What are the other zeros of the function #g(x) = x^4+x^3+10x^2+16x-96#, given that one is #x=2# ?

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Answer 1 · 2017-11-03T07:38:12

The zeros are #2#, #3#, #4i# and #-4i#

Given:

#g(x) = x^4+x^3+10x^2+16x-96#

Since #x=2# is a zero, #(x-2)# must be a factor.

So separate it out as a factor and find:

#x^4+x^3+10x^2+16x-96 = (x-2)(x^3+3x^2+16x+48)#

Note that in the remaining cubic factor the ratio between the first and second term is the same as that between the third and fourth term.

So this cubic will factor by grouping:

#x^3+3x^2+16x+48 = (x^3+3x^2)+(16x+48)#

#color(white)(x^3+3x^2+16x+48) = x^2(x+3)+16(x+3)#

#color(white)(x^3+3x^2+16x+48) = (x^2+16)(x+3)#

The linear factor #(x+3)# gives us another real zero #x=-3#

The remaining quadratic factor is positive for any real values of #x#.

It has zeros #x=+-4i# where #i# is the imaginary unit, satisfying #i^2=-1#, since we find:

#(x-4i)(x+4i) = x^2-(4i)^2 = x^2-4^2 i^2 = x^2-(16)(-1) = x^2+16#