How do you find all the zeros of #f(x) = x^4 + 2x^3 + 22x^2 + 50x - 75#?

Question

3868 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-26T10:25:47

Zeros of #x^4+2x^3+22x^2+50x-75# are #-3# and #1#

To find all the zeros of #P(X)=x^4+2x^3+22x^2+50x-75#, first find a factor of the independent term #75# such as

#{1, -1, 3, -3, 5, -5, 15, -15, ....}# for which #P(x)=0#.

It is apparent that for #x=1# as well as for #x=-3#, #P(x)=0#. Hence #1# and #-3# are two zeros of #x^4−4x^3−35x^2+6x+144#

Hence, #(x-1)(x+3)# or #x^2+2x-3# divides #P(x)#. Dividing latter by former, we get

#(x^4+2x^3+22x^2+50x-75)/(x^2+2x-3)# = #x^2+25#

As the RHS cannot be factorized in real factors there are no further zeros.

Hence zeros of #x^4+2x^3+22x^2+50x-75# are #-3# and #1#