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

3 Answers

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

Feb 27, 2016

Answer:

#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

Answer:

#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.