Let S=x^3+ax^2+bx+c:a,b,c N and a,b,c 20 be a set of polynomials.

Question

Let $S={{x}^{3}+a{x}^{2}+bx+c:a,b,c\in N$ and $a,b,c\le 20}$ be a set of polynomials. Then the number of polynomials in S , which are divisible by ${x}^{2}+2$, is

Accepted Answer

$S={{x}^{3}+a{x}^{2}+bx+c:a,b,c\in Na,b,c\le 20}a,b,c\in {1,2,3,\ldots 20}$

${x}^{3}+a{x}^{2}+bx+c$ is divisible by ${x}^{2}+2$

$	herefore $ ${x}^{3}+a{x}^{2}+bx+c=({x}^{2}+2)L(x)$

Let $L(x)=x+n$ because L.H.S is cubic have leading coefficient 1 .

$\begin{gathered} {x}^{3}+a{x}^{2}+bx+c =({x}^{2}+2)(x+n) \ ={x}^{3}+{x}^{2}n+2x+2n \end{gathered}$

Compare $n=a,b=2,2n=c\implies c=2a$

so, $c=2a$ and $b=2$

Finding the Number of Possible Polynomials

For $b$ : There is only $1$ choice $(b=2)$, which is within the range $[1,20]$.

For $a$ and $c$ : Since $c=2a$, we must ensure $c\le 20$.

If $a=1,c=2$

If $a=2,c=4$

\ldots

If $a=10,c=20$

There are 10 possible values for the pair ( $a,c$ ).

Total polynomials $=($ Choices for $b)	imes ($ Choices for $a,c$ pairs $)=1	imes 10=10$

Solutions · 1