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
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, we can divide g(x) by (x+3):
x^2 - 6x + 8
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x + 3 | x^3 - 3x^2 - 10x + 24
- (x^3 + 3x^2)
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- 6x - 10x
- (-6x - 18)
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8x + 24
- (8x + 24)
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0
The quotient is x^2 - 6x + 8, so g(x) can be factored as (x+3)(x^2 - 6x + 8).
To use the Factor Theorem to factor the given polynomial, we will first determine if −3 is a zero of g(x)=x^3−3x^2−10x+24.
According to the Factor Theorem, if a polynomial has a factor (x−k), where k is a zero of the polynomial, then the polynomial can be divided evenly by (x−k). In other words, if g(k) = 0, then (x−k) is a factor of the polynomial.
Now, let's substitute −3 into g(x) and check if we get 0:
g(-3) = (-3)^3 − 3(-3)^2 − 10(-3) + 24
= -27 - 27 + 30 + 24
= 0
Since g(-3) equals zero, −3 is indeed a zero of g(x). Therefore, (x+3) is a factor of the polynomial.
To find the remaining factors, we will divide g(x) by (x+3) using long division or synthetic division. Since long division is easier to explain step-by-step, let's use that method.
Divide x^3−3x^2−10x+24 by x+3:
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x+3 | x^3 -3x^2 -10x + 24
- (x^3 + 3x^2)
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- 6x^2
We continue the division process:
_______________________
x+3 | x^3 -3x^2 -10x + 24
- (x^3 + 3x^2)
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- 6x^2
+ 6x^2
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- 16x
+ 16x
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40
The division process results in a remainder of 40. Therefore, the factored form of g(x) is:
g(x) = (x+3) * (x^2 - 6x + 16)
And there you have it! The polynomial g(x)=x^3−3x^2−10x+24 can be factored into a product of linear factors as (x+3) * (x^2 - 6x + 16).