Solution:
Cross-multiplying gives (x+7)P(2x)=8(x+7)P(x+1)−56P(x+1)=8(x+7)P(x+1)−56P(x+1). But more simply, rearrange:
P(x+1)P(2x)=8−x+756
⇒P(x+1)P(2x)=x+78(x+7)−56=x+78x+56−56=x+78x
So:
P(x+1)P(2x)=x+78x
Cross-multiplied:
(x+7)P(2x)=8xP(x+1)
Suppose P(x) has degree n and leading coefficient c. Then P(2x) has leading term c(2x)n=c2nxn, and P(x+1) has leading term cxn. So the left side's leading term is x⋅c2nxn=c2nxn+1, and the right side's leading term is 8x⋅cxn=8cxn+1. Equate:
c2n=8c⟹2n=8⟹n=3.
So P(x) is cubic.
Now, if x=0, the right side is 0, so the left side is 7P(0)=0⟹P(0)=0.
So P(x)=xQ(x) for some quadratic Q(x).
Plug into the original equation:
P(2x)=2xQ(2x), P(x+1)=(x+1)Q(x+1)
So:
(x+1)Q(x+1)2xQ(2x)=x+78x
Cross-multiplied:
(x+7)2xQ(2x)=8x(x+1)Q(x+1)
If x=0, both sides are 0.
Divide both sides by x (for x=0):
2(x+7)Q(2x)=8(x+1)Q(x+1)
Or:
(x+7)Q(2x)=4(x+1)Q(x+1)
Now, try x=−1:
(−1+7)Q(−2)=4(0)Q(0)⟹6Q(−2)=0⟹Q(−2)=0
So Q(x) has root at x=−2, so Q(x)=(x+2)R(x) for some linear R(x).
Now, Q(2x)=(2x+2)R(2x), Q(x+1)=(x+3)R(x+1)
Plug into the previous equation:
(x+7)(2x+2)R(2x)=4(x+1)(x+3)R(x+1)
Divide both sides by 2:
(x+7)(x+1)R(2x)=2(x+1)(x+3)R(x+1)
If x=−3:
(−3+7)(−3+1)R(−6)=2(−3+1)(−3+3)R(−2)
(4)(−2)R(−6)=2(−2)(0)R(−2)=0
So R(−6)=0
So R(x) has root at x=−6, so R(x)=(x+6)S(x), S(x) constant.
Therefore,
P(x)=x(x+2)(x+6)S
Since P(x) is cubic, S is a constant.
Given P(1)=1:
P(1)=1⋅3⋅7⋅S=21S=1⟹S=211
Therefore,
P(x)=211x(x+2)(x+6)
So,
P(−1)=211⋅(−1)⋅1⋅5=21−5