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(A) Given that $g(x)$ is a polynomial of degree $1$,let $g(x) = ax + b$.Then $f(g(x)) = f(ax + b) = 8x^2 - 2x$. Since $f$ is of degree $2$,let $f(x) =

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$18$

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(A) Given that $g(x)$ is a polynomial of degree $1$,let $g(x) = ax + b$.Then $f(g(x)) = f(ax + b) = 8x^2 - 2x$. Since $f$ is of degree $2$,let $f(x) = px^2 + qx + r$.Substituting $g(x)$ into $f(x)$,we get $p(ax+b)^2 + q(ax+b) + r = 8x^2 - 2x$.Comparing the coefficients of $x^2$,we have $pa^2 = 8$. Since $g(f(x)) = 4x^2 + 6x + 1$,the leading coefficient of $g(f(x))$ is $a \cdot p = 4$.Dividing $pa^2 = 8$ by $ap = 4$,we get $a = 2$. Then $p(2)^2 = 8 \implies 4p = 8 \implies p = 2$.Now,$f(g(x)) = 2(2x+b)^2 + q(2x+b) + r = 2(4x^2 + 4bx + b^2) + 2qx + qb + r = 8x^2 + (8b + 2q)x + (2b^2 + qb + r) = 8x^2 - 2x$.Comparing coefficients: $8b + 2q = -2$ and $2b^2 + qb + r = 0$.From $g(f(x)) = a(px^2 + qx + r) + b = 2(2x^2 + qx + r) + b = 4x^2 + 2qx + 2r + b = 4x^2 + 6x + 1$.Comparing coefficients: $2q = 6 \implies q = 3$. Then $8b + 2(3) = -2 \implies 8b = -8 \implies b = -1$.Thus,$g(x) = 2x - 1$ and $f(x) = 2x^2 + 3x + r$. Using $2b^2 + qb + r = 0$,we get $2(-1)^2 + 3(-1) + r = 0 \implies 2 - 3 + r = 0 \implies r = 1$.So,$f(x) = 2x^2 + 3x + 1$.Now,$f(2) = 2(2)^2 + 3(2) + 1 = 8 + 6 + 1 = 15$ and $g(2) = 2(2) - 1 = 3$.Therefore,$f(2) + g(2) = 15 + 3 = 18$.

Let $f(x)$ and $g(x)$ be two real polynomials of degree $2$ and $1$ respectively. If $f(g(x)) = 8x^2 - 2x$ and $g(f(x)) = 4x^2 + 6x + 1$,then the value of $f(2) + g(2)$ is

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