Question

y=(-3x^2+2x-5)(x+3)
how many soloutions does this equation have
what type of solutions are they
how do you know the number of soloutions and the type
what are the soloutions

Answer 1

To analyze the equation \( y = (-3x^2 + 2x - 5)(x + 3) \), we'll need to set it to zero to find the solutions:

\[
-3x^2 + 2x - 5 = 0
\]
and
\[
x + 3 = 0
\]

Let’s first solve the equation \( x + 3 = 0 \):

\[
x = -3
\]

Now, we will solve the quadratic equation \( -3x^2 + 2x - 5 = 0 \). We can use the quadratic formula:

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

where \( a = -3 \), \( b = 2 \), and \( c = -5 \).

**Step 1: Calculate the discriminant \( b^2 - 4ac \)**

\[
b^2 - 4ac = 2^2 - 4(-3)(-5) = 4 - 60 = -56
\]

**Step 2: Analyze the discriminant**

Since the discriminant is negative \((-56 < 0)\), the quadratic equation has no real solutions. It has two complex solutions.

**Total number of solutions:**

1. From the linear factor \( x + 3 = 0 \), we have **one real solution** \( x = -3 \).
2. From the quadratic part \( -3x^2 + 2x - 5 = 0 \), we have **two complex solutions**.

Thus, there is a total of **one real solution** and **two complex solutions**.

### Summary:

- **Number of solutions**: 1 real solution, 2 complex solutions
- **Type of solutions**:
- 1 real solution \( x = -3 \)
- 2 complex solutions (from the quadratic equation) which can be explicitly calculated but will not be real values.

To find the complex solutions, we proceed with the quadratic formula:

\[
x = \frac{-2 \pm \sqrt{-56}}{-6} = \frac{-2 \pm \sqrt{56}i}{-6}
\]
\[
x = \frac{-2 \pm 2\sqrt{14}i}{-6} = \frac{1 \pm \sqrt{14}i}{3}
\]

### Final Solutions:

- Real solution: \( x = -3 \)
- Complex solutions: \( x = \frac{1 + \sqrt{14}i}{3} \) and \( x = \frac{1 - \sqrt{14}i}{3} \)