Question

finding x-intercepts in unfactored form
this is the only question in this section.
question
factor to find all x-intercepts of the function.
$f(x) = 4x^4 + 12x^3 - 16x^2$
answer attempt 1 out of 2
additional solutions no solution
$x = $

Explanation:

Step1: Factor out the GCF

The greatest common factor (GCF) of \(4x^4\), \(12x^3\), and \(-16x^2\) is \(4x^2\). Factoring it out:
\(f(x) = 4x^2(x^2 + 3x - 4)\)

Step2: Factor the quadratic

Factor \(x^2 + 3x - 4\). We need two numbers that multiply to \(-4\) and add to \(3\). Those numbers are \(4\) and \(-1\):
\(x^2 + 3x - 4 = (x + 4)(x - 1)\)

Step3: Combine factors

Now, the fully factored form of \(f(x)\) is:
\(f(x) = 4x^2(x + 4)(x - 1)\)

Step4: Find x-intercepts

Set \(f(x) = 0\):
\(4x^2(x + 4)(x - 1) = 0\)
Using the zero - product property, we set each factor equal to zero:

• For \(4x^2 = 0\), we have \(x = 0\) (with multiplicity 2).
• For \(x + 4 = 0\), we have \(x=-4\).
• For \(x - 1 = 0\), we have \(x = 1\).

Answer:

The x - intercepts are \(x = 0\), \(x=-4\), and \(x = 1\) (where \(x = 0\) has multiplicity 2). In the box (if we list all distinct x - intercepts), the values are \(-4\), \(0\), \(1\) (usually written as \(x=-4\), \(x = 0\), \(x = 1\) or in the box as \(-4,0,1\) depending on the format required. But since the problem says "x - intercepts" and the box is for x values, we can list the distinct ones: \(-4\), \(0\), \(1\))