Question

Factor the high-degree polynomial 2x^5−4x^3+2x to solve the polynomial equation 2x^5−4x^3+2x=0

Answer 1

To factor the polynomial equation \(2x^5 - 4x^3 + 2x = 0\), we can start by factoring out the common term.

First, observe that each term in the polynomial has a common factor:

\[
2x^5 - 4x^3 + 2x = 2x(x^4 - 2x^2 + 1)
\]

Now, we are left with the factorization:

\[
2x(x^4 - 2x^2 + 1) = 0
\]

Next, we need to solve for \(x\) by setting each factor equal to zero:

1. From \(2x = 0\):
\[
x = 0
\]

2. Now, we solve \(x^4 - 2x^2 + 1 = 0\). Let's make a substitution to simplify this. Let \(y = x^2\). Then the equation becomes:

\[
y^2 - 2y + 1 = 0
\]

This factors as:

\[
(y - 1)^2 = 0
\]

Hence, we have:

\[
y - 1 = 0 \implies y = 1
\]

Substituting back \(y = x^2\), we get:

\[
x^2 = 1
\]

This gives us two solutions:

\[
x = 1 \quad \text{or} \quad x = -1
\]

Now, we can summarize the solutions for the original equation:

\[
x = 0, \quad x = 1, \quad x = -1
\]

Thus, the complete factored form of the polynomial is:

\[
2x(x - 1)^2(x + 1)^2 = 0
\]

The solutions are:

\[
x = 0, \quad x = 1, \quad x = -1
\]