Question

1. find all complex zeros for the function ( h(x) = 3x^3 - 12x^2 - 36x + 144 ). be sure to give exact answers.

Explanation:

Step1: Factor out the common factor

First, we can factor out a 3 from the polynomial \( h(x) = 3x^3-12x^2 - 36x + 144\). So we get \(h(x)=3(x^3 - 4x^2-12x + 48)\)

Step2: Group the terms in the cubic polynomial

Group the first two terms and the last two terms in \(x^3 - 4x^2-12x + 48\):
\(x^2(x - 4)-12(x - 4)\)

Step3: Factor out the common binomial factor

We can factor out \((x - 4)\) from the above expression:
\((x - 4)(x^2-12)\)

Step4: Find the roots of the factored form

Now we set \(h(x) = 0\), then \(3(x - 4)(x^2-12)=0\)
Since \(3
eq0\), we have two cases:
Case 1: \(x - 4=0\), which gives \(x = 4\)
Case 2: \(x^2-12=0\), then \(x^2=12\), so \(x=\pm\sqrt{12}=\pm2\sqrt{3}\)

Answer:

The complex zeros (in this case, all zeros are real, which are also complex numbers with imaginary part 0) of the function \(h(x)\) are \(x = 4\), \(x = 2\sqrt{3}\) and \(x=- 2\sqrt{3}\)