Factor completely. 16x^4 - 81.

Please show all work.

Answer 1

`\large\boxed{16x^4-81=(2x-3)(2x+3)(4x^2+9)}`

Explanation:

`Use\\\\ (a^n)^m=a^(nm)\\\\a^2-b^2=(a-b)(a+b)\\------------------------\\\\16x^4-81=4^2(x^2)^2-9^2=(4x^2)^2-9^2=(4x^2-9)(4x^2+9)\\\\=(2^2x^2-3^2)(4x^2+9)=\left[(2x)^2-3^2\right](4x^2+9)\\\\=(2x-3)(2x+3)(4x^2+9)`

Answer 2

(4x^2 + 9)(2x + 3)(2x - 3)

Explanation:

The first step in factoring is to check for a common factor in all terms. There is no common factor for 16x^4 and 81 other than 1.

Now we look at what we have. We have two terms. The first term, 16x^4, is the square of 4x^2. The second term, 81, is the square of 9. We have a difference of two squares. A difference of two squares factors like this:

a^2 - b^2 = (a + b)(a - b)

We apply that factoring to 16x^4 - 81.

16x^4 - 81 = (4x^2 + 9)(4x^2 - 9)

Factoring means factoring completely, so now we look at what we have, and we see if we can factor it further. We have two binomials with the terms 4x^2 and 9 and 4x^2 and -9. 4x^2 is the square of 2x. 9 is the square of 3. We have a sum of two squares and a difference of two squares. A sum of two squares is not factorable. The difference of two square, 4x^2 - 9, is factorable using the same factoring shown above.

16x^4 - 81 =

= (4x^2 + 9)(4x^2 - 9)

= (4x^2 + 9)(2x + 3)(2x - 3)

Every factor is fully factored, so the answer is:

(4x^2 + 9)(2x + 3)(2x - 3)