Animated Solution for Mathematics - Quadratic Equations: Let $S=\{x^3+a{x}^2+bx+c:a,b,c\in \mathbb{N},a,b,c\le 20\}$ be a set of polynomials. Then the number of polynomials in S, which are divisible by $x^2+2$, is

Welcome, fellow traveler on the path to JEE mastery! Today, we are going to dissect a problem that, at first glance, looks like a simple counting exercise, but underneath, it is a beautiful dance of algebraic structure.

We are dealing with a set of cubic polynomials $S=\{x^3+a{x}^2+bx+c:a,b,c\in \mathbb{N},a,b,c\le 20\}$. Our mission is to find how many of these polynomials are perfectly divisible by the quadratic $x^2+2$.

When we say a polynomial $P(x)$ is divisible by a divisor $D(x)$, we are essentially saying that $P(x)=D(x)\cdot Q(x)$, where $Q(x)$ is the quotient.

In our case, $P(x)=x^3+a{x}^2+bx+c$ and $D(x)=x^2+2$. Since $P(x)$ is a cubic polynomial (degree $3$) and $D(x)$ is a quadratic (degree $2$), the quotient $Q(x)$ must be a linear polynomial of degree $1$.

Because the leading coefficient of $x^3$ is $1$, the quotient must take the form $(x+k)$. This is our master key:

Now, let us expand the right-hand side of our identity. Take a breath and expand carefully:

Now, we equate this to our original polynomial $x^3+a{x}^2+bx+c$. By comparing the coefficients of the corresponding powers of $x$, we get a system of equations:

1. Coefficient of $x^2$: $a=k$ 2. Coefficient of $x$: $b=2$ 3. Constant term: $c=2k$

Look at the elegance of this! We have completely determined $b$, and we have linked $a$ and $c$ through the parameter $k$. Specifically, since $a=k$, we can write $c=2a$.

This is where the JEE examiners test your attention to detail. We are given that $a,b,c\in \mathbb{N}$ and $a,b,c\le 20$.

We have already found $b=2$, which satisfies the condition $2\le 20$. Now, we must satisfy the condition for $c$. Since $c=2a$, the constraint $c\le 20$ becomes:

Since $a$ must be a natural number, $a$ can take any integer value from $1$ to $10$. For each of these $10$ values of $a$, $b$ is fixed at $2$, and $c$ is uniquely determined as $2a$.

Thus, we have exactly 10 valid polynomials. Never fear the complexity of a problem; often, the most complex-looking expressions hide the most elegant, simple solutions.