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

  • Context: MHB 
  • Thread starter Thread starter Monoxdifly
  • Start date Start date
  • Tags Tags
    Polynomial
Click For Summary
SUMMARY

The polynomial f(x) = 2x^3 - 5x^2 + ax + 18 is confirmed to be divisible by (x - 3) when a = -9, leading to f(3) = 0. However, when dividing f(x) by (x - 1), the result is 2x^2 - 3x - 12 with a remainder of 6, indicating that f(x) is not divisible by (x - 1). The confusion arises from the interpretation of the polynomial's divisibility, as the calculations demonstrate that f(1) ≠ 0, confirming the polynomial's non-divisibility by (x - 1).

PREREQUISITES
  • Understanding of polynomial functions and their properties
  • Knowledge of synthetic division techniques
  • Familiarity with the Remainder Theorem
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the Remainder Theorem in detail
  • Practice synthetic division with various polynomials
  • Explore polynomial long division techniques
  • Learn about polynomial roots and their implications on divisibility
USEFUL FOR

Students preparing for national level exams, educators teaching polynomial functions, and anyone interested in mastering polynomial division and related concepts.

Monoxdifly
MHB
Messages
288
Reaction score
0
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?
 
Mathematics news on Phys.org
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.
 

Similar threads

  • · Replies 48 ·
2
Replies
48
Views
4K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 18 ·
Replies
18
Views
3K
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K