Question

Factor the high-degree polynomial 2x5−4x3+2x to solve the polynomial equation 2x5−4x3+2x=0.(1 point)%0D%0AResponses%0D%0A%0D%0Ax=2%0D%0Ax equals 2%0D%0A%0D%0Ax=−1, x=1%0D%0Ax equals negative 1 , x equals 1%0D%0A%0D%0A x=0, x=1%0D%0A x equals 0 , x equals 1%0D%0A%0D%0Ax=−1, x=0, x=1

Answer 1

To factor the polynomial \(2x^5 - 4x^3 + 2x\), we start by factoring out the greatest common factor.

1. We can see that each term contains at least a \(2x\):
\[
2x^5 - 4x^3 + 2x = 2x(x^4 - 2x^2 + 1)
\]

2. Now, we need to factor the quadratic \(x^4 - 2x^2 + 1\). We can treat it as a quadratic in terms of \(y = x^2\):
\[
y^2 - 2y + 1 = (y - 1)^2
\]
This means:
\[
x^4 - 2x^2 + 1 = (x^2 - 1)^2
\]

3. Next, we recognize that \( (x^2 - 1) \) can further be factored as:
\[
x^2 - 1 = (x - 1)(x + 1)
\]

4. Thus, we can write:
\[
(x^2 - 1)^2 = ((x - 1)(x + 1))^2 = (x - 1)^2(x + 1)^2
\]

5. Combining this with the earlier factorization, we have:
\[
2x(x^4 - 2x^2 + 1) = 2x(x - 1)^2(x + 1)^2
\]

Now, setting our equation equal to zero:
\[
2x(x - 1)^2(x + 1)^2 = 0
\]

To find the solutions, we set each factor equal to zero:
1. \(2x = 0 \Rightarrow x = 0\)
2. \((x - 1)^2 = 0 \Rightarrow x - 1 = 0 \Rightarrow x = 1\) (with multiplicity 2)
3. \((x + 1)^2 = 0 \Rightarrow x + 1 = 0 \Rightarrow x = -1\) (with multiplicity 2)

Thus, the complete set of solutions is:
\[
x = 0, x = 1, x = -1
\]

Given the response options:

- **x = 0, x = 1** (This is partially correct since it doesn't include -1)
- **x = -1, x = 0, x = 1** (This includes all solutions)

The correct response is: **x = -1, x = 0, x = 1**.