To find all the zeros of the polynomial function [tex]\( P(x) = x^5 - 6x^3 + 5x \)[/tex], we follow these steps:
1. Factor out the Greatest Common Factor (GCF):
The given polynomial [tex]\( P(x) = x^5 - 6x^3 + 5x \)[/tex] has a common factor of [tex]\( x \)[/tex]. We can factor [tex]\( x \)[/tex] out of the polynomial:
[tex]\[
P(x) = x(x^4 - 6x^2 + 5)
\][/tex]
This gives us one of the zeros directly:
[tex]\[
x = 0
\][/tex]
2. Solve the remaining polynomial:
Now we need to solve the polynomial inside the parentheses:
[tex]\[
x^4 - 6x^2 + 5 = 0
\][/tex]
Let's make a substitution to simplify this equation. Let [tex]\( y = x^2 \)[/tex]. Then, the equation becomes:
[tex]\[
y^2 - 6y + 5 = 0
\][/tex]
3. Solve the quadratic equation:
We solve the quadratic equation [tex]\( y^2 - 6y + 5 = 0 \)[/tex] using the quadratic formula [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -6 \)[/tex], and [tex]\( c = 5 \)[/tex]:
[tex]\[
y = \frac{6 \pm \sqrt{36 - 20}}{2} = \frac{6 \pm \sqrt{16}}{2} = \frac{6 \pm 4}{2}
\][/tex]
This gives us two solutions for [tex]\( y \)[/tex]:
[tex]\[
y = \frac{6 + 4}{2} = 5 \quad \text{and} \quad y = \frac{6 - 4}{2} = 1
\][/tex]
4. Convert back to [tex]\( x \)[/tex]:
Recall that [tex]\( y = x^2 \)[/tex], so we have two equations to solve:
[tex]\[
x^2 = 5 \quad \text{and} \quad x^2 = 1
\][/tex]
Solving these, we get:
[tex]\[
x = \pm \sqrt{5} \quad \text{and} \quad x = \pm 1
\][/tex]
5. List all zeros:
Combining all the solutions, the zeros of the polynomial [tex]\( P(x) = x^5 - 6x^3 + 5x \)[/tex] are:
[tex]\[
x = 0, \pm 1, \pm \sqrt{5}
\][/tex]
Thus, the zeros of the polynomial [tex]\( P(x) = x^5 - 6x^3 + 5x \)[/tex] are [tex]\( 0, \pm 1, \pm \sqrt{5} \)[/tex].
Therefore, the correct answer is:
[tex]\[ \boxed{0 ; \pm 1 ; \pm \sqrt{5}} \][/tex]
