If X = 2i Is One Of The Root, Find All Possible Roots Of The Polynomial F(x) = 2x4-4x3+7x2-4x-4.
Understanding how to find all roots of a polynomial when given a specific root is a fundamental skill in algebra and polynomial theory. This problem involves a quartic polynomial, which can seem complex at first glance. However, with systematic application of algebraic principles, factorization techniques, and complex conjugate root properties, we can determine all roots efficiently.
In this comprehensive guide, we will analyze the given polynomial, leverage the knowledge that complex roots come in conjugate pairs, and methodically find all roots of the polynomial \( F(x) = 2x^4 - 4x^3 + 7x^2 - 4x - 4 \). The problem specifies that \( x = 2i \) is one of the roots, and from there, we will explore the implications and proceed step-by-step toward the complete solution.
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Understanding the Given Polynomial and Roots
Before delving into calculations, it's important to understand the nature of the polynomial and the properties of its roots.
Key Features of the Polynomial
- The polynomial is of degree 4, indicating there are four roots in total (including complex roots), counting multiplicities.
- The coefficients are real numbers, which has important implications for the roots:
- Complex roots occur in conjugate pairs. If \( a + bi \) is a root, then \( a - bi \) is also a root.
- Given that \( x = 2i \) is a root, its conjugate \( x = -2i \) must also be a root.
Implications of the Given Root
- Since \( 2i \) is a root, the polynomial \( F(x) \) must be divisible by \( (x - 2i) \).
- Similarly, because the coefficients are real, \( (x + 2i) \) must also be a factor.
Step 1: Expressing Factors Corresponding to Known Roots
Given the roots \( 2i \) and \( -2i \), we can form their corresponding quadratic factor:
\[
(x - 2i)(x + 2i) = x^2 - (2i)^2 = x^2 - (-4) = x^2 + 4
\]
This quadratic factor accounts for the roots \( 2i \) and \( -2i \).
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Step 2: Polynomial Division to Find Remaining Roots
Our goal is to factor the original quartic polynomial \( F(x) \) into:
\[
F(x) = (x^2 + 4) \times Q(x)
\]
where \( Q(x) \) is a quadratic polynomial. Once found, solving \( Q(x) = 0 \) will yield the remaining roots.
Performing Polynomial Division
We will divide \( F(x) = 2x^4 - 4x^3 + 7x^2 - 4x - 4 \) by \( x^2 + 4 \).
Method: Polynomial long division or synthetic division (if applicable).
Division process:
- Divide the leading term: \( 2x^4 \) by \( x^2 \) gives \( 2x^2 \).
- Multiply: \( 2x^2 \times (x^2 + 4) = 2x^4 + 8x^2 \).
- Subtract:
(2x^4 - 4x^3 + 7x^2) - (2x^4 + 8x^2) = -4x^3 - x^2
\]
- Bring down remaining terms: Now consider \( -4x^3 - x^2 - 4x - 4 \).
- Divide the leading term: \( -4x^3 \) by \( x^2 \) gives \( -4x \).
- Multiply: \( -4x \times (x^2 + 4) = -4x^3 - 16x \).
- Subtract:
(-4x^3 - x^2 - 4x - 4) - (-4x^3 - 16x) = 15x - 4
\]
- Next division: \( 15x \) divided by \( x^2 \) gives \( 15/x \), which is not a polynomial term, indicating the division process ends here.
However, since the degree of the remainder is less than the divisor, the quotient is:
\[
Q(x) = 2x^2 - 4x + \text{(remaining term)}
\]
Let me perform the division more precisely:
Detailed Polynomial Division:
- Step 1: Divide \( 2x^4 \) by \( x^2 \): quotient term \( 2x^2 \).
- Multiply: \( 2x^2 \times (x^2 + 4) = 2x^4 + 8x^2 \).
- Subtract:
\[
(2x^4 - 4x^3 + 7x^2) - (2x^4 + 8x^2) = -4x^3 - x^2
\]
- Step 2: Divide \( -4x^3 \) by \( x^2 \): quotient term \( -4x \).
- Multiply: \( -4x \times (x^2 + 4) = -4x^3 - 16x \).
- Subtract:
\[
(-4x^3 - x^2 - 4x - 4) - (-4x^3 - 16x) = 15x - 4
\]
- Step 3: Divide \( 15x \) by \( x^2 \): since degree is less, division stops here.
Thus, the division yields:
\[
F(x) = (x^2 + 4)(2x^2 - 4x) + \text{remainder}
\]
But to find the exact quadratic factor, it's better to perform polynomial division explicitly. Let's do that properly:
Full Polynomial Division:
Dividend: \( 2x^4 - 4x^3 + 7x^2 - 4x - 4 \)
Divisor: \( x^2 + 4 \)
Division steps:
- Leading term division: \( 2x^4 \div x^2 = 2x^2 \).
- Multiply divisor by \( 2x^2 \):
\[
2x^2 \times (x^2 + 4) = 2x^4 + 8x^2
\]
- Subtract:
\[
(2x^4 - 4x^3 + 7x^2) - (2x^4 + 8x^2) = -4x^3 - x^2
\]
- Bring down remaining terms: \( -4x - 4 \).
- Next division: \( -4x^3 \div x^2 = -4x \).
- Multiply divisor by \( -4x \):
\[
-4x \times (x^2 + 4) = -4x^3 - 16x
\]
- Subtract:
\[
(-4x^3 - x^2 - 4x - 4) - (-4x^3 - 16x) = 15x - 4
\]
- Now, degree of remainder (1) is less than divisor (2), so division terminates.
Thus, the quotient is \( 2x^2 - 4x \), and the remainder is \( 15x - 4 \).
But since the remainder is not zero, it indicates that \( x^2 + 4 \) does not divide \( F(x) \) evenly, which conflicts with the earlier assumption that \( 2i \) is a root leading to \( x^2 + 4 \) being a factor.
Wait: This suggests that our initial assumption might need correction. Let's verify whether \( x = 2i \) is a root by plugging into \( F(x) \):
\[
F(2i) = 2(2i)^4 -4(2i)^3 + 7(2i)^2 - 4(2i) - 4
\]
Calculate step-by-step:
- \( (2i)^4 = (2^4)(i^4) = 16 \times i^4 \). Since \( i^4 = 1 \), this equals 16.
- \( (2i)^3 = (2^3) i^3 = 8 \times i^3 \). Since \( i^3 = -i \), this equals \( 8 \times -i = -8i \).
- \( (2i)^2 = 4 i^2 = 4 \times -1 = -4 \).
Now, substitute:
\[
F(2i) = 2 \times 16 - 4 \
