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

##### 2 Answers

#### 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,

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#