How do you find all zeros with multiplicities of #f(x)=x^3-7x^2+x-7#?

1 Answer
Sep 17, 2017

Answer:

The zeros of #f(x)# are #7#, #i#, #-i#, all with multiplicity #1#.

Explanation:

Given:

#f(x) = x^3-7x^2+x-7#

Note that the ratio between the first and second terms is the same as that between the third and fourth terms.

So this cubic will factor by grouping:

#x^3-7x^2+x-7 = (x^3-7x^2)+(x-7)#

#color(white)(x^3-7x^2+x-7) = x^2(x-7)+1(x-7)#

#color(white)(x^3-7x^2+x-7) = (x^2+1)(x-7)#

Note that #x^2+1# has no linear factors with real coefficients since #x^2+1 > 0# for all real values of #x#. We can factor it as a difference of squares using complex coefficients:

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

So the zeros of #f(x)# are #7#, #i#, #-i#, all with multiplicity #1#.