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$f$ is a polynomial satisfying $$f(x+2)−f(x)=(6x+4)^2$$ and $f(0)=−16$. Determine $f(5)$.

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Let $f(x)=a_0+a_1x+\cdots+a_nx^n$. Then $f(x+2)=a_0+a_1x+2a_1+\cdots+a_nx^n+2na_nx^{n-1} +\cdots+2^na_n$. So $$f(x+2)-f(x)=2a_1+4a_2+\cdots + 2^na_n + (2a_2+12a_3+\cdots+2^{n-1}a_n)x+\cdots + 2na_nx^{n-1}.$$ As this has to be $(6x+4)^2=36x^2+48x+16$ we see that $f(x)$ is a polynomial of degree $3$. So we have \begin{eqnarray*} & & f(x)=a_0+a_1x+a_2x^2+a_3x^3\\ & & f(x+2)-f(x)=2a_1+4a_2+8a_3 + x(4a_2+12a_3) + 6a_3x^2\\ &\Leftrightarrow & 6a_3=36 \wedge 4a_2+12a_3=48 \wedge 2a_1+4a_2+8a_3=16\\ &\Leftrightarrow & a_3=6 \wedge a_2=\frac{48-72}{4}=-6 \wedge a_1=\frac{16-48+24}{2}=-4 \end{eqnarray*} Finally we know that $f(0)=a_0=-16$. So $f(x)=6x^3-6x^2-4x-16$, and $f(5)=564$.

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Just assume that $f$ is a polynomial of degree 3:

$$ f(x) = ax^3 +bx^2+cx+d, $$

then, substitute it in your equation, and find the unknown coefficients $a,b,c,d$.

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