A polynomial P(x) has degree 3 and leading coefficient -6. Two of its roots are -42 and 1. Find a possible formula for P(x). You do not need to simplify/distribute your answer. P(x) =
A polynomial P(x) has degree 3 and leading coefficient -6. Two of its roots are -42 and 1. Find a possible formula for P(x). You do not need to simplify/distribute your answer. P(x) =
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter5: Inverse, Exponential, And Logarithmic Functions
Section: Chapter Questions
Problem 28RE
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![**Problem Statement:**
*A polynomial \( P(x) \) has degree 3 and leading coefficient \(-6\). Two of its roots are \(-4i\) and 1.*
*Find a possible formula for \( P(x) \). You do not need to simplify/distribute your answer.*
**Solution:**
To find the polynomial \( P(x) \), we start with the given roots and leading coefficient. Since \( P(x) \) is a polynomial of degree 3, it must have three roots. The given roots are \(-4i\) and 1.
For complex roots, their conjugates also must be roots of the polynomial if the coefficients are real. Hence, the conjugate of \(-4i\) is \(4i\).
So, the roots of the polynomial \( P(x) \) are:
1. \(1\)
2. \(-4i\)
3. \(4i\)
Using these roots and the fact that the leading coefficient is \(-6\), we can construct the polynomial. If a polynomial with roots \( a, b, \) and \( c \) is factored, it looks like this \( k(x - a)(x - b)(x - c) \), where \( k \) is the leading coefficient.
Therefore, a possible polynomial \( P(x) \) can be written as:
\[ P(x) = -6 (x - 1) \left(x - (-4i)\right) (x - 4i) \]
This yields:
\[ P(x) = -6 (x - 1) (x + 4i) (x - 4i) \]
Using the difference of squares on the imaginary numbers:
\[ (x + 4i)(x - 4i) = x^2 - (4i)^2 = x^2 + 16 \]
Thus, the polynomial can be written as:
\[ P(x) = -6 (x - 1)(x^2 + 16) \]
This is the factored form of the polynomial \( P(x) \) with the given properties.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcae97c47-da87-4dbe-a1b5-55efe44ed72d%2F145c77a4-4e79-439d-85ea-7486ae1141e0%2Fnr7j4y_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
*A polynomial \( P(x) \) has degree 3 and leading coefficient \(-6\). Two of its roots are \(-4i\) and 1.*
*Find a possible formula for \( P(x) \). You do not need to simplify/distribute your answer.*
**Solution:**
To find the polynomial \( P(x) \), we start with the given roots and leading coefficient. Since \( P(x) \) is a polynomial of degree 3, it must have three roots. The given roots are \(-4i\) and 1.
For complex roots, their conjugates also must be roots of the polynomial if the coefficients are real. Hence, the conjugate of \(-4i\) is \(4i\).
So, the roots of the polynomial \( P(x) \) are:
1. \(1\)
2. \(-4i\)
3. \(4i\)
Using these roots and the fact that the leading coefficient is \(-6\), we can construct the polynomial. If a polynomial with roots \( a, b, \) and \( c \) is factored, it looks like this \( k(x - a)(x - b)(x - c) \), where \( k \) is the leading coefficient.
Therefore, a possible polynomial \( P(x) \) can be written as:
\[ P(x) = -6 (x - 1) \left(x - (-4i)\right) (x - 4i) \]
This yields:
\[ P(x) = -6 (x - 1) (x + 4i) (x - 4i) \]
Using the difference of squares on the imaginary numbers:
\[ (x + 4i)(x - 4i) = x^2 - (4i)^2 = x^2 + 16 \]
Thus, the polynomial can be written as:
\[ P(x) = -6 (x - 1)(x^2 + 16) \]
This is the factored form of the polynomial \( P(x) \) with the given properties.
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