Question

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)?

Answer 1

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)`

Answer 2

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`