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).
To use the Factor Theorem, we know that if a polynomial function g(x) has a zero at x = c, then (x - c) is a factor of g(x).
In this case, we are given that -3 is a zero of g(x), so (x + 3) is a factor of g(x).
We can now use synthetic division to divide g(x) by (x + 3):
-3 | 1 -3 -10 24
| -3 18 -24
_______________
1 -6 8 0
The result of the division is x^2 - 6x + 8. Therefore, we have factored g(x) as:
g(x) = (x + 3)(x^2 - 6x + 8)
To use the factor theorem to factor \(g(x)\) and find the other linear factors, you need to know that if \(c\) is a zero of a polynomial, then \((x - c)\) is a factor of the polynomial.
Given that \(-3\) is a zero of \(g(x)\), we can use the factor theorem to conclude that \((x - (-3))\) or \((x + 3)\) is a factor of \(g(x)\).
To factor \(g(x)\) into a product of linear factors, we can use long division or synthetic division to divide \(g(x)\) by \((x + 3)\). Let's use synthetic division:
```
1 -3 -10 24
-3 | 1 -3 -10 24
| -3 18 -24
-------------
1 15 8
```
The result of the synthetic division is \(x^2 + 15x + 8\). Hence, we can write the factored form of \(g(x)\) as:
\[g(x) = (x + 3)(x^2 + 15x + 8)\]
Therefore, the factored form of \(g(x)\) into a product of linear factors is \((x + 3)(x^2 + 15x + 8)\).