How do you find all the real and complex roots of #x^3 + 9x^2 + 19x - 29 = 0#?

1 Answer
mason m
Jan 22, 2016

#x=1,-5+-2i#

Explanation:

The potential real roots of a polynomial are the factors of #p/q#, where #p# is the constant and #q# is the coefficient of the term with the highest degree.

Since #p=29# and #q=1#, the possible real root(s) must be factors of #29/1=29#. This makes the problem significantly easier since the #29# is prime, so the only possibilities for the roots are #+-1,+-29#.

Use synthetic division or polynomial long division to find that #(x-1)# is a factor, which means that #color(red)1# is a root:

#(x^3+9x^2+19x-29)/(x-1)=x^2+10x+29#

The other two roots can be found through applying the quadratic formula on #x^2+10x+29#:

#x=(-b+-sqrt(b^2-4ac))/(2a)=(-10+-sqrt(-16))/2=color(red)(-5+-2i#

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