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:
- From the linear factor \( x + 3 = 0 \), we have one real solution \( x = -3 \).
- 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 \( x = \frac{1 - \sqrt{14}i}{3} \)