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Magazine Chapter 11: Problem 39
Solve. \(y^{3}+9 y-y^{2}-9=0\)
Short Answer
Expert verified
The solutions are \(y = 1\), \(y = 3i\), \(y = -3i\).
Step by step solution
01
Rearrange the equation
Write the given equation in a standard form of a polynomial equation: \[ y^3 - y^2 + 9y - 9 = 0 \]. Here, we simply rearranged the terms to follow the order of decreasing powers of \(y\).
02
Apply the Rational Root Theorem
According to the Rational Root Theorem, any rational root \( \frac{p}{q} \) of the polynomial, where \(y^3 - y^2 + 9y - 9 = 0\), must have \(p\) as a factor of the constant term \(-9\) and \(q\) as a factor of the leading coefficient \(1\). Possible rational roots are \( \pm 1, \pm 3, \pm 9 \).
03
Test the possible rational roots
Test the possible rational roots by substituting them into the polynomial to see which value satisfies the equation. Start with \(y = 1\):\[ 1^3 - 1^2 + 9(1) - 9 = 1 - 1 + 9 - 9 = 0 \]. Therefore, \(y = 1\) is a root of the polynomial.
04
Perform synthetic division
Use synthetic division to divide \(y^3 - y^2 + 9y - 9\) by \(y - 1\) since \(y = 1\) is a root. The coefficients of the polynomial are \(1, -1, 9, -9\). After performing synthetic division, the result is \(y^2 + 9\).
05
Solve the quadratic equation
Now solve the quadratic equation \(y^2 + 9 = 0\). Move the constant to the other side: \(y^2 = -9\). Taking the square root of both sides, we find \(y = \pm\sqrt{-9} = \pm 3i\), where \(i\) is the imaginary unit.
06
List all solutions
The solutions to the equation \(y^3 - y^2 + 9y - 9 = 0\) are \(y = 1\), \(y = 3i\), and \(y = -3i\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Rational Root Theorem
The Rational Root Theorem is a handy tool for finding potential rational solutions to polynomial equations.
It gives us a systematic way to list all possible rational roots to test.
This theorem applies to polynomials with integer coefficients, like the one we are dealing with: \( y^3 - y^2 + 9y - 9 = 0 \).
We test these values by substituting them into the polynomial until we find one that makes the polynomial equal to zero.
In this problem, that solution is \( y = 1 \).
It gives us a systematic way to list all possible rational roots to test.
This theorem applies to polynomials with integer coefficients, like the one we are dealing with: \( y^3 - y^2 + 9y - 9 = 0 \).
- According to the theorem, any rational root \( \frac{p}{q} \) must have \( p \) as a factor of the constant term, which is the last term in the polynomial (here, -9),
- and \( q \) as a factor of the leading coefficient, which is the first term's coefficient (here, 1).
We test these values by substituting them into the polynomial until we find one that makes the polynomial equal to zero.
In this problem, that solution is \( y = 1 \).
Synthetic Division
Synthetic division is a simplified division process used to divide polynomials by linear factors of the form \( y - c \).
It鈥檚 quicker and more concise than long division and is perfect for verifying that a potential root is truly a root.
In our polynomial, since we found \( y = 1 \) to be a root, we use synthetic division to divide \( y^3 - y^2 + 9y - 9 \) by \( y - 1 \).
This quadratic equation will then help us find the remaining roots of the original polynomial.
It鈥檚 quicker and more concise than long division and is perfect for verifying that a potential root is truly a root.
In our polynomial, since we found \( y = 1 \) to be a root, we use synthetic division to divide \( y^3 - y^2 + 9y - 9 \) by \( y - 1 \).
- We focus on the coefficients of the polynomial: \( 1, -1, 9, -9 \).
- As you organize these for synthetic division, remember: bring down the first coefficient, multiply it by the root \( c = 1 \), add it to the next coefficient, and repeat.
This quadratic equation will then help us find the remaining roots of the original polynomial.
Imaginary Numbers
Imaginary numbers are numbers that include the imaginary unit \( i \), where \( i = \sqrt{-1} \).
They arise when we take the square root of a negative number, which cannot be done within the realm of real numbers.
In the quadratic equation \( y^2 + 9 = 0 \), to find \( y \), we isolate \( y^2 \) and get \( y^2 = -9 \).
Understanding imaginary numbers is essential for solving polynomials that don鈥檛 neatly factor into real-number solutions.
They arise when we take the square root of a negative number, which cannot be done within the realm of real numbers.
In the quadratic equation \( y^2 + 9 = 0 \), to find \( y \), we isolate \( y^2 \) and get \( y^2 = -9 \).
- Taking the square root of both sides gives \( y = \sqrt{-9} \).
- This simplifies to \( y = \pm 3i \), where \( i \) denotes an imaginary unit.
Understanding imaginary numbers is essential for solving polynomials that don鈥檛 neatly factor into real-number solutions.
