How do you find the roots of #x^3-5x^2+4x+10=0#?

1 Answer
Nov 14, 2016

Answer:

The roots are: #-1# and #3+-i#

Explanation:

#f(x) = x^3-5x^2+4x+10#

By the reational roots theorem, any rational zeros of #f(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #10# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#+-1, +-2, +-5, +-10#

We find:

#f(-1) = -1-5-4+10 = 0#

So #x=-1# is a zero and #(x+1)# a factor:

#x^3-5x^2+4x+10 = (x+1)(x^2-6x+10)#

We can factor the remaining quadratic by completing the square and using the difference of squares identity:

#a^2-b^2 = (a-b)(a+b)#

with #a=(x-3)# and #b=i# as follows:

#x^2-6x+10 = x^2-6x+9+1#

#color(white)(x^2-6x+10) = (x-3)^2-i^2#

#color(white)(x^2-6x+10) = ((x-3)-i)((x-3)+i)#

#color(white)(x^2-6x+10) = (x-3-i)(x-3+i)#

Hence the other two zeros are: #x = 3+-i#