Question

find the intercepts of the function.
$f(x) = x^3 - 3x^2 - 4x + 12$
answer
attempt 1 out of 2

Explanation:

Step1: Factor by grouping

Group the terms: \((x^3 - 3x^2) + (-4x + 12)\)
Factor out common terms: \(x^2(x - 3) - 4(x - 3)\)
Now factor out \((x - 3)\): \((x - 3)(x^2 - 4)\)

Step2: Factor the difference of squares

Notice that \(x^2 - 4\) is a difference of squares, so \(x^2 - 4=(x - 2)(x + 2)\)
So the factored form is \(f(x)=(x - 3)(x - 2)(x + 2)\)

Step3: Find the x-intercepts

Set \(f(x) = 0\): \((x - 3)(x - 2)(x + 2)=0\)
Using the zero - product property, we have \(x - 3 = 0\) or \(x - 2 = 0\) or \(x+2 = 0\)
Solving these equations gives \(x = 3\), \(x = 2\), \(x=-2\)

Answer:

The x - intercepts of the function \(f(x)=x^{3}-3x^{2}-4x + 12\) are \(x=-2\), \(x = 2\) and \(x = 3\)