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

3 Answers
Feb 27, 2016

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

Feb 27, 2016

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)

Feb 27, 2016

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.