Question

The value of $a{x}^2+bx+c$ is 5 when$x=0$. The remainder is 6 when divided by $x-1$ and 10 when divided by $x+1$ then find value of 5a - 2b + 5c

Answer 1

Let $f\left(x\right)=a{x}^2+bx+c$
Given $f\left(0\right)=c=5$ and $f\left(1\right)=$ $a{\left(1\right)}^2+b\left(1\right)+5=6$
$\Rightarrow a+b=\mathrm{1......}\left(1\right)$
Also given $f(-1)=10,$ $a{(-1)}^2+b(-1)+5=10$
$\Rightarrow a-b=\mathrm{5.....}\left(2\right)$
Solving (1) and (2) we have $2a=6$
$\Rightarrow a=3$ and b$=-2$
$\therefore$ Value of $5a-2b+5c=5\left(3\right)-2(-2)+5\left(5\right)=44$