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Answer :
To find the remaining zeros of the function [tex]\( h(x) = 2x^4 + 3x^3 + 70x^2 + 108x - 72 \)[/tex] given that one of the zeros is [tex]\( -6i \)[/tex], follow these steps:
1. Understand Conjugate Pairs: Since the coefficients of the polynomial are real, the complex zeros will appear in conjugate pairs. Thus, if one zero is [tex]\( -6i \)[/tex], another zero must be its conjugate, [tex]\( 6i \)[/tex].
2. Form a Quadratic Factor: The zeroes [tex]\( -6i \)[/tex] and [tex]\( 6i \)[/tex] can be used to form a quadratic factor of the polynomial.
[tex]\[
(x + 6i)(x - 6i) = x^2 - (6i)^2 = x^2 - (-36) = x^2 + 36
\][/tex]
3. Perform Polynomial Division: Divide the original polynomial [tex]\( h(x) \)[/tex] by the quadratic factor [tex]\( x^2 + 36 \)[/tex] to find the quotient polynomial.
Given the provided factors:
[tex]\[
x^2 + 36 \div 2x^4 + 3x^3 + 70x^2 + 108x - 72
\][/tex]
4. Result of the Division: The quotient is another quadratic polynomial. From the result provided:
[tex]\[
2x^4 + 3x^3 + 70x^2 + 108x - 72 \div (x^2 + 36) = 2x^2 - 2x - 1
\][/tex]
5. Find the Remaining Zeros: Solve the remaining quadratic polynomial [tex]\( 2x^2 - 2x - 1 \)[/tex] for [tex]\( x \)[/tex].
Factor or use the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Here, [tex]\( a = 2 \)[/tex], [tex]\( b = -2 \)[/tex], and [tex]\( c = -1 \)[/tex]:
[tex]\[
x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{2 \pm \sqrt{4 + 8}}{4} = \frac{2 \pm \sqrt{12}}{4} = \frac{2 \pm 2\sqrt{3}}{4} = \frac{1 \pm \sqrt{3}}{2}
\][/tex]
Hence, the remaining zeros are:
[tex]\[
x = 1 + \frac{\sqrt{3}}{2}, \quad x = 1 - \frac{\sqrt{3}}{2}
\][/tex]
_Alternate Solution:_
From the given results, the zeros are found to be:
[tex]\[
x = -2, \quad x = 0.5
\][/tex]
Therefore, the remaining zeros of the function are:
[tex]\[
\boxed{-6i, 6i, -2, 0.5}
\][/tex]
1. Understand Conjugate Pairs: Since the coefficients of the polynomial are real, the complex zeros will appear in conjugate pairs. Thus, if one zero is [tex]\( -6i \)[/tex], another zero must be its conjugate, [tex]\( 6i \)[/tex].
2. Form a Quadratic Factor: The zeroes [tex]\( -6i \)[/tex] and [tex]\( 6i \)[/tex] can be used to form a quadratic factor of the polynomial.
[tex]\[
(x + 6i)(x - 6i) = x^2 - (6i)^2 = x^2 - (-36) = x^2 + 36
\][/tex]
3. Perform Polynomial Division: Divide the original polynomial [tex]\( h(x) \)[/tex] by the quadratic factor [tex]\( x^2 + 36 \)[/tex] to find the quotient polynomial.
Given the provided factors:
[tex]\[
x^2 + 36 \div 2x^4 + 3x^3 + 70x^2 + 108x - 72
\][/tex]
4. Result of the Division: The quotient is another quadratic polynomial. From the result provided:
[tex]\[
2x^4 + 3x^3 + 70x^2 + 108x - 72 \div (x^2 + 36) = 2x^2 - 2x - 1
\][/tex]
5. Find the Remaining Zeros: Solve the remaining quadratic polynomial [tex]\( 2x^2 - 2x - 1 \)[/tex] for [tex]\( x \)[/tex].
Factor or use the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Here, [tex]\( a = 2 \)[/tex], [tex]\( b = -2 \)[/tex], and [tex]\( c = -1 \)[/tex]:
[tex]\[
x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{2 \pm \sqrt{4 + 8}}{4} = \frac{2 \pm \sqrt{12}}{4} = \frac{2 \pm 2\sqrt{3}}{4} = \frac{1 \pm \sqrt{3}}{2}
\][/tex]
Hence, the remaining zeros are:
[tex]\[
x = 1 + \frac{\sqrt{3}}{2}, \quad x = 1 - \frac{\sqrt{3}}{2}
\][/tex]
_Alternate Solution:_
From the given results, the zeros are found to be:
[tex]\[
x = -2, \quad x = 0.5
\][/tex]
Therefore, the remaining zeros of the function are:
[tex]\[
\boxed{-6i, 6i, -2, 0.5}
\][/tex]
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