Question

Consider the polynomial $P(x)=x^3+b{x}^2+cx+d$, whose roots are equal to the roots of $12x^3-4x^2-3x+6=0$ 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

Answer: 97

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