Question

Let function $f\left(x\right)=x^{4}+ax^{3}+bx^{2}+cx+d$ such that $f\left(- 1\right)=100,f\left(- 2\right)=200,f\left(- 3\right)=300$ then the value $\frac{f \left(10\right) + f \left(- 14\right)}{16}$ equals

Answer 1

$f\left(x\right)=\left(x + 1\right)\left(x + 2\right)\left(x + 3\right)\left(x + t\right)-100x$ for some $t$
$P=\frac{\left(11 \times 12 \times 13 \left(10 + t\right) - 1000\right) + \left(- 13 \times - 12 \times - 11 \times \left(t - 14\right) + 1400\right)}{16}$
$=\frac{11 \times 12 \times 13 \times 24 + 400}{16}=2599$