SOLUTION: Evaluate A = (a^2 − 3a +1)^2 − (1− a)(−2a +1)(1− 3a) . Then WITHOUT doing any further calculation write down the value of B = (a^2 + 3a +1)^2 −

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Evaluate A = (a^2 − 3a +1)^2 − (1− a)(−2a +1)(1− 3a) . Then WITHOUT doing any further calculation write down the value of B = (a^2 + 3a +1)^2 −      Log On


   

Question 823760: Evaluate A = (a^2 − 3a +1)^2 − (1− a)(−2a +1)(1− 3a) . Then WITHOUT doing any
further calculation write down the value of B = (a^2 + 3a +1)^2 − (1+ a)(2a +1)(1+ 3a) .
Found 2 solutions by richwmiller, KMST:
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
a^4 for both A and B

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Without doing any calculation we can see that
A%28a%29 and B%28a%29 are both polynomial functions of a with degree 4.
Without doing much calculation we can see that

and B%28-a%29=A%28a%29 .
So, the fact that A%28-a%29=B%28a%29 makes calculations easier.
We do not even need to calculate to see that
A%280%29=A%28-0%29=B%280%29 .
We can calculate the value of the functions for
a=-1 , a=-0.5 , a=0.5 , a=1 .




In sum
A%280%29=B%280%29=0
A%281%29=A%28-1%29=B%281%29=B%28-1%29=1
A%280.5%29=A%28-0.5%29=B%280.5%29=B%28-0.5%29=0.0625
We know that the polynomial of degree 4 P%28a%29=a%5E4 has
P%280%29=0
P%281%29=P%28-1%29=1
P%280.5%29=P%28-0.5%29=0.0625
and since 5 points uniquely determine a degree 4 polynomial,
just like 2 points uniquely determine a line,
A%28a%29=B%28a%29=a%5E4