How do you factor # z^3 + 7z + 7^2 + 7#?

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Answer 1 · 2017-06-13T04:54:45

It can be factored in two different ways. Check out the Explanation! :)

Apply the power function when it appears on a constant value:
#z^3 + 7z + 7^2 + 7 = z^3 + 7z + 49 + 7 = z^3 + 7z + 56#.

We can now do two different factorings:

#-># by #z#:
#z(z^2 + 7) + 56#

or

#-># by #7#:
#z^3 + 7(z + 8)#

Answer 2 · 2017-07-21T19:10:46

#z^3+7z+z^2+7 = (z+1)(z^2+7)#

#color(white)(z^3+7z+z^2+7) = (z+1)(z-sqrt(7)i)(z+sqrt(7)i)#

I suspect the question has been mistranscribed somewhere along the line. A more plausible cubic that we can factor by grouping would be:

#z^3+7z+z^2+7 = (z^3+7x)+(z^2+7)#

#color(white)(z^3+7z+z^2+7) = z(z^2+7)+1(z^2+7)#

#color(white)(z^3+7z+z^2+7) = (z+1)(z^2+7)#

This can only be factored further using complex coefficients, since #z^2+7 > 0# for any real values of #z# ...

#color(white)(z^3+7z+z^2+7) = (z+1)(z^2-(sqrt(7)i)^2)#

#color(white)(z^3+7z+z^2+7) = (z+1)(z-sqrt(7)i)(z+sqrt(7)i)#