Question

How do you factor the expression `64x^2 + 81`?

Answer 1

`(8x+9) (8x-9)`

Answer 2

`64x^2+81 = (8x-9i)(8x+9i)`

Explanation:

If `x` is any Real number then `x^2 >= 0`. Hence `64x^2+81 >= 81 > 0`.

As a result `64x^2+81` has no linear factors with Real coefficients.

We can do something with Complex coefficients...

First notice that `64x^2 = (8x)^2` and `81 = 9^2`

The difference of squares identity can be written:

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

The imaginary unit `i` has the property that `i^2 = -1`.

Hence we find:

`64x^2+81`

`=(8x)^2+9^2`

`=(8x)^2-(9i)^2`

`=(8x-9i)(8x+9i)`

If we want to, we can write a "sum of squares" identity:

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

Answer 3

`64x^2+81=(8x-9i)(8x+9i)`

(there are no factors with only Real components)

Explanation:

Remember that
`color(white)("XXX")(color(red)(a)^2-color(blue)(b)^2)` can be factored as `(color(red)(a)-color(blue)(b))(color(red)(a)+color(blue)(b))`

`64x^2+81 = (color(red)((8x))^2- color(blue)((9i))^2)`

`color(white)("XXXXxX")=(color(red)(8x)-color(blue)(9i))(color(red)(8x)+color(blue)(9i))`

If there were any factors with only Real components the equation
`color(white)("XXX")64x^2+81=0` would have Real solutions;
however we can tell from the fact that the discriminant
`color(white)("XXX")(b^2-4ac) = (0^2-4(64)(81)`) is less than zero that there are no Real roots.