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To find all the roots of the polynomial [tex]\( f(x) = x^3 + 10x^2 - 25x - 250 \)[/tex], given that one of the roots is [tex]\( x = -10 \)[/tex]:
1. Factoring out the known root:
Since [tex]\( x = -10 \)[/tex] is a root, [tex]\( (x + 10) \)[/tex] must be a factor of [tex]\( f(x) \)[/tex]. We can perform polynomial division to factor [tex]\( f(x) \)[/tex] by [tex]\( (x + 10) \)[/tex].
2. Divide [tex]\( f(x) \)[/tex] by [tex]\( (x + 10) \)[/tex] using synthetic division or long division:
We need to divide [tex]\( x^3 + 10x^2 - 25x - 250 \)[/tex] by [tex]\( x + 10 \)[/tex].
3. Set up the synthetic division:
- The coefficients of [tex]\( f(x) \)[/tex] are: 1, 10, -25, and -250.
- We'll use [tex]\( -10 \)[/tex] as the root.
```
[-10 | 1 10 -25 -250]
| -10 0 250
----------------------
1 0 -25 0
```
This gives us a quotient of [tex]\( x^2 - 25 \)[/tex] and a remainder of 0.
4. Solve the quadratic equation:
Now that we have factored [tex]\( f(x) \)[/tex] as [tex]\( (x + 10)(x^2 - 25) \)[/tex], we need to solve [tex]\( x^2 - 25 = 0 \)[/tex].
[tex]\[ x^2 - 25 = 0 \][/tex]
This simplifies to:
[tex]\[ (x - 5)(x + 5) = 0 \][/tex]
5. Find the roots from the quadratic:
The solutions to [tex]\( (x - 5)(x + 5) = 0 \)[/tex] are:
[tex]\[ x = 5 \quad \text{and} \quad x = -5 \][/tex]
6. Combine all roots:
Including the given root [tex]\( x = -10 \)[/tex], the complete set of roots of the polynomial [tex]\( f(x) \)[/tex] are:
[tex]\[ x = -10, x = -5, x = 5 \][/tex]
Therefore, the roots of [tex]\( f(x) = x^3 + 10x^2 - 25x - 250 \)[/tex] are [tex]\(\boxed{x = -10, x = -5, x = 5}\)[/tex].
1. Factoring out the known root:
Since [tex]\( x = -10 \)[/tex] is a root, [tex]\( (x + 10) \)[/tex] must be a factor of [tex]\( f(x) \)[/tex]. We can perform polynomial division to factor [tex]\( f(x) \)[/tex] by [tex]\( (x + 10) \)[/tex].
2. Divide [tex]\( f(x) \)[/tex] by [tex]\( (x + 10) \)[/tex] using synthetic division or long division:
We need to divide [tex]\( x^3 + 10x^2 - 25x - 250 \)[/tex] by [tex]\( x + 10 \)[/tex].
3. Set up the synthetic division:
- The coefficients of [tex]\( f(x) \)[/tex] are: 1, 10, -25, and -250.
- We'll use [tex]\( -10 \)[/tex] as the root.
```
[-10 | 1 10 -25 -250]
| -10 0 250
----------------------
1 0 -25 0
```
This gives us a quotient of [tex]\( x^2 - 25 \)[/tex] and a remainder of 0.
4. Solve the quadratic equation:
Now that we have factored [tex]\( f(x) \)[/tex] as [tex]\( (x + 10)(x^2 - 25) \)[/tex], we need to solve [tex]\( x^2 - 25 = 0 \)[/tex].
[tex]\[ x^2 - 25 = 0 \][/tex]
This simplifies to:
[tex]\[ (x - 5)(x + 5) = 0 \][/tex]
5. Find the roots from the quadratic:
The solutions to [tex]\( (x - 5)(x + 5) = 0 \)[/tex] are:
[tex]\[ x = 5 \quad \text{and} \quad x = -5 \][/tex]
6. Combine all roots:
Including the given root [tex]\( x = -10 \)[/tex], the complete set of roots of the polynomial [tex]\( f(x) \)[/tex] are:
[tex]\[ x = -10, x = -5, x = 5 \][/tex]
Therefore, the roots of [tex]\( f(x) = x^3 + 10x^2 - 25x - 250 \)[/tex] are [tex]\(\boxed{x = -10, x = -5, x = 5}\)[/tex].
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