MHB [ASK] polynomial f(x) divided by (x - 1)

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The polynomial f(x) = 2x^3 - 5x^2 + ax + 18 is divisible by (x - 3), leading to the conclusion that a must equal -9. When substituting a into f(x), the polynomial becomes 2x^3 - 5x^2 - 9x + 18. Dividing this by (x - 1) results in a quotient of 2x^2 - 3x - 12 with a remainder of 6, indicating that f(x) is not divisible by (x - 1). The discussion highlights a potential error in the original question, as the expected results do not align with the calculations performed.
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A polynomial f(x) = [math]2x^3-5x^2+ax+18[/math] is divisible by (x - 3). The result of that polynomial f(x) divided by (x - 1) is ...
A. [math]2x^2-7x+2[/math]
B. [math]2x^2+7x-2[/math]
C. [math]2x^2-7x-2[/math]
D. [math]x^2-6x-2[/math]
E. [math]x^2-6x+3[/math]

I got a + 3 = -6 and so a = -9 and f(x) = [math]2x^3-5x^2-9x+18[/math], but when I divided it with x - 1 I got [math]x^2-3x+6[/math] with the remainder 24. I'm quite sure that the question isn't wrong since it's in a national level exam, but where did I go wrong?
 
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Monoxdifly said:
A polynomial f(x) = [math]2x^3-5x^2+ax+18[/math] is divisible by (x - 3).

If a polynomial is divisible by $(x-k)$, then $f(k) = 0$

$f(3) = 0 \implies a = -9 \implies f(1) = 6$

$f(1) \ne 0 \implies f(x)$ is not divisible by $(x-1)$.

My opinion is there is an error in the question somewhere.
 
Synthetic Division does some useful things.

Code:
3 |  2  -5    a       18
         6    3     3a + 9
-------------------------
     2   1  a+3      3a+27 = 0 ==> a = -9

1 |  2  -5    -9      18
         2    -3     -12
-------------------------
     2  -3   -12      6

And thus we see: $2x^{2} - 3x - 12 + \dfrac{6}{x-1}$

How did you get only x^2, instead of 2x^2? Something wrong there.
 
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