How do you factor #x^4+81# ?

#x^4+81# ?

1 Answer
George C.
Jan 27, 2018

#x^4+81 = (x^2-3sqrt(2)x+9)(x^2+3sqrt(2)x+9)#

Explanation:

Here's one method:

Note that:

#A^2-B^2=(A-B)(A+B)#

Use this with #A=(x^2+9)# and #B=3sqrt(2)x# as follows:

#x^4+81 = (x^2+9)^2-18x^2#

#color(white)(x^4+81) = (x^2+9)^2-(3sqrt(2)x)^2#

#color(white)(x^4+81) = ((x^2+9)-3sqrt(2)x)((x^2+9)+3sqrt(2)x)#

#color(white)(x^4+81) = (x^2-3sqrt(2)x+9)(x^2+3sqrt(2)x+9)#

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