How do you find all the zeros of #f(x)=x^3-x^2+49x-49#?

1 Answer
George C.
Aug 19, 2016

#f(x)# has zeros #+-7i# and #1#

Explanation:

This cubic factors by grouping, then by using the difference of squares identity:

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

with #a=x# and #b=7i# as follows:

#f(x) = x^3-x^2+49x-49#

#=(x^3-x^2)+(49x-49)#

#=x^2(x-1)+49(x-1)#

#=(x^2+49)(x-1)#

#=(x^2-(7i)^2)(x-1)#

#=(x-7i)(x+7i)(x-1)#

Hence #f(x)# has zeros #+-7i# and #1#

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