Find polynomials $p(x) $ such that $(x+10)p(2x)=(8x-32)p(x+6) $ when $p(1)=210$

Question

Hi I have polynomials as below form.

$(x+10)p(2x)=(8x-32)p(x+6)$, when $p(1)=210$

I try to assign some $x$ eg. $0.5$ , $-6$ but can not find the correct relation. Any advice or guidance on this would be greatly appreciated, Thanks.

Answer 1

Let $S$ be the set of zeroes of $p$.

If $x \in S$, $x \ne -4$ then $$\underbrace{(x+4)}_{\ne 0}p(2x-12) = 8(x-10)p((x-6)+6) = 8(x-10)p(x) = 0$$ so $2x-12 \in S$.

Notice that $8 \in S$ since plugging in $x = 4$ gives $14p(8) = 8(4-4)p(10) = 0$.

Therefore $4 = 2\cdot 8 -12 \in S$ and then $-4 = 2\cdot 4 - 12 \in S$ so $\{8,-4,4\} \subseteq S$.

@Vasily Mitch noticed that $\deg p = 3$ so the only solution is $S = \{8,-4,4\}$ or $$p(x) = a(x-8)(x-4)(x+4)$$ for some $a \in \mathbb{R}$. From $p(1) = 210$ it follows $a = 2$.

Answer 2

If $p(x)=a_0x^n+\ldots+a_{n-1}x+a_n$, then considering highest order, we conclude: $$ a_0(2x^{n})\times x = a_0x^{n}\times8x,\\ 2^n = 8,\qquad n=3. $$ So it's a polynomial of order 3: $p(x)=a_0x^3+a_1x^2+a_2x+a_3$.

Then you have to write the expression for polynomial: $$ (x+10)(8a_0x^3+4a_1x^2+2a_2x+a_3) = (8x+32)(a_0(x+6)^3+a_1(x+6)^2+a_2(x+6)+a_3), $$ express every $a_i$ in terms of $a_0$, and finally use $p(1)=210$ to find the $a_0$.