Problem 95 - Algebra for College Students

Question

Exploration. a) Given that \(P(x)=x^{4}+6 x^{2}-27,\) find \(P(3 i)\) 8 \(P(-3 i), P(\sqrt{3}),\) and \(P(-\sqrt{3})\) b) What can you conclude about the values \(3 i,-3 i, \sqrt{3}\) and \(-\sqrt{3}\) and their relationship to each other?

Step-by-Step Solution

Answer

The values \(3i, -3i, \root 3, and -\root 3\) are roots of the polynomial \(P(x) = x^4 + 6x^2 - 27\).

Step 1 - Evaluate \(P(3i)\)

Substitute \(3i\) into the polynomial \(P(x) = x^4 + 6x^2 - 27\): \(P(3i) = (3i)^4 + 6(3i)^2 - 27 = 81i^4 + 54i^2 - 27\). Since \(i^4 = 1\) and \(i^2 = -1\), this results in \(81 \times 1 + 54 \times (-1) - 27 = 81 - 54 - 27 = 0\). So, \(P(3i) = 0\).

Step 2 - Evaluate \(P(-3i)\)

Substitute \(-3i\) into the polynomial \(P(x) = x^4 + 6x^2 - 27\): \(P(-3i) = (-3i)^4 + 6(-3i)^2 - 27 = 81i^4 + 54i^2 - 27\). Using the same identities as above, \(81 \times 1 + 54 \times (-1) - 27 = 81 - 54 - 27 = 0\). So, \(P(-3i) = 0\).

Step 3 - Evaluate \(P(\root 3 )\)

Substitute \(\root 3\) into the polynomial \(P(x) = x^4 + 6x^2 - 27\): \(P(\root 3) = (\root 3)^4 + 6(\root 3)^2 - 27 = 9 + 18 - 27 = 0\). So, \(P(\root 3) = 0\).

Step 4 - Evaluate \(P(-\root 3)\)

Substitute \(\root 3\) into the polynomial \(P(x) = x^4 + 6x^2 - 27\): \(P(\root 3) = (\root 3)^4 + 6(\root 3)^2 - 27 = 9 + 18 - 27 = 0\). So, \(P(\root 3) = 0\).

Step 5: Conclusion about values

From the evaluations, \(3i, -3i, \root 3, and -\root 3\) are all roots of the polynomial \(P(x)\). Therefore, these values have a relationship in that they all solve or satisfy the polynomial equation \(P(x) = 0\).

Key Concepts

Complex Numbers

Complex numbers extend the idea of one-dimensional number line to the complex plane by using two orthogonal axes. Typically, these numbers are composed of a real part and an imaginary part. The imaginary unit, denoted as \(i\), satisfies the equation \(i^2 = -1\). For example, the complex number \(3i\) has a real part of 0 and an imaginary part of 3.

In the exercise, we see the substitution of \(3i\) and \(-3i\) into the polynomial \(P(x) = x^4 + 6x^2 - 27\). By replacing \(x\) with \(3i\), we understand how to manipulate and evaluate polynomials using complex numbers. We recall that \(i^4 = 1\) and \(i^2 = -1\), which helps simplify the polynomial evaluation.

• This emphasizes that complex numbers are not just abstract constructs but play a crucial role in solving polynomial equations.
• It also shows how to handle higher powers of \(i\) by knowing the fundamental identity \(i^2 = -1\).

Evaluating Polynomials

Polynomial evaluation means substituting a value into a polynomial and simplifying to get a result. For our exercise, the polynomial given is \(P(x) = x^4 + 6x^2 - 27\).

Here's a brief breakdown:

• Substitute the value into the polynomial.
• Follow the rules of arithmetic operations and exponents.

For instance, when substituting \(3i\):
\[P(3i) = (3i)^4 + 6(3i)^2 - 27 = 81i^4 + 54i^2 - 27\] Using the identities \(i^4 = 1\) and \(i^2 = -1\), it simplifies to: \[81 - 54 - 27 = 0\] Evaluation shows how different values (real or complex) interact within the polynomial and yield a result.

Roots of Polynomials

The roots of a polynomial are values for which the polynomial equals zero. To determine if a value is a root, substitute it into the polynomial and see if the result is zero.

In our exercise, we find:

• \(P(3i) = 0\)
• \(P(-3i) = 0\)
• \(P(\root 3) = 0\)
• \(P(-\root 3) = 0\)

All of these values are roots.
This shows that they solve \(P(x) = 0\), which signifies they are solutions to the polynomial equation. Identifying roots is key in understanding the behavior of polynomials, as the roots define where the polynomial crosses the x-axis in a graph.

Algebra

Algebra is the branch of mathematics that uses symbols and rules to manipulate those symbols, helping us to model and solve real-world problems. The exercise on polynomial roots is a prime example of applying algebraic methods.

Here, the polynomial \(P(x) = x^4 + 6x^2 - 27\) was factored and evaluated using algebraic transformations and principles.

• By substituting different values, we see the polynomial's behavior.
• Complex numbers illustrate the polynomial's versatility in algebra.

In essence, mastering algebra enables understanding and solving diverse mathematical problems.