One of the solutions to the equation x3+5x2+10x+50=0 is −5 . Find the full solution set. Which of the following options correctly describes the solution set?

Question

Option #1: It has one real solution and two non-real solutions.

Option #2: It has one non-real solution and two real solutions.

Option #3: It has three real solutions.

Answer 1

Given the cubic equation

\[ x^3 + 5x^2 + 10x + 50 = 0 \]

we know that one of the roots is \( x = -5 \). To find the full solution set, we can use polynomial long division to divide the cubic polynomial by \( (x + 5) \).

Step 1: Polynomial Division

We divide \( x^3 + 5x^2 + 10x + 50 \) by \( x + 5 \):

1. Divide the leading term: \( x^3 \div x = x^2 \).

2. Multiply \( x + 5 \) by \( x^2 \):

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

3. Subtract this from the original polynomial:

   \[ (x^3 + 5x^2 + 10x + 50) - (x^3 + 5x^2) = 10x + 50 \]

4. Next, divide the leading term: \( 10x \div x = 10 \).

5. Multiply \( x + 5 \) by \( 10 \):

   \[ 10(x + 5) = 10x + 50 \]

6. Subtract:

   \[ (10x + 50) - (10x + 50) = 0 \]

The division is exact, meaning we have:

\[ x^3 + 5x^2 + 10x + 50 = (x + 5)(x^2 + 10) \]

Step 2: Find the Remaining Roots

Now, we solve \( x^2 + 10 = 0 \):

\[ x^2 = -10 \] \[ x = \pm \sqrt{-10} = \pm i\sqrt{10} \]

Step 3: Solution Set

The solutions to the equation \( x^3 + 5x^2 + 10x + 50 = 0 \) are:

1. \( x = -5 \) (real root)
2. \( x = i\sqrt{10} \) (non-real root)
3. \( x = -i\sqrt{10} \) (non-real root)

Conclusion

The solution set consists of one real solution \( (-5) \) and two non-real solutions \( (i\sqrt{10}, -i\sqrt{10}) \).

Correct Option

Thus, the appropriate option that describes the solution set is:

Option #1: It has one real solution and two non-real solutions.