Can You Solve This?

Algebra Level 3

$P(x)=Ax^{3}+Bx^{2}+Cx+D$

What is $A-B-C-D$ when

$P(0)=1$

$P(1)=8$

$P(2)=17$

$P(3)=31$

1 solution

Terry Yu
Apr 19, 2017

$P(0)=D=1$

$P(2)=8A+4B+2C+1=17$ so $8A+4B+2C=16$

$P(3)=27A+9B+3C+1=31$ so $27A+9B+3C=30$ and $9A+3B+C=10$

So now we have

So now you subtract the second equation from the first and get $A-B-C=-6$

Since you know $D=1$ , so $A-B-C-D=\boxed{-7}$

Cool Facts : $-7$ is one of my lucky numbers, along with $7$ and $49$