The number of polynomial of form $x^3+ax^2+bx+c$ which are divisible by $x^2+1$ and where a,b,c belongs to {1,2,...10}, is?

Question

How exactly should I approach this problem? There could be many possible values a, b and c from the set is there any simple method?

All polynomials you can build where $x^2+1$ is a factor will qualify. How many are those? — mathreadler, Apr 19 '17 at 18:04

score 2 · Accepted Answer · answered Apr 19 '17 at 18:09

Since $x^2+1$ divides the plynomial, $f(\pm i)=0$

$$(i)^3+a(i)^2+b(i)+c=0$$ $$\implies -i-a+bi+c=0 $$

$$\implies i (b-1)+(c-a)=0 \implies b=1~,~a=c$$

Total possibilities $b=1 (1 ~\text{possibility})$, $a=c ~(10 ~\text{possibilities})$

$x=-i$ will lead to same result.

score 1 · Answer 2 · answered Apr 19 '17 at 18:05

1

If $x^3+ax^2+bx+c$ is divisible by $x^2+1$, then we must have $1=b$ and $a=c$. (Why?)

Therefore, there are $10\times1\times1=10$ polynomials.

answered Apr 19 '17 at 18:05

DHMO

3,014
12
20

score 1 · Answer 3 · answered Apr 19 '17 at 18:08

Denote the polynomial by $f$. We have that the polynomials are divisible by $x^2+1$ if and only if both of $\pm i$ are roots. Since the coefficients are real, $f(i) = 0 \iff f(-i) = 0$.

Note $f(i) = -i -a +bi + c = 0 \iff b = 1$, $c = a$ since $a, b, c \in \mathbb{R}$.

So there are exactly 10, corresponding to the one degree of freedom (say, $a$).

The number of polynomial of form $x^3+ax^2+bx+c$ which are divisible by $x^2+1$ and where a,b,c belongs to {1,2,...10}, is?

3 Answers3