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

##### 2 Answers

#### Answer:

#### Explanation:

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

As the given polynomial

If the domain for

However, if domain is extended to complex numbers,

#### Answer:

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

If all of the factors are non-zero, then their product

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#

The last factor

First note that it has no Real zeros, since for any Real value of

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#