Question

Find the number of polynomials of the form ${{x}^{3}}+a{{x}^{2}}+bx+c$ that are divisible by ${{x}^{2}}+1$, where a,b,c\in \left\\{ 1,2,3.....10 \right\\}.

Answer 1

Hint: The polynomial is said to be divisible to another polynomial when we get the remainder as zero when we put the zeroes of the polynomial in the polynomial whose divisibility is to be determined. Thus, in the above question, we have to find the values of a, b and c or relation between them.

Complete step-by-step answer:
First we have to obtain the zeros of the polynomial ${{x}^{2}}+1$. When these zeroes will be put in place of $x$in the polynomial ${{x}^{3}}+a{{x}^{2}}+bx+c$, we will get its value as zero. So the zeros of polynomial ${{x}^{2}}+1$ is given by:

& \to {{x}^{2}}+1=0 \\\ & \\\ & \to {{x}^{2}}=-1 \\\ & \\\ & \to x=\pm \sqrt{-1} \\\ & \\\ & \to x=\pm i \\\ \end{aligned}$$ Thus the zeroes of the polynomial $${{x}^{2}}+1$$ are $$i$$ and $$-i$$. We will put these values ( $$i$$ and $$-i$$) in the polynomial in place of $$x$$ one by one to obtain different numbers of polynomials. So we are considering two cases here: Case 1: $$x$$= $$i$$. Now put the value of $$x$$ in the polynomial $${{x}^{3}}+a{{x}^{2}}+bx+c$$. Thus after putting the value of $$x$$in the polynomial will become zero because the polynomial $${{x}^{3}}+a{{x}^{2}}+bx+c$$ is divisible by $${{x}^{2}}+1$$. So after putting the value we get, $$\begin{aligned} & \to {{(i)}^{3}}+a{{(i)}^{2}}+b(i)+c=0 \\\ & \\\ & \to (-i)-a+b(i)+c=0 \\\ \end{aligned}$$ The above equation will be zero only when the real and imaginary parts of the equation will be separately zero. Thus, we have to equate the real and imaginary part of the above equation to zero. Imaginary part = -1+b Real part = - a + c Now, equating then to zero. After equating we get, $$\begin{aligned} & \to -1+b=0 \\\ & \\\ & \to b=1.....(i) \\\ & \\\ & \to -a+c=0 \\\ & \\\ & \to a=c.......(ii) \\\ & \\\ \end{aligned}$$ So, we can conclude that we can only have one value while a (or c) can have 10 values. So the total number of polynomial obtained are $$10*1=10$$. Case 2: $$x=-i$$ This case is very similar to case 1. Instead of $$x=i$$, we are using $$x=-i$$. So, now will put $$x=-i$$ in the polynomial $${{x}^{3}}+a{{x}^{2}}+bx+c$$. So, after doing this, we get, $$\begin{aligned} & \to {{(i)}^{3}}+a(){{i}^{2}}+b(i)+c=0 \\\ & \\\ & \to (-i)-a+b(i)+c=0 \\\ \end{aligned}$$ Now, the imaginary part and real part are separately equal to zero. So, we get $$\begin{aligned} & -1+b=0 \\\ & \\\ & b=1.....(iii) \\\ & \\\ & -a+c=0 \\\ & \\\ & a=c......(iv) \\\ \end{aligned}$$ Here, we can see that the polynomials formed in this case will coincide with the polynomials formed in case 1. So, the overall total number of polynomials formed are=10. Note: Instead of forming two cases, we could have done it as shown below: - $$p(x)={{x}^{3}}+a{{x}^{2}}+bx+c=x({{x}^{2}}+b)+a{{x}^{2}}+c$$ Now putting the value of $${{x}^{2}}$$ in above equation equal to zero, $$x(-1+b)-a+c=0$$ In this case also, we get the same equation.