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

2 Answers
Jun 15, 2016

Answer:

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#.

Jun 15, 2016

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 #(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#

#color(white)()#
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#