Answer :
Let's solve the given polynomial \( P(x) = x^3 + 3x^2 + 3x \).
### Part (a): Finding the Zeros of \( P \)
To find the zeros of the polynomial \( P(x) \), we set the polynomial equal to zero and solve for \( x \):
[tex]\[ x^3 + 3x^2 + 3x = 0 \][/tex]
First, factor out an \( x \) from the polynomial:
[tex]\[ x(x^2 + 3x + 3) = 0 \][/tex]
This gives us one immediate zero:
[tex]\[ x = 0 \][/tex]
Next, we solve the quadratic equation \( x^2 + 3x + 3 \) for the remaining zeros. We use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, \( a = 1 \), \( b = 3 \), and \( c = 3 \). Plugging in these values, we get:
[tex]\[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1} = \frac{-3 \pm \sqrt{9 - 12}}{2} = \frac{-3 \pm \sqrt{-3}}{2} \][/tex]
Since \(\sqrt{-3}\) involves an imaginary number, \(\sqrt{-3} = i\sqrt{3}\), we get:
[tex]\[ x = \frac{-3 \pm i\sqrt{3}}{2} \][/tex]
This simplifies to two complex zeros:
[tex]\[ x = -\frac{3}{2} + \frac{\sqrt{3}i}{2}, \quad x = -\frac{3}{2} - \frac{\sqrt{3}i}{2} \][/tex]
So, the zeros of \( P(x) \) are:
[tex]\[ x = 0, -\frac{3}{2} + \frac{\sqrt{3}i}{2}, -\frac{3}{2} - \frac{\sqrt{3}i}{2} \][/tex]
### Part (b): Factoring \( P \) Completely
We will now factor \( P(x) \) completely. As we noted earlier, we can factor out an \( x \) initially:
[tex]\[ P(x) = x(x^2 + 3x + 3) \][/tex]
The quadratic \( x^2 + 3x + 3 \) cannot be factored further using real numbers, but we recognize the form from our quadratic solution. Thus, the polynomial is factored as:
[tex]\[ P(x) = x \left(x + \frac{3}{2} + \frac{\sqrt{3}i}{2}\right) \left(x + \frac{3}{2} - \frac{\sqrt{3}i}{2}\right) \][/tex]
In simpler form, keeping the quadratic term as is, we present:
[tex]\[ P(x) = x(x^2 + 3x + 3) \][/tex]
This is the fully factored form of the polynomial \( P(x) \).
### Summary of Solutions
(a) The zeros of \( P(x) = x^3 + 3x^2 + 3x \) are:
[tex]\[ x = 0, -\frac{3}{2} + \frac{\sqrt{3}i}{2}, -\frac{3}{2} - \frac{\sqrt{3}i}{2} \][/tex]
(b) The completely factored form of \( P(x) \) is:
[tex]\[ P(x) = x(x^2 + 3x + 3) \][/tex]
### Part (a): Finding the Zeros of \( P \)
To find the zeros of the polynomial \( P(x) \), we set the polynomial equal to zero and solve for \( x \):
[tex]\[ x^3 + 3x^2 + 3x = 0 \][/tex]
First, factor out an \( x \) from the polynomial:
[tex]\[ x(x^2 + 3x + 3) = 0 \][/tex]
This gives us one immediate zero:
[tex]\[ x = 0 \][/tex]
Next, we solve the quadratic equation \( x^2 + 3x + 3 \) for the remaining zeros. We use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, \( a = 1 \), \( b = 3 \), and \( c = 3 \). Plugging in these values, we get:
[tex]\[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1} = \frac{-3 \pm \sqrt{9 - 12}}{2} = \frac{-3 \pm \sqrt{-3}}{2} \][/tex]
Since \(\sqrt{-3}\) involves an imaginary number, \(\sqrt{-3} = i\sqrt{3}\), we get:
[tex]\[ x = \frac{-3 \pm i\sqrt{3}}{2} \][/tex]
This simplifies to two complex zeros:
[tex]\[ x = -\frac{3}{2} + \frac{\sqrt{3}i}{2}, \quad x = -\frac{3}{2} - \frac{\sqrt{3}i}{2} \][/tex]
So, the zeros of \( P(x) \) are:
[tex]\[ x = 0, -\frac{3}{2} + \frac{\sqrt{3}i}{2}, -\frac{3}{2} - \frac{\sqrt{3}i}{2} \][/tex]
### Part (b): Factoring \( P \) Completely
We will now factor \( P(x) \) completely. As we noted earlier, we can factor out an \( x \) initially:
[tex]\[ P(x) = x(x^2 + 3x + 3) \][/tex]
The quadratic \( x^2 + 3x + 3 \) cannot be factored further using real numbers, but we recognize the form from our quadratic solution. Thus, the polynomial is factored as:
[tex]\[ P(x) = x \left(x + \frac{3}{2} + \frac{\sqrt{3}i}{2}\right) \left(x + \frac{3}{2} - \frac{\sqrt{3}i}{2}\right) \][/tex]
In simpler form, keeping the quadratic term as is, we present:
[tex]\[ P(x) = x(x^2 + 3x + 3) \][/tex]
This is the fully factored form of the polynomial \( P(x) \).
### Summary of Solutions
(a) The zeros of \( P(x) = x^3 + 3x^2 + 3x \) are:
[tex]\[ x = 0, -\frac{3}{2} + \frac{\sqrt{3}i}{2}, -\frac{3}{2} - \frac{\sqrt{3}i}{2} \][/tex]
(b) The completely factored form of \( P(x) \) is:
[tex]\[ P(x) = x(x^2 + 3x + 3) \][/tex]
