Use the Factor Theorem to factor g(x)=x^3−3x^2−10x+24 into a product of linear factors, given that −3 is a zero of g(x).
g(x)=
To use the Factor Theorem, we know that if -3 is a zero of g(x), then (x+3) is a factor of g(x).
Using long division or synthetic division, divide g(x) by (x+3):
-3 | 1 -3 -10 24
|_______-3____0____-30
1 0 -10 -6
The quotient is 1x^2 + 0x - 10, which can be factored further if possible.
Now, we have the factored form of g(x):
g(x) = (x+3)(x^2 - 10)
To factor the polynomial g(x) using the Factor Theorem, we first need to determine the linear factor associated with the given zero, which is x = -3.
The Factor Theorem states that if a polynomial has a factor (x - a), then a is a zero of the polynomial. In this case, since -3 is a zero of g(x), it means that (x + 3) is a factor of g(x).
To find the other factor, we can perform long division or synthetic division of g(x) by (x + 3):
```
-3 │ x^3 - 3x^2 - 10x + 24
- (x^3 + 3x^2)
-------------------
-6x^2 - 10x
+ (6x^2 + 18x)
-------------------
8x + 24
- (8x + 24)
-------------------
0
```
The result of the division is 0, which means that (x + 3) is a factor of g(x).
Therefore, we can write g(x) as a product of linear factors:
g(x) = (x + 3)(x^2 - 6x + 8)
To factor the quadratic factor x^2 - 6x + 8 further, we can use factoring or the quadratic formula. By factoring, we find:
x^2 - 6x + 8 = (x - 2)(x - 4)
Therefore, the factored form of g(x) is:
g(x) = (x + 3)(x - 2)(x - 4)