Question

Consider the polynomial $P ( x )= x ^3+ bx ^2+ cx + d$, whose roots are equal to the roots of $12 x ^3-4 x ^2-3 x +6=0$ multiplied by $k$. If $k$ is the smallest number which will make $b, c$ and $d$ integers, then find the value of $(b+c+d)$.

Answer 1

Given $P = x ^3+ bx ^2+ cx + d$ ....(1)
and $Q=12 x^3-4 x^2-3 x+6=0$....(2)
Let the roots of (2) be $x$ and roots of (1) be $y$, then
$y=k x \Rightarrow x=\frac{y}{k}(y$ is the root of $P(x)$ and $x$ is the root of $Q(x))$
$12\left(\frac{ y }{ k }\right)^3-4\left(\frac{ y }{ k }\right)^2-3\left(\frac{ y }{ k }\right)+6=0 $ (replacing $y \rightarrow x$ and multiplying $k ^3$ )
$12 y ^3-4 y ^2 k -3 yk ^2+6 k ^3=0$ must be identical to $x ^3+ bx ^2+ cx + d =0$
Hence, $\frac{12}{1}=\frac{-4 k }{ b }=\frac{-3 k ^2}{ c }=\frac{6 k ^3}{ d }$
$b =-\frac{ k }{3} ; c =-\frac{ k ^2}{4} ; d =\frac{ k ^3}{2}$
minimum value of $k$ is 6
$b=-\frac{6}{3}=-2 ; c=-\frac{6}{4}=-9 ; d=\frac{6 \times 6 \times 6}{2}=108$
$b+c+d=108-11=97$