Question

If $p\left(x\right)=a{x}^2+bx$ and $q\left(x\right)=l{x}^2+mx+n$ with $p\left(1\right)=q\left(1\right);p\left(2\right)-q\left(2\right)=1$ and $p\left(3\right)=q\left(3\right)=4$, then $p\left(4\right)-q\left(4\right)$ be $m/n$.Find $m-(8\ast n)$?

Answer 1

$p\left(x\right)=a{x}^2+bx$ and $q\left(x\right)=l{x}^2+mx+n$
It is given $p\left(1\right)=q\left(1\right)$
$\Rightarrow (a+b)-(l+m+n)=0$
$\Rightarrow (a-l)+(b-m)-n=0.....\left(1\right)$
It is given $p\left(2\right)-q\left(2\right)=1$
$\Rightarrow (4a+2b)-(4l+2m+n)=1$
$\Rightarrow 4(a-l)+2(b-m)-n=1.....\left(2\right)$
It is given $p\left(3\right)-q\left(3\right)=4$
$\Rightarrow (9a+3b)-(9l+3m+n)=4$
$\Rightarrow 9(a-l)+3(b-m)-n=4.....\left(3\right)$
Solving these equations, we get
$a-l=\frac{4}{5},b-m=-1,n=\frac{1}{5}$
Now consider, $p\left(4\right)-q\left(4\right)$
$=(16a+4b)-(16l+4m+n)$
$=16(a-l)+4(b-m)-n$
$\Rightarrow p\left(4\right)-q\left(4\right)=\frac{43}{5}$
$\Rightarrow m-8\ast n=3$