Question

I am irritated with this question , please help me

Answer 1

The cubic polynomial P(x) satisfies the condition (x-1)^2 is a factor of P(x) +2 and (x+1)^2 is a factor of P(x) -2 . Then P(3) is

Answer 2

@Kainui @sidsiddhartha

Answer 3

@SolomonZelman

Answer 4

Just tell me what to do , i have already spent a considerable amount of time on this problem

Answer 5

@ganeshie8

Answer 6

I sort of ran out of paper and am tired so I didn't finish it, but you might find this useful:

\[\LARGE 1) \ P(x)+2=n(x-1)^2 \\ \LARGE 2) \ P(x)-2=m(x+1)^2 \]
I started here and realized we can take this several different routes:
\[\LARGE 3) \ P(-x)+2=n(x+1)^2 \\ \LARGE 4) \ P(-x)-2=m(x-1)^2 \]
So you might want to combine these together somehow, my idea was to link all 4 of them together like this by starting with equation (1), then solving for (x-1)^2 in eq (4), then solving for P(-x) in eq (3), and then solving for (x+1)^2 in eq (2) to link them all together. I wasn't able to finish this.

The other route I was interested in taking was simply to add eq(1) and (2) together and subtract eq(2) from (1) to get: \[\LARGE 5) \ 2P(x)=n(x-1)^2+m(x+1)^2 \\ \LARGE 6) \ 4=n(x-1)^2-m(x+1)^2 \]
Now taking equation (5) and (6) we can algebraically rearrange them to get:

\[\LARGE 5') \ P(x)=\frac{n+m}{2}x^2+(n-m)x+\frac{n+m}{2}\] Since we know P(x) is cubic, I decided to replace (n+m)/2=kx to get: \[\LARGE 5'') \ P(x)=kx*x^2+(n-m)x+kx\] Then using eq(6) I got \[\LARGE 6') \ (n-m)=4 \frac{1+kx^2}{1+x^2}\] and plugged in to get \[\LARGE P(x)= kx^3+4 \frac{1+kx^2}{1+x^2}x+kx\]

Now I imagine if we do something similar with the two P(-x) formulas eq (3) and (4) we might be able to figure something out... Yeah sorry this isn't what you wanted but I don't know the answer so... that's the breaks I guess lol.