Mathematics: One factor of f(x) = x3 − 12x2 + 47x − 60 is (x − 5). what are the zeros of the function? 5, −4, −3 5, −4, 3 5, 4, −3 5, 4, 3

One factor of f(x) = x3 − 12x2 + 47x − 60 is - id3122476053388845, mckinleesmomp6qj1e

viktoria1198zz

07.01.2022

The correct option is D) 5, 4, 3

Step-by-step explanation:

Consider the provided function.

f(x) = x^3 -12x^2 + 47x - 60

It is given that one factor is (x-5) and we need to find the zeros of the function.

That means x-5 will completely divide the provided polynomial.

$\frac{x^3 -12x^2 + 47x - 60}{x-5}$

The long division is shown in figure below.

$\frac{x^3-12x^2+47x-60}{x-3}=x^2-7x+12$

Simplify the expression x^2-7x+12

x^2-4x-3x+12

x(x-4)-3(x-4)

(x-4)(x-3)

Therefore, the required polynomial can be written as: f(x)=(x-3)(x-4)(x-5)

Now find the zeros by substituting f(x)=0.

(x-3)(x-4)(x-5)=0

$x-3=0\ or\ x-4=0\ or\ x-5=0$

$x=3\ or\ x=4\ or\ x=5$

Hence, the correct option is D) 5, 4, 3

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