 Find all exact (unrounded, not decimal approximations) complex (real & imaginary) solutions to the equation
    8x4 – 12x3 + 10x2 - 3x + 2 = 0.   (Hint:  one root is  (1/2)i )

We are given the polynomial equation:

\[
8x^4 - 12x^3 + 10x^2 - 3x + 2 = 0
\]
with one of the roots being \( \frac{1}{2}i \).

### Step 1: Use the Conjugate Root Theorem
Since the coefficients of the polynomial are real, the complex roots must occur in conjugate pairs. Therefore, if \( \frac{1}{2}i \) is a root, its conjugate \( -\frac{1}{2}i \) is also a root.

### Step 2: Form a Quadratic Factor from the Complex Roots
The roots \( \frac{1}{2}i \) and \( -\frac{1}{2}i \) give us the quadratic factor:
\[
(x - \frac{1}{2}i)(x + \frac{1}{2}i) = x^2 + \left( \frac{1}{2}i \right)^2 = x^2 + \frac{1}{4}.
\]
So, one factor of the polynomial is \( x^2 + \frac{1}{4} \).

### Step 3: Perform Polynomial Division
We now divide the original polynomial \( 8x^4 - 12x^3 + 10x^2 - 3x + 2 \) by \( x^2 + \frac{1}{4} \) using synthetic or long division.

To simplify the division, it's easier to multiply the factor \( x^2 + \frac{1}{4} \) by 4, resulting in \( 4x^2 + 1 \), and then divide the original polynomial by this scaled factor.

We now divide \( 8x^4 - 12x^3 + 10x^2 - 3x + 2 \) by \( 4x^2 + 1 \).

#### Division:

1. Divide \( 8x^4 \) by \( 4x^2 \), which gives \( 2x^2 \).
2. Multiply \( 2x^2 \) by \( 4x^2 + 1 \) to get \( 8x^4 + 2x^2 \).
3. Subtract \( 8x^4 + 2x^2 \) from the original polynomial: 
   \[
   (8x^4 - 12x^3 + 10x^2 - 3x + 2) - (8x^4 + 2x^2) = -12x^3 + 8x^2 - 3x + 2.
   \]
4. Divide \( -12x^3 \) by \( 4x^2 \), which gives \( -3x \).
5. Multiply \( -3x \) by \( 4x^2 + 1 \) to get \( -12x^3 - 3x \).
6. Subtract \( -12x^3 - 3x \) from \( -12x^3 + 8x^2 - 3x + 2 \): 
   \[
   (-12x^3 + 8x^2 - 3x + 2) - (-12x^3 - 3x) = 8x^2 + 2.
   \]
7. Divide \( 8x^2 \) by \( 4x^2 \), which gives \( 2 \).
8. Multiply \( 2 \) by \( 4x^2 + 1 \) to get \( 8x^2 + 2 \).
9. Subtract \( 8x^2 + 2 \) from \( 8x^2 + 2 \): 
   \[
   (8x^2 + 2) - (8x^2 + 2) = 0.
   \]

The quotient from the division is \( 2x^2 - 3x + 2 \), and the remainder is zero. Therefore, we can factor the polynomial as:
\[
(4x^2 + 1)(2x^2 - 3x + 2) = 0.
\]

### Step 4: Solve for Roots
We now solve both quadratic equations.

#### 1. Solve \( 4x^2 + 1 = 0 \):
\[
4x^2 = -1 \quad \Rightarrow \quad x^2 = -\frac{1}{4} \quad \Rightarrow \quad x = \pm \frac{1}{2}i.
\]
These are the roots \( \frac{1}{2}i \) and \( -\frac{1}{2}i \), which were already known.

#### 2. Solve \( 2x^2 - 3x + 2 = 0 \) using the quadratic formula:
\[
x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(2)}}{2(2)} = \frac{3 \pm \sqrt{9 - 16}}{4} = \frac{3 \pm \sqrt{-7}}{4} = \frac{3 \pm i\sqrt{7}}{4}.
\]

Thus, the solutions from this quadratic are:
\[
x = \frac{3 + i\sqrt{7}}{4}, \quad x = \frac{3 - i\sqrt{7}}{4}.
\]

### Final Answer:
The exact complex solutions to the equation \( 8x^4 - 12x^3 + 10x^2 - 3x + 2 = 0 \) are:
\[
x = \frac{1}{2}i, \quad x = -\frac{1}{2}i, \quad x = \frac{3 + i\sqrt{7}}{4}, \quad x = \frac{3 - i\sqrt{7}}{4}.
\]

Would you like to dive deeper into any step of this solution?

### Relative Questions:
1. How do you apply synthetic division in polynomial division?
2. Why do complex roots occur in conjugate pairs for polynomials with real coefficients?
3. How is the quadratic formula derived?
4. What is the significance of the discriminant in quadratic equations?
5. How do you interpret the result of polynomial division in factorization?

**Tip:** Always check for possible complex conjugates when working with polynomials that have real coefficients!

Find all exact (unrounded, not decimal approximations) complex (real & imaginary) solutions to the equation 8x4 – 12x3 + 10x2 - 3x + 2 = 0. (Hint: one root is (1/2)i)

Learn how to solve the polynomial equation 8x^4 - 12x^3 + 10x^2 - 3x + 2 = 0 using the Conjugate Root Theorem and quadratic formula. This step-by-step solution shows how to factor the polynomial and find the complex roots, including (1/2)i. Suitable for students in Grades 11-12.

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