I'm trying to find the real zeros and put it in factored form, but I can't get it right. A\N is f(x) = (x-1)(3x-1)(x-i)(x+i). How do I get (3x-1)?

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2 Answers
Dec 11, 2017

The first two terms are correct. Then, you got #f(x)=(x-1)(x-1/3)(3x^2+3)#
#=>f(x)=(x-1)(x-1/3)( **3** (x-i)(x+i))# You forgot about the three while factoring.

This then becomes #f(x)=3(x-1)(x-1/3)(x-i)(x+i)=(x-1)(3*(x-1/3))(x-i)(x+i)=(x-1)(3x-1)(x-i)(x+i)#

Dec 11, 2017

The two answers are the same:

#3*(x-1/3)=3*0#
#-> (3x-1)=0#

Though the one without the fraction is the conventional way to present the root.

Explanation:

In terms of a solution of the zeros of a function, the factors #(3x-1)# and #(x-1/3)# are equivalent, so your answer is correct. Which one to write is a matter of convention - not wanting to have a fraction when it is possible to have an integer. But when you solve either one you get the same answer:

#(x-1/3)=0#
#-> x=1/3#

or

#(3x-1)=0#
#-> x=1/3#

The way to think about how to get one from the other is to think about multiplying both sides by a common factor. Since one side of the equation is zero, it has no effect overall.

#3*(x-1/3)=3*0#
#-> (3x-1)=0#