Question

How do you find all the zeros of `f(x)= (x-1)(3x-4)(x^2+3)`?

Answer 1

If the domain is real numbers, zeros are `x=1` and `x=4/3`. If domain is extended to complex numbers, we have two additional zeros `x=isqrt3` and `x=-isqrt3`.

Explanation:

Zeros of a polynomial are those values of variable, for which value of the polynomial becomes zero.

As the given polynomial `f(x)=(x-1)(3x-4)(x^2+3_` is already written as product of binomials, `f(x)` will be zero if

`x-1=0` or `3x-4=0` or `x^2+3=0`.

If the domain for `x` is real numbers, the two zeros are given by first two binomials i.e. `x=1` and `x=4/3` (as in such a case `x^2+3` will always be non-zero.

However, if domain is extended to complex numbers, `x^2+3=0` gives us two more zeros given by `x=isqrt3` and `x=-isqrt3`.

Answer 2

The zeros, which correspond to the factors are:

`x = 1`, `x = 4/3`, `x = +-sqrt(3)i`

Explanation:

`f(x) = (x-1)(3x-4)(x^2+3)`

Note that if any one of the factors `(x-1)`, `(3x-4)` or `(x^2+3)` is zero, then their product `f(x)` is zero.

If all of the factors are non-zero, then their product `f(x)` is also non-zero.

In symbols we could write:

`f(x) = 0 <=> ((x-1) = 0 vv (3x-4) = 0 vv (x^2+3) = 0)`

Looking at each factor in turn:

`(x-1) = 0` if and only if `x = 1`

`(3x-4) = 0` if and only if `x = 4/3`

In general, note that each linear factor corresponds to a zero:

`(x-a) = 0` if and only if `x = a`

`(qx-p) = 0` if and only if `x = p/q`

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The last factor `(x^2+3)` is slightly more complicated.

First note that it has no Real zeros, since for any Real value of `x` we have `x^2 >= 0`, so `(x^2+3) >= 3`.

If we look at Complex numbers, we find:

`(sqrt(3)i)^2 = (sqrt(3))^2 * i^2 = 3*(-1) = -3`

So:

`x^2+3 = x^2-(-3) = x^2-(sqrt(3)i)^2 = (x-sqrt(3)i)(x+sqrt(3)i)`

We now have linear factors, corresponding to zeros:

`x = sqrt(3)i" "` and `" "x = -sqrt(3)i`