Answer :
Certainly! Let's solve the given polynomial step by step.
The polynomial \( P(x) \) is given by:
[tex]\[ P(x) = x^3 + 216 \][/tex]
### (a) Finding all zeros of \( P \), real and complex:
To find the zeros of the polynomial, we set \( P(x) \) equal to zero:
[tex]\[ x^3 + 216 = 0 \][/tex]
First, we can rewrite this equation as:
[tex]\[ x^3 = -216 \][/tex]
To solve for \( x \), we can take the cube root of both sides. The cube root of \(-216\) provides us with one real root and two complex roots. The real root is:
[tex]\[ x = -6 \][/tex]
For the complex roots, we need the complex cube roots of \(-216\). These roots can be expressed as:
[tex]\[ x = 3( \sqrt[3]{-1} ) = 3 \left( -1 \right) = -3 \][/tex]
However, since cube roots involve complex numbers, we consider the unit roots in the complex plane.
The complex roots include:
[tex]\[ 3 \left( e^{i (\frac{2\pi}{3} + \frac{4\pi n}{3})} \right) \][/tex]
where \( n \) is an integer.
These roots simplify to:
[tex]\[ x = 3 - 3\sqrt{3}i \quad \text{and} \quad x = 3 + 3\sqrt{3}i \][/tex]
So, the complete list of zeros is:
[tex]\[ x = -6, \ x = 3 - 3\sqrt{3}i, \ x = 3 + 3\sqrt{3}i \][/tex]
Therefore, the zeros of the polynomial \( P(x) = x^3 + 216 \) are:
[tex]\[ x = -6, 3 - 3\sqrt{3}i, 3 + 3\sqrt{3}i \][/tex]
### (b) Factor \( P \) completely:
Given that we have found all the zeros, we can factor the polynomial using these roots. The polynomial can be written as:
[tex]\[ P(x) = (x + 6) (x^2 - 6x + 36) \][/tex]
Here’s the reasoning:
1. We know that \( x = -6 \) is a zero, so \( (x + 6) \) is a factor.
2. Dividing \( x^3 + 216 \) by \( (x + 6) \) results in the quadratic \(x^2 - 6x + 36 \).
Thus, factoring completely, we get:
[tex]\[ P(x) = (x + 6) (x^2 - 6x + 36) \][/tex]
Therefore, the fully factored form of \( P(x) \) is:
[tex]\[ P(x) = (x + 6)(x^2 - 6x + 36) \][/tex]
The polynomial \( P(x) \) is given by:
[tex]\[ P(x) = x^3 + 216 \][/tex]
### (a) Finding all zeros of \( P \), real and complex:
To find the zeros of the polynomial, we set \( P(x) \) equal to zero:
[tex]\[ x^3 + 216 = 0 \][/tex]
First, we can rewrite this equation as:
[tex]\[ x^3 = -216 \][/tex]
To solve for \( x \), we can take the cube root of both sides. The cube root of \(-216\) provides us with one real root and two complex roots. The real root is:
[tex]\[ x = -6 \][/tex]
For the complex roots, we need the complex cube roots of \(-216\). These roots can be expressed as:
[tex]\[ x = 3( \sqrt[3]{-1} ) = 3 \left( -1 \right) = -3 \][/tex]
However, since cube roots involve complex numbers, we consider the unit roots in the complex plane.
The complex roots include:
[tex]\[ 3 \left( e^{i (\frac{2\pi}{3} + \frac{4\pi n}{3})} \right) \][/tex]
where \( n \) is an integer.
These roots simplify to:
[tex]\[ x = 3 - 3\sqrt{3}i \quad \text{and} \quad x = 3 + 3\sqrt{3}i \][/tex]
So, the complete list of zeros is:
[tex]\[ x = -6, \ x = 3 - 3\sqrt{3}i, \ x = 3 + 3\sqrt{3}i \][/tex]
Therefore, the zeros of the polynomial \( P(x) = x^3 + 216 \) are:
[tex]\[ x = -6, 3 - 3\sqrt{3}i, 3 + 3\sqrt{3}i \][/tex]
### (b) Factor \( P \) completely:
Given that we have found all the zeros, we can factor the polynomial using these roots. The polynomial can be written as:
[tex]\[ P(x) = (x + 6) (x^2 - 6x + 36) \][/tex]
Here’s the reasoning:
1. We know that \( x = -6 \) is a zero, so \( (x + 6) \) is a factor.
2. Dividing \( x^3 + 216 \) by \( (x + 6) \) results in the quadratic \(x^2 - 6x + 36 \).
Thus, factoring completely, we get:
[tex]\[ P(x) = (x + 6) (x^2 - 6x + 36) \][/tex]
Therefore, the fully factored form of \( P(x) \) is:
[tex]\[ P(x) = (x + 6)(x^2 - 6x + 36) \][/tex]
