Question

Find all the zeros. Write the answer in exact form.
C(x)=2x^4-5x^3-20x^2+35x-12

Answer 1

To find the zeros of the polynomial \( C(x) = 2x^4 - 5x^3 - 20x^2 + 35x - 12 \), we can start by trying to factor the polynomial or apply the Rational Root Theorem to identify possible rational roots.

### Step 1: Rational Root Theorem
According to the Rational Root Theorem, the possible rational roots are of the form \(\frac{p}{q}\), where \(p\) is a factor of the constant term (-12), and \(q\) is a factor of the leading coefficient (2).

**Factors of -12:** ±1, ±2, ±3, ±4, ±6, ±12
**Factors of 2:** ±1, ±2

**Possible rational roots:** ±1, ±2, ±3, ±4, ±6, ±12, ±\(\frac{1}{2}\), ±\(\frac{3}{2}\), ±\(\frac{6}{2}\)

This gives us the candidates:
\(\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12, \pm \frac{1}{2}, \pm \frac{3}{2}\)

### Step 2: Testing Possible Roots
We will evaluate \(C(x)\) at these possible rational roots.

1. **Testing \(x = 1\)**:
\[
C(1) = 2(1)^4 - 5(1)^3 - 20(1)^2 + 35(1) - 12 = 2 - 5 - 20 + 35 - 12 = 0
\]
Thus, \(x = 1\) is a root.

### Step 3: Polynomial Division
Now, we will perform synthetic division of \(C(x)\) by \(x - 1\):

\[
\begin{array}{r|rrrrr}
1 & 2 & -5 & -20 & 35 & -12 \\
& & 2 & -3 & -23 & 12 \\
\hline
& 2 & -3 & -23 & 12 & 0 \\
\end{array}
\]

So, the quotient is \(2x^3 - 3x^2 - 23x + 12\).

### Step 4: Finding Zeros of the Quotient
Now, we need to find the zeros of \(P(x) = 2x^3 - 3x^2 - 23x + 12\).

#### Step 4.1: Testing for more rational roots:
Continuing with possible rational roots on \(P(x)\):

2. **Testing \(x = 2\)**:
\[
P(2) = 2(2)^3 - 3(2)^2 - 23(2) + 12 = 16 - 12 - 46 + 12 = -30 \quad (\text{not a root})
\]

3. **Testing \(x = 3\)**:
\[
P(3) = 2(3)^3 - 3(3)^2 - 23(3) + 12 = 54 - 27 - 69 + 12 = -30 \quad (\text{not a root})
\]

4. **Testing \(x = 4\)**:
\[
P(4) = 2(4)^3 - 3(4)^2 - 23(4) + 12 = 128 - 48 - 92 + 12 = 0
\]
Thus, \(x = 4\) is another root.

### Step 5: Synthetic Division Again
Now perform synthetic division of \(P(x)\) by \(x - 4\):

\[
\begin{array}{r|rrrr}
4 & 2 & -3 & -23 & 12 \\
& & 8 & 20 & -12 \\
\hline
& 2 & 5 & -3 & 0 \\
\end{array}
\]

This gives us \(2x^2 + 5x - 3\).

### Step 6: Solving the Quadratic
Now, we need to find the roots of \(2x^2 + 5x - 3 = 0\) using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \(a = 2\), \(b = 5\), and \(c = -3\):
\[
x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 2 \cdot (-3)}}{2 \cdot 2} = \frac{-5 \pm \sqrt{25 + 24}}{4} = \frac{-5 \pm \sqrt{49}}{4} = \frac{-5 \pm 7}{4}
\]

Calculating the two possible values:
1. \(x = \frac{-5 + 7}{4} = \frac{2}{4} = \frac{1}{2}\)
2. \(x = \frac{-5 - 7}{4} = \frac{-12}{4} = -3\)

### Final Roots
Thus, the roots of the polynomial \( C(x) = 2x^4 - 5x^3 - 20x^2 + 35x - 12 \) are:
\[
x = 1, \quad x = 4, \quad x = \frac{1}{2}, \quad x = -3
\]
These are the zeros in exact form.