Solve the following polynomial equation.

z4+32z2−144=0

We are given the polynomial equation:

\[
z^4 + 32z^2 - 144 = 0
\]

To solve this equation, we can use substitution. Let \( u = z^2 \). Then, the equation becomes:

\[
u^2 + 32u - 144 = 0
\]

This is now a quadratic equation in \( u \), and we can solve it using the quadratic formula:

\[
u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For the equation \( u^2 + 32u - 144 = 0 \), the coefficients are:
- \( a = 1 \)
- \( b = 32 \)
- \( c = -144 \)

Now, substituting these values into the quadratic formula:

\[
u = \frac{-32 \pm \sqrt{32^2 - 4(1)(-144)}}{2(1)}
\]

\[
u = \frac{-32 \pm \sqrt{1024 + 576}}{2}
\]

\[
u = \frac{-32 \pm \sqrt{1600}}{2}
\]

\[
u = \frac{-32 \pm 40}{2}
\]

This gives two solutions for \( u \):

1. \( u = \frac{-32 + 40}{2} = \frac{8}{2} = 4 \)
2. \( u = \frac{-32 - 40}{2} = \frac{-72}{2} = -36 \)

Now, we recall that \( u = z^2 \), so we solve for \( z \) in both cases:

1. For \( u = 4 \):

\[
z^2 = 4 \quad \Rightarrow \quad z = \pm 2
\]

2. For \( u = -36 \):

\[
z^2 = -36 \quad \Rightarrow \quad z = \pm 6i
\]

Thus, the solutions for \( z \) are:

\[
z = \pm 2 \quad \text{or} \quad z = \pm 6i
\]

So, the complete solution is:

\[
z = 2, -2, 6i, -6i
\]

### Do you need more details or have any questions?

Here are 5 related questions:
1. How does substitution simplify solving higher-degree polynomials?
2. How does the quadratic formula work for non-real solutions?
3. Why do we get imaginary numbers when solving for \( u = -36 \)?
4. Can we solve a similar equation without using substitution?
5. What are some common uses of polynomial equations in real life?

**Tip:** Always check for extraneous solutions when solving equations, especially in cases involving square roots or complex numbers.

Solve the following polynomial equation: z^4 + 32z^2 - 144 = 0

Discover how to solve the polynomial equation z^4 + 32z^2 - 144 using substitution and the quadratic formula. This detailed guide explores algebraic techniques and complex solutions, ideal for students in Grades 9-12.

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