Question 1186707
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Find the constant C such that the denominator will divides evenly into the numerator.
x^4-x^3-3x^2-Cx-3/x-3
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<pre>
The linear binomial (x-3) in the denominator divides the polynomial  f(x) = x^4 - x^3 - 3x^2 - Cx - 3  of the numerator evenly 

if and only if  f(3) = 0   (the Remainder Theorem).


From condition  f(3) = 0  find the value of C


    3^4 - 3^3 - 3*3^2  - C*3 - 3 = 0.


It gives


    3C = 3^4 - 3^3 - 3*3^2 - 3 = 81 - 27 - 3*9 - 3 = 24.


Hence,  C = 24/3 = 8.    <U>ANSWER</U>


<U>CHECK</U>.  f(3) = 3^4 - 3^3 - 3*3^2 - 8*3 - 3 = 81 - 27 - 27 - 24 - 3 = 0.
</pre>

Solved.



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The response by @josgarithmetic, &nbsp;giving the answer  &nbsp;&nbsp;{{{highlight(C=26)}}}  &nbsp;&nbsp;is &nbsp;INCORRECT.



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After my noticing, @josgarithmetic fixed his erroneous answer,


but his calculations (starting from the setup equation) remained INCORRECT.



NEVER trust his solutions; ALWAYS avoid his posts, for your safety.


This person makes errors everywhere and always.