Problem 73 - Algebra A Combined Function

Question

Solve. $$ y^{3}+9 y-y^{2}-9=0 $$

Step-by-Step Solution

Answer

The solutions are \(y = 1,\) \(y = 3i,\) and \(y = -3i\).

Step 1: Reorder and Group Terms

Reorder the given terms from highest to lowest exponent:\[ y^3 - y^2 + 9y - 9 = 0 \]

Step 2: Use Synthetic Division or Factoring Techniques

Attempt to factor by grouping or trial roots. For this equation, let's factor by grouping.Group terms as follows: \( (y^3 - y^2) + (9y - 9) \).Factor each group: \( y^2(y - 1) + 9(y - 1) \).

Step 3: Factor Out the Common Term

Notice \(y - 1\) is a common factor. Factor it out:\[ (y - 1)(y^2 + 9) = 0 \]

Step 4: Solve for Each Factor

Solve each factor for \(y\):\( y - 1 = 0 \) leads to \( y = 1 \).\( y^2 + 9 = 0 \) implies \( y^2 = -9 \), which has solutions \( y = 3i \) and \( y = -3i \), where \( i \) is the imaginary unit.

Step 5: Write the Complete Solution

The solutions to the original equation are:\[ y = 1, \, y = 3i, \, y = -3i \]

Key Concepts

Factoring Polynomials

Factoring polynomials is a core skill in algebra that involves breaking down a polynomial into simpler components, known as factors, that when multiplied together give back the original polynomial. This technique is essential because it simplifies problems, making them easier to solve. In our exercise, we have a cubic polynomial, and the strategy used is factoring by grouping.
In the case of grouping, you look for terms that can pair together to reveal a common factor. Here, we re-order the terms and group them as:

• \((y^3 - y^2)\)
• \((9y - 9)\)

Next, we factor out the greatest common factor from each group.

• For \((y^3 - y^2)\), we factor out \(y^2\) to get \(y^2(y - 1)\).
• For \((9y - 9)\), we factor out \(9\) to obtain \(9(y - 1)\).

Now, the original polynomial is expressed as \(y^2(y - 1) + 9(y - 1)\). Since \((y - 1)\) is a common factor in both terms, it's further factored out, resulting in the factored form: \((y - 1)(y^2 + 9)\). This step reveals that some roots may be derived immediately from the factors.

Synthetic Division

Synthetic division is a shortcut method of polynomial division, especially when dividing by a linear factor. While this method was not explicitly employed in our exercise solution, it's a beneficial ally in factoring and finding polynomial roots.
To use synthetic division, one typically needs a suspected root from the polynomial. As we were factoring \((y - 1)\), this suggests the root could be \(y = 1\). By performing synthetic division with \(y = 1\), we confirm whether \((y - 1)\) is indeed a factor.
Here’s a simplified procedure for synthetic division:

• Write down the coefficients of the polynomial.
• Use the suspected root (in this case \(1\)).
• Perform synthetic division to obtain a new set of coefficients which fleshes out the remaining factor of the polynomial.

This process gives a rapid way to assess factorizability and confirm roots, reinforcing the polynomial's structure for additional solvable factors.

Complex Numbers

Complex numbers are numbers that have both a real part and an imaginary part. They are written in the form \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit satisfying \(i^2 = -1\).
In solving polynomial equations, especially when factoring leads to terms like \(y^2 + 9 = 0\), complex numbers become essential. Solving \(y^2 + 9 = 0\) requires manipulating real numbers into a form that incorporates the imaginary unit.
To solve this equation, we take the square root of both sides:

• \(y^2 = -9\)
• \(y = \pm\sqrt{-9}\)

By recognizing \(\sqrt{-9} = \sqrt{9} \cdot \sqrt{-1}\), and knowing \(\sqrt{-1} = i\), we derive \(y = \pm 3i\). This incorporation of \(i\) highlights why complex numbers are significant in polynomial solutions.

Imaginary Unit

The imaginary unit, denoted by \(i\), is vital in extending the concept of numbers beyond the real line. Defined by the property \(i^2 = -1\), it allows the square root of negative numbers to be expressed as a multiple of \(i\). This concept is fundamental in engineering, physics, and mathematics for solving equations and expressing numbers not limited to the real line.
When solving polynomials where a negative under the square root arises, such as \(y^2 + 9 = 0\), the imaginary unit comes into play. Here, \(y^2 = -9\) leads to the realization that:

• \(y = \pm\sqrt{-9} = \pm 3i\)

In this way, the imaginary unit makes complex solutions possible. It's crucial for interpretative and computational purposes, enabling mathematical equations to have broader applications and utility in various fields.