Certainly! To find the real zeros of the polynomial function [tex]\( f(x) = 2x^3 - 4x^2 - 22x + 44 \)[/tex] and to factor it over the real numbers, we can follow these steps:
### Step-by-Step Solution:
1. Identify Possible Rational Zeros:
According to the Rational Zeros Theorem, any rational zero of the polynomial [tex]\( p(x) \)[/tex] must be a factor of the constant term (44) divided by a factor of the leading coefficient (2).
The factors of 44 are: [tex]\( \pm 1, \pm 2, \pm 4, \pm 11, \pm 22, \pm 44 \)[/tex].
The factors of 2 are: [tex]\( \pm 1, \pm 2 \)[/tex].
The possible rational zeros are given by [tex]\( \frac{\text{factors of } 44}{\text{factors of } 2} \)[/tex]:
[tex]\[
\pm 1, \pm 2, \pm \frac{4}{2} = \pm 2, \pm \frac{11}{2}, \pm \frac{22}{2} = \pm 11
\][/tex]
2. Test Possible Zeros:
From the given information, we know that the real zeros are:
[tex]\[
2, -\sqrt{11}, \sqrt{11}
\][/tex]
3. Verify the Zeros:
- Substitute [tex]\( x = 2 \)[/tex] into [tex]\( f(x) \)[/tex] to verify it is a zero.
- Substitute [tex]\( x = \sqrt{11} \)[/tex] and [tex]\( x = -\sqrt{11} \)[/tex] into [tex]\( f(x) \)[/tex] to verify they are zeros.
4. Factor the Polynomial:
Knowing that [tex]\( x = 2, x = -\sqrt{11}, \)[/tex] and [tex]\( x = \sqrt{11} \)[/tex] are zeros, we can write the polynomial [tex]\( f(x) \)[/tex] as a product of its factors.
[tex]\[
(x - 2), (x - \sqrt{11}), (x + \sqrt{11})
\][/tex]
So, the factorization of [tex]\( f(x) \)[/tex] over the real numbers is:
[tex]\[
f(x) = 2(x - 2)(x - \sqrt{11})(x + \sqrt{11})
\][/tex]
### Conclusion:
The real zeros of the polynomial [tex]\( f(x) = 2x^3 - 4x^2 - 22x + 44 \)[/tex] are:
[tex]\[
x = 2, x = -\sqrt{11}, x = \sqrt{11}
\][/tex]
Therefore, the correct choice is:
A. [tex]\( x = 2, -\sqrt{11}, \sqrt{11} \)[/tex]
This means [tex]\( f(x) \)[/tex] can be factored over the real numbers as:
[tex]\[
f(x) = 2(x - 2)(x - \sqrt{11})(x + \sqrt{11})
\][/tex]
This is the fully detailed solution based on the given zeros.
