+0

# find a quadratic polynomial such that

0
120
2
+4

I have seen Alan answer this type of question in an old post(https://web2.0calc.com/questions/find-a-quadratic-polynomial-p-x-such-that-p-0-1-p-1-5-p-2-11), but I had a hard time figuring out one of his steps.

My specific question is Find quadratic polynomial P such that

P(o)=-3, P(-2) = 13, P(3) = 3

I have done most of the work up until when Alan said "Multiply (1) by 2 and subtract from (2) to get:    2 = 2a  or a = 1"

the numbers aren't the same on my question, and I am confused as to where he got the number that he multiplied (1) by.

Do I always multiply by 2, or is there information missing?

idk if that made sense, please lmk what I should clarify if you don't understand me.

Oct 13, 2020
edited by uvula  Oct 13, 2020
edited by uvula  Oct 13, 2020
edited by uvula  Oct 13, 2020

#1
+31512
+2

The "multiply by 2" was specific to that question,. The idea was to make the first equation have the same number of b's as the second, so that when (1) was subtracted from (2) the b's would cancel out and there would only be a's and constants left.

Oct 13, 2020
#2
+31512
+3

Here :

P(x) = ax2 + bx + c

-3 = a*(0)2 + b*(0) + c  so c = -3 immediately

P(x) = ax2 + bx -3

13 = a*(-2)2 + b*(-2) - 3  or  4a - 2b  = 16     Divide all terms by 2    2a - b = 8    (1)

3 = a*(3)2 + b*(3) - 3      or  9a + 3b  = 6      Divide all terms by 3     3a + b = 2   (2)

Now subtract (1) from (2) to find a, then substitute the result back into either (1) or (2) to get b.

Oct 14, 2020