Question

Use synthetic division to find the real zeroes of f left parenthesis x right parenthesis equals 2 x cubed plus 11 x squared plus 12 x minus 9.
(1 point)
Responses

negative 1 comma 3 halves comma 9
Image with alt text: negative 1 comma 3 halves comma 9

negative 3 comma one half
Image with alt text: negative 3 comma one half

negative one half comma 1 comma 3
Image with alt text: negative one half comma 1 comma 3

negative 9 halves comma 3
Image with alt text: negative 9 halves comma 3

Answer 1

To find the real zeroes of the polynomial \( f(x) = 2x^3 + 11x^2 + 12x - 9 \) using synthetic division, we'll first look for potential rational roots using the Rational Root Theorem. The possible rational roots (factors of the constant term divided by factors of the leading coefficient) are:

- Factors of -9: ±1, ±3, ±9
- Factors of 2: ±1, ±2

This gives us the possible rational roots: ±1, ±3, ±9, ±1/2.

We can test these values in synthetic division to see if they yield a remainder of 0.

Let's test \( x = 1 \):
```
1 | 2 11 12 -9
| 2 13 25
---------------------
2 13 25 16
```
The remainder is 16, so \( x = 1 \) is not a root.

Next, test \( x = -1 \):
```
-1 | 2 11 12 -9
| -2 -9 -3
-----------------------
2 9 3 -12
```
The remainder is -12, so \( x = -1 \) is not a root.

Now test \( x = 3 \):
```
3 | 2 11 12 -9
| 6 51 192
-----------------------
2 17 63 183
```
The remainder is 183, so \( x = 3 \) is not a root.

Next, let’s test \( x = -3 \):
```
-3 | 2 11 12 -9
| -6 -15 9
------------------------
2 5 -3 0
```
The remainder is 0, so \( x = -3 \) is a root.

Now, we have reduced the polynomial by synthetic division:
The result from synthetic division is: \( 2x^2 + 5x - 3 \).

Next, we need to factor \( 2x^2 + 5x - 3 \).

We can use the quadratic formula to find the remaining roots:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 2, b = 5, c = -3 \).

Calculating the discriminant:
\[
D = b^2 - 4ac = 5^2 - 4 \cdot 2 \cdot (-3) = 25 + 24 = 49
\]
Now, applying the quadratic formula:
\[
x = \frac{-5 \pm \sqrt{49}}{2 \cdot 2} = \frac{-5 \pm 7}{4}
\]
This gives us two solutions:
\[
x = \frac{2}{4} = \frac{1}{2} \quad \text{and} \quad x = \frac{-12}{4} = -3
\]

Thus, the real zeroes of \( f(x) \) are:
1. \( x = -3 \) (which we found through synthetic division)
2. \( x = \frac{1}{2} \) (from quadratic formula)

So the final real zeroes are \( x = -3 \) and \( x = \frac{1}{2} \).

Looking at the options provided, the answer is:

**negative 3, one half**