Question 1209677
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<font color=red>There isn't enough info given.</font>


Consider 
g(x) = x^3+6x^2-11x-35
h(x) = x^2-x-6


Use polynomial long division to determine that,
g/h = (x+7) remainder (2x+7)
We fulfill the requirement of dividing a function g(x) over h(x) = x^2-x-6 where it yields remainder 2x+7.
There are many online calculators that will perform polynomial long division (some even provide the step-by-step guide).
I used the <a href="https://geogebra.github.io/docs/manual/en/commands/Division/">division tool in GeoGebra</a>  
Then,
g(x) = x^3+6x^2-11x-35
g(8) = (8)^3+6*(8)^2-11*(8)-35
g(8) = 773


Now consider,
g(x) = x^4-6x^3+12x^2+19x-71
h(x) = x^2-x-6
The first function has been changed, but the h(x) function is the same as before.
You should find that,
g/h = (x^2-5x+13) remainder (2x+7)
We get the same remainder as before.
However,
g(x) = x^4-6x^3+12x^2+19x-71
g(8) = (8)^4-6*(8)^3+12*(8)^2+19*(8)-71
g(8) = 1873
This contradicts the previous result g(8) = 773
This means we simply <font color=red>do <u>not</u> have enough information to determine g(8)</font>


I have a feeling that your teacher made a typo when asking about g(8)


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Despite not being able to find g(8) we can find either g(-2) or g(3)
Note that -2 and 3 are roots of x^2-x-6 = (x+2)(x-3) = 0
The fact we get 0 is important.


Here's the scratch work to finding g(3)
g(x) = unknown
h(x) = x^2-x-6
h(3) = 3^2-3-6 = 0
g(x)/h(x) = quotient + remainder/h(x)
g(x) = h(x)*quotient + remainder
g(x) = h(x)*quotient + 2x+7
g(3) = <font color=blue>h(3)*quotient</font> + 2(3)+7
g(3) = <font color=blue>0*quotient</font> + 2(3)+7
g(3) = <font color=blue>0</font> + 2(3)+7
g(3) = 13


Pay close attention to the fact that the <font color=blue>h(3)*quotient</font> portion goes to 0 since h(3) = 0.
This allows us to ignore the quotient entirely.
This only happens for x = -2 or x = 3. Unfortunately it does not happen when x = 8.


Through similar steps, we can find g(-2) = 3
This assumes that the x^2-x-6 and 2x+7 portions aren't typos.
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