Consider the equation z4−3z3−2z2+112z+192=0*z*4−3*z*3−2*z*2+112*z*+192=0, which has z=−3*z*=−3 and z=−2*z*=−2 as solutions.

Give the (complex and real) solutions to z4−3z3−2z2+112z+192=0*z*4−3*z*3−2*z*2+112*z*+192=0.

Give your solutions as a comma-separated list z1,z2,…*z*1​,*z*2​,… 

Include solutions as often as their multiplicity.

We are given the polynomial equation:

\[
z^4 - 3z^3 - 2z^2 + 112z + 192 = 0
\]

with the known solutions \( z = -3 \) and \( z = -2 \). The task is to find all the real and complex solutions to this polynomial, including any multiplicities.

### Step 1: Factorizing the polynomial

Since we know two solutions \( z = -3 \) and \( z = -2 \), we can factor the polynomial using these roots. This implies that \( z + 3 \) and \( z + 2 \) are factors of the polynomial. So we can express the polynomial as:

\[
(z + 3)(z + 2)Q(z) = z^4 - 3z^3 - 2z^2 + 112z + 192
\]

where \( Q(z) \) is a quadratic polynomial. We will now divide the given quartic polynomial by \( (z + 3)(z + 2) \) to determine \( Q(z) \).

### Step 2: Perform polynomial division

We divide \( z^4 - 3z^3 - 2z^2 + 112z + 192 \) by \( (z + 3)(z + 2) \), which will give us the remaining quadratic factor \( Q(z) \).

Performing the division (either manually or using synthetic division), we find that:

\[
Q(z) = z^2 - 8z + 32
\]

### Step 3: Solve the quadratic equation

Now we solve the quadratic equation:

\[
z^2 - 8z + 32 = 0
\]

Using the quadratic formula:

\[
z = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(32)}}{2(1)} = \frac{8 \pm \sqrt{64 - 128}}{2} = \frac{8 \pm \sqrt{-64}}{2}
\]

\[
z = \frac{8 \pm 8i}{2}
\]

\[
z = 4 \pm 4i
\]

Thus, the two complex solutions are \( z = 4 + 4i \) and \( z = 4 - 4i \).

### Step 4: List all solutions

We now have four solutions to the quartic equation:

- \( z = -3 \) (real root)
- \( z = -2 \) (real root)
- \( z = 4 + 4i \) (complex root)
- \( z = 4 - 4i \) (complex root)

### Final Answer:
The solutions are:

\[
z_1 = -3, z_2 = -2, z_3 = 4 + 4i, z_4 = 4 - 4i
\]

Would you like any details or have any questions?

Here are 5 related questions to explore:
1. How do you perform polynomial division manually?
2. What is the significance of the quadratic formula in solving polynomials?
3. How do complex roots come in conjugate pairs for polynomials with real coefficients?
4. Can the factor theorem be used to find all roots of any polynomial?
5. How does the discriminant help in predicting the nature of solutions?

**Tip**: Complex roots in polynomials with real coefficients always appear in conjugate pairs!

Consider the equation z^4 - 3z^3 - 2z^2 + 112z + 192 = 0, which has z = -3 and z = -2 as solutions. Give the (complex and real) solutions to the equation. Give your solutions as a comma-separated list z1, z2, ... Include solutions as often as their multiplicity.

Discover how to solve the quartic equation z^4 - 3z^3 - 2z^2 + 112z + 192 = 0, which includes real and complex solutions. The equation is factored using known roots, and the quadratic formula is applied. Suitable for students in Grades 11-12.

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