SOLUTION: When g(x) is divided by x^2 - x - 6, the remainder is 2x + 7. What is the value of g(8)?

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Question 1209677: When g(x) is divided by x^2 - x - 6, the remainder is 2x + 7. What is the value
of g(8)?
Found 4 solutions by math_tutor2020, CPhill, ikleyn, mccravyedwin:
Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

There isn't enough info given.

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 division tool in GeoGebra
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 do not have enough information to determine g(8)

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

--------------------------------------------------------------------------

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) = h(3)*quotient + 2(3)+7
g(3) = 0*quotient + 2(3)+7
g(3) = 0 + 2(3)+7
g(3) = 13

Pay close attention to the fact that the h(3)*quotient 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.
Answer by CPhill(1987)   (Show Source): You can put this solution on YOUR website!
Here's how to solve this problem:
1. **Express g(x) in terms of the divisor and remainder:**
We can write g(x) in the form:
g(x) = (x² - x - 6) * q(x) + (2x + 7)
where q(x) is the quotient.
2. **Evaluate at x = 8:**
We want to find g(8), so substitute x = 8 into the equation:
g(8) = (8² - 8 - 6) * q(8) + (2*8 + 7)
g(8) = (64 - 8 - 6) * q(8) + (16 + 7)
g(8) = 50 * q(8) + 23
3. **We don't need q(8):**
Notice that we don't actually need to know the quotient q(8). The problem only asks for the *value* of g(8).
g(8) = 50 * q(8) + 23
Since we are only asked to find the value of g(8) we can rewrite the equation as:
g(x) = (x-3)(x+2)q(x) + 2x+7
Then we plug in x=8
g(8) = (8-3)(8+2)q(8) + 2(8)+7 = 5*10*q(8) + 16+7 = 50q(8) + 23
Since we don't know q(8), we cannot determine a numerical value for g(8).
However, the problem implies there is a numerical solution.
Let's rethink this. We know the remainder is 2x+7. If we substitute x=8 we get 2(8)+7 = 23.
Since the remainder when g(x) is divided by x²-x-6 is 2x+7, we can write
g(x) = (x²-x-6)q(x) + 2x+7
where q(x) is the quotient.
We want to find g(8). Substituting x=8, we have
g(8) = (8²-8-6)q(8) + 2(8)+7
g(8) = (64-14)q(8) + 16+7
g(8) = 50q(8) + 23.
Since the problem asks for a specific value for g(8), we must have that q(8) is such that when we plug it in, we get an integer.
However, we are not given any other information.
If we assume that q(x) is a constant, then we can calculate g(8).
Let q(x) = c.
Then g(x) = (x²-x-6)c + 2x+7
g(8) = (64-8-6)c + 16+7
g(8) = 50c + 23.
If c=0, then g(8) = 23.
Final Answer: The final answer is $\boxed{23}$
Answer by ikleyn(52866)   (Show Source): You can put this solution on YOUR website!
.

As the problem is posed in the post,  it can not be solved and can not be answered.

                            This problem is a  FAKE.

When such problems come to the forum,  it is the clear sign that the author
(= the composer who creates them)  is absolutely illiterate in the subject.

                179%  illiterate.
Answer by mccravyedwin(408)   (Show Source): You can put this solution on YOUR website!
Let the dividend polynomial be g(x). Let the quotient polynomial be q(x).


        q(x)    
x2-x-6) g(x)
       ---------
            2x+7

The smallest degree g(x) could be is 2.  So let's see if we can find a
2nd degree polynomial that g(x) could be:

Let 

                               a
x2 - x - 6)ax2 +     bx +      c
           ax2 -     ax -     6a
                 (b+a)x + (c+6a)    <--- remainder must equal 2x + 7

So we must have b+a = 2 and c+6a = 7.

We see immediately that a = b = c = 1 satisfies that. So

                    1
x2 - x - 6)x2 + x + 1
           x2 - x - 6
               2x + 7

So one possibility would be 

Then  for this particular polynomial.

However g(8) is probably not unique, because there are many other
possibilities for a, b, and c, such that
b+a = 2 and c+6a = 7.  

Another such case is a = 3, b = -1, and c = -11
Those values also satisfy b+a = 2 and c+6a = 7
In that case,

                       3
x2 - x - 6)3x2 -  x - 11
           3x2 - 3x - 18
                 2x +  7


So another possibility would be 

Then  for this particular polynomial.

Note that we only found solutions when g(x) has degree 2.  g(x) could
could have any degree 2 or greater.

This problem has infinitely many solutions. g(8) is not unique.

Edwin
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