To use the Factor Theorem, we know that if x = a is a zero of g(x), then (x - a) is a factor of g(x).
Given that -3 is a zero of g(x), we can say that (x + 3) is a factor of g(x).
So, we can rewrite g(x) as a polynomial of degree 2:
g(x) = (x + 3)(x^2 + bx + c)
Expanding this, we get:
g(x) = x^3 + bx^2 + cx + 3x^2 + 3bx + 3c
Simplifying this by combining like terms, we get:
g(x) = x^3 + (b + 3)x^2 + (c + 3b)x + 3c
Comparing this with the original polynomial g(x) = x^3 - 3x^2 - 10x + 24, we can see that:
b + 3 = -3 => b = -6
c + 3b = -10 => c = 12
3c = 24 => c = 8
Therefore, we have:
g(x) = (x + 3)(x^2 - 6x + 8)
So, g(x) can be factored into the product of linear factors as (x + 3)(x^2 - 6x + 8).
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)=
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