Question

Which of the following represents the zeros of f(x) = x3 − 12x2 + 47x − 60?

Answer 1

This actually should be reported. There are no choices. However there are ways of finding out what the zeros are and I will do that first to begin with. 

The first is a bit of a cheat. My calculator will do it, so I'm just going to get the answer from it. 

x can equal 3
x can be 5
x can be 4

So f(x)= a (x - 3)(x-5 )(x -4) is the basic equation. You need another point other than the zeros to determine the value of a.

Method 2
Get the graph. 
I've done that below. Basically it agrees with what my calculator says.

Method 3
You could factor it, but simple factoring methods don't work with this question.

Method 4
You could do this in a method similar to a quadratic, but I can't do that. It is on my to do list to lean though.

Answer 2

Answer:

x=3, 4, or 5

Step-by-step explanation:

$x^3 − 12x^2 + 47x − 60=0$

First we have to determine one value of x by hit and trial method for which , the above polynomial becomes 0

on trying random values of x , we sees that for x = 3 , the polynomial becomes 0

Hence (x-3) is one of the factor

Hence now let us move with the factorization part

The above polynomial can be re written as

$x^2(x-3)-9x(x-3)+20(x-3)$

$(x-3)(x^2-9x+20)$

$(x-3)(x^2-5x-4x+20)$

$(x-3)(x(x-5)-4(x-5))$

$(x-3)(x-5)(x-4)$

hence we have our $f(x) = (x-3)(x-5)(x-4)$

In order to find the zeroes we put f(x) =0

Hence

[tex](x-3)(x-5)(x-4)=0

Therefore

if (x-3)=0 , x=3

if (x-5) , x= 5

if (x-4), x=4