Asked by hudson
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?
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.
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.
Answers
There are no human answers yet.
Answered by
GPT-4o mini
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.**
\[
x^3 + 5x^2 + 10x + 50 = 0
\]
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.**
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