How do you find all the real and complex roots of #2x^4 - 5x^3 + 53x^2 - 125x + 75 = 0#?

1 Answer
Jan 14, 2016

Answer:

Check the sum of the coefficients to find root #x=1#, then divide by #(x-1)# and factor by grouping to find the remaining roots #x=3/2# and #x=+-5i#.

Explanation:

#f(x) = 2x^4-5x^3+53x^2-125x+75#

First note that the sum of the coefficients is zero.

That is: #2-5+53-125+75 = 0#

So #f(1) = 0# and #(x-1)# is a factor.

Use your favourite method to divide #f(x)# by #(x-1)#. Here I long divide the coefficients:

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Then factor by grouping...

#f(x) = (x-1)(2x^3-3x^2+50x-75)#

#=(x-1)((2x^3-3x^2)+(50x-75))#

#=(x-1)(x^2(2x-3)+25(2x-3))#

#=(x-1)(x^2+25)(2x-3)#

The remaining quadratic factor has no linear factors with Real coefficients since #x^2+25 >= 25 != 0# for any Real values of #x#, but it does have zeros #+-5i#, since #(5i)^2 = (-5i)^2 = -25#

#=(x-1)(x-5i)(x+5i)(2x-3)#

So #f(x)# has zeros #x=1#, #x=3/2#, #x=+-5i#