f(x)=ax^2+bx+c is an even function and g(x)=p | vedclass

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(C) Given that $f(x) = ax^2 + bx + c$ is an even function, $f(x) = f(-x)$.$ax^2 + bx + c = a(-x)^2 + b(-x) + c = ax^2 - bx + c$.Comparing coefficients

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$4a+2b+c$

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(C) Given that $f(x) = ax^2 + bx + c$ is an even function, $f(x) = f(-x)$.$ax^2 + bx + c = a(-x)^2 + b(-x) + c = ax^2 - bx + c$.Comparing coefficients, we get $b = 0$.Given that $g(x) = px^3 + qx^2 + rx$ is an odd function, $g(-x) = -g(x)$.$p(-x)^3 + q(-x)^2 + r(-x) = -(px^3 + qx^2 + rx)$.$-px^3 + qx^2 - rx = -px^3 - qx^2 - rx$.Comparing coefficients, we get $q = 0$.Now, $h(x) = f(x) + g(x) = ax^2 + bx + c + px^3 + qx^2 + rx$.Since $b = 0$ and $q = 0$, $h(x) = px^3 + ax^2 + rx + c$.We are given $h(-2) = 0$.$p(-2)^3 + a(-2)^2 + r(-2) + c = 0$.$-8p + 4a - 2r + c = 0$.$4a + c = 8p + 2r$.Since $b = 0$, $4a + 2b + c = 4a + 2(0) + c = 4a + c$.Thus, $8p + 2r = 4a + 2b + c$.

$f(x)=ax^2+bx+c$ is an even function and $g(x)=px^3+qx^2+rx$ is an odd function. If $h(x)=f(x)+g(x)$ and $h(-2)=0$, then $8p+4q+2r=$

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