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To find the zeros of the function [tex]\( h(x) = x^3 + 7x^2 + 4x + 28 \)[/tex], we need to find the values of [tex]\( x \)[/tex] where [tex]\( h(x) = 0 \)[/tex]. These values are often referred to as the roots of the equation [tex]\( x^3 + 7x^2 + 4x + 28 = 0 \)[/tex].
Let's perform a step-by-step analysis to find these zeros:
### Step 1: Identify possible rational roots
We use the Rational Root Theorem, which states that any rational solution, expressed as a fraction in simplest form [tex]\( \frac{p}{q} \)[/tex], must have [tex]\( p \)[/tex] as a factor of the constant term (28) and [tex]\( q \)[/tex] as a factor of the leading coefficient (1, in this case).
Factors of 28: [tex]\( \pm 1, \pm 2, \pm 4, \pm 7, \pm 14, \pm 28 \)[/tex]
Factors of 1: [tex]\( \pm 1 \)[/tex]
Possible rational roots: [tex]\( \pm 1, \pm 2, \pm 4, \pm 7, \pm 14, \pm 28 \)[/tex]
### Step 2: Evaluate possible rational roots
We can evaluate each of these possible roots to see if any of them are zeros of the polynomial:
[tex]\( h(x) = x^3 + 7x^2 + 4x + 28 \)[/tex]
- For [tex]\( x = -7 \)[/tex]:
[tex]\[ h(-7) = (-7)^3 + 7(-7)^2 + 4(-7) + 28 = -343 + 343 - 28 + 28 = 0 \][/tex]
So, [tex]\( x = -7 \)[/tex] is a root.
### Step 3: Factor out [tex]\( x + 7 \)[/tex] from [tex]\( h(x) \)[/tex]
Since [tex]\( x = -7 \)[/tex] is a root, we can perform polynomial division or synthetic division to factor [tex]\( x + 7 \)[/tex] out of [tex]\( h(x) \)[/tex].
Upon factoring, we get:
[tex]\[ h(x) = (x + 7)(x^2 + 4) \][/tex]
### Step 4: Solve for the remaining quadratic equation
Now, we need to find the zeros of the quadratic equation [tex]\( x^2 + 4 = 0 \)[/tex]:
[tex]\[ x^2 + 4 = 0 \][/tex]
[tex]\[ x^2 = -4 \][/tex]
[tex]\[ x = \pm \sqrt{-4} \][/tex]
Using the property of imaginary numbers, we get:
[tex]\[ x = \pm 2i \][/tex]
### Conclusion
Thus, the zeros of the function [tex]\( h(x) = x^3 + 7x^2 + 4x + 28 \)[/tex] are:
[tex]\[ x = -7, x = 2i, x = -2i \][/tex]
So, the zeros of the function are [tex]\(-7\)[/tex], [tex]\(2i\)[/tex], and [tex]\(-2i\)[/tex]. There are no multiplicities greater than 1 for these roots.
Let's perform a step-by-step analysis to find these zeros:
### Step 1: Identify possible rational roots
We use the Rational Root Theorem, which states that any rational solution, expressed as a fraction in simplest form [tex]\( \frac{p}{q} \)[/tex], must have [tex]\( p \)[/tex] as a factor of the constant term (28) and [tex]\( q \)[/tex] as a factor of the leading coefficient (1, in this case).
Factors of 28: [tex]\( \pm 1, \pm 2, \pm 4, \pm 7, \pm 14, \pm 28 \)[/tex]
Factors of 1: [tex]\( \pm 1 \)[/tex]
Possible rational roots: [tex]\( \pm 1, \pm 2, \pm 4, \pm 7, \pm 14, \pm 28 \)[/tex]
### Step 2: Evaluate possible rational roots
We can evaluate each of these possible roots to see if any of them are zeros of the polynomial:
[tex]\( h(x) = x^3 + 7x^2 + 4x + 28 \)[/tex]
- For [tex]\( x = -7 \)[/tex]:
[tex]\[ h(-7) = (-7)^3 + 7(-7)^2 + 4(-7) + 28 = -343 + 343 - 28 + 28 = 0 \][/tex]
So, [tex]\( x = -7 \)[/tex] is a root.
### Step 3: Factor out [tex]\( x + 7 \)[/tex] from [tex]\( h(x) \)[/tex]
Since [tex]\( x = -7 \)[/tex] is a root, we can perform polynomial division or synthetic division to factor [tex]\( x + 7 \)[/tex] out of [tex]\( h(x) \)[/tex].
Upon factoring, we get:
[tex]\[ h(x) = (x + 7)(x^2 + 4) \][/tex]
### Step 4: Solve for the remaining quadratic equation
Now, we need to find the zeros of the quadratic equation [tex]\( x^2 + 4 = 0 \)[/tex]:
[tex]\[ x^2 + 4 = 0 \][/tex]
[tex]\[ x^2 = -4 \][/tex]
[tex]\[ x = \pm \sqrt{-4} \][/tex]
Using the property of imaginary numbers, we get:
[tex]\[ x = \pm 2i \][/tex]
### Conclusion
Thus, the zeros of the function [tex]\( h(x) = x^3 + 7x^2 + 4x + 28 \)[/tex] are:
[tex]\[ x = -7, x = 2i, x = -2i \][/tex]
So, the zeros of the function are [tex]\(-7\)[/tex], [tex]\(2i\)[/tex], and [tex]\(-2i\)[/tex]. There are no multiplicities greater than 1 for these roots.
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