Question

Factor the polynomial to solve the polynomial equation x ^ 3 - 2x ^ 2 - 25x = - 50 You must show thorough, handwritten work. You may only use scratch paper, a writing utensil, and a calculator that does not have a computer algebra system.

Answer 1

To solve the polynomial equation \( x^3 - 2x^2 - 25x = -50 \), we first rewrite it in standard form. By moving \(-50\) to the left side, we obtain:

\[
x^3 - 2x^2 - 25x + 50 = 0
\]

Next, we'll factor this polynomial. Let's denote the polynomial as:

\[
f(x) = x^3 - 2x^2 - 25x + 50
\]

To factor this polynomial, we can apply the Rational Root Theorem to test for potential rational roots. The Rational Root Theorem suggests that any rational solution \( \frac{p}{q} \) must have \( p \) as a factor of the constant term (50) and \( q \) as a factor of the leading coefficient (1). The factors of 50 are:

\[
\pm 1, \pm 2, \pm 5, \pm 10, \pm 25, \pm 50
\]

We will test these possible rational roots in \( f(x) \).

1. Testing \( x = 1 \):
\[
f(1) = 1^3 - 2(1^2) - 25(1) + 50 = 1 - 2 - 25 + 50 = 24 \quad (\text{not a root})
\]

2. Testing \( x = -1 \):
\[
f(-1) = (-1)^3 - 2(-1)^2 - 25(-1) + 50 = -1 - 2 + 25 + 50 = 72 \quad (\text{not a root})
\]

3. Testing \( x = 2 \):
\[
f(2) = 2^3 - 2(2^2) - 25(2) + 50 = 8 - 8 - 50 + 50 = 0 \quad (\text{is a root})
\]

Since \( x = 2 \) is a root, we can now factor \( f(x) \) using synthetic division by \( x - 2 \).

**Synthetic Division by \(x = 2\)**:

We set up the synthetic division with the coefficients of \( f(x) \):

```
2 | 1 -2 -25 50
| 2 0 -50
-----------------------
1 0 -25 0
```

This division gives us the quotient polynomial:

\[
x^2 - 25
\]

We now have:

\[
f(x) = (x - 2)(x^2 - 25)
\]

Next, we can factor \( x^2 - 25 \) further as it is a difference of squares:

\[
x^2 - 25 = (x - 5)(x + 5)
\]

Putting it all together, we have:

\[
f(x) = (x - 2)(x - 5)(x + 5)
\]

Now we need to set each factor to zero to find the roots of the polynomial equation:

1. \( x - 2 = 0 \) → \( x = 2 \)
2. \( x - 5 = 0 \) → \( x = 5 \)
3. \( x + 5 = 0 \) → \( x = -5 \)

Thus, the solutions to the polynomial equation \( x^3 - 2x^2 - 25x + 50 = 0 \) are:

\[
x = 2, \quad x = 5, \quad x = -5
\]